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# Knowledge base: Warsaw University of Technology

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## Research of the impact of water on the efficiency of aviation fuel combustion process in the flow of the stabilizer

### Jakub Kowara

#### Abstract

The introduction presents the reasons, the history, and effects of the use of water in the combustion process. Afterwards, the reactions of formation of exhaust components were briefly described, focusing primarily on nitrogen oxides and carbon monoxide. It also presents the most commonly used methods of water supply and their effects confirmed by tests. At the end contains principles and purpose of the study. The second chapter is devoted to the calculation of the necessary data required for the experiment and for a comparative analysis of the results. It includes also the structure of the station, scheme of it's working and how it is controlled. The next chapter describes the tests carried out for the combustion of the fuel itself, modification of the combustion chamber and ultimately the use of direct injection of water. It also presents graphs with the results of the study. Then analyzes the measurements of temperature and emission of exhaust gas and comparing them with calculated values contained in the second chapter. Finally, conclusions drawn from the conducted experiments are presented.
Diploma type
Engineer's / Bachelor of Science
Diploma type
Engineer's thesis
Author
Jakub Kowara (FPAE) Jakub Kowara,, Faculty of Power and Aeronautical Engineering (FPAE)
Title in Polish
Badania wpływu wody na efektywność procesu spalania paliwa lotniczego w przepływie za stabilizatorem korytkowym
Supervisor
Marian Gieras (FPAE/IHE) Marian Gieras,, The Institute of Heat Engineering (FPAE/IHE)Faculty of Power and Aeronautical Engineering (FPAE)
Certifying unit
Faculty of Power and Aeronautical Engineering (FPAE)
Affiliation unit
The Institute of Heat Engineering (FPAE/IHE)
Study subject / specialization
, Lotnictwo i Kosmonautyka
Language
(pl) Polish
Status
Finished
Defense Date
02-11-2016
Issue date (year)
2016
Pages
66
Internal identifier
MEL; PD-3883
Reviewers
Mateusz Żbikowski (FPAE/IHE) Mateusz Żbikowski,, The Institute of Heat Engineering (FPAE/IHE)Faculty of Power and Aeronautical Engineering (FPAE) Marian Gieras (FPAE/IHE) Marian Gieras,, The Institute of Heat Engineering (FPAE/IHE)Faculty of Power and Aeronautical Engineering (FPAE) Bartosz Borowiak (FPAE/IHE) Bartosz Borowiak ,, The Institute of Heat Engineering (FPAE/IHE)Faculty of Power and Aeronautical Engineering (FPAE)
Keywords in Polish
proces spalania, wtrysk wody, stabilizator, badanie
Keywords in English
combustion process, water injection, stabilizer, research
Abstract in Polish
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Identyfikator pracy APD: 14247

Uniform Resource Identifier
https://repo.pw.edu.pl/info/bachelor/WUTcc8e5912e92a4f6ea5366e2abee6fedc/
URN
urn:pw-repo:WUTcc8e5912e92a4f6ea5366e2abee6fedc

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# Series R, L, and C

## Chapter 5 - Reactance And Impedance -- R, L, And C

Let’s take the following example circuit and analyze it:

Example series R, L, and C circuit.

### Solving for Reactance

The first step is to determine the reactance (in ohms) for the inductor and the capacitor.

The next step is to express all resistances and reactances in a mathematically common form: impedance. (Figure below)

Remember that an inductive reactance translates into a positive imaginary impedance (or an impedance at +90°), while a capacitive reactance translates into a negative imaginary impedance (impedance at -90°). Resistance, of course, is still regarded as a purely “real” impedance (polar angle of 0°):

Example series R, L, and C circuit with component values replaced by impedances.

#### Tabulate Results:

Now, with all quantities of opposition to electric current expressed in a common, complex number format (as impedances, and not as resistances or reactances), they can be handled in the same way as plain resistances in a DC circuit.

This is an ideal time to draw up an analysis table for this circuit and insert all the “given” figures (total voltage, and the impedance of the resistor, inductor, and capacitor).

Unless otherwise specified, the source voltage will be our reference for phase shift, and so will be written at an angle of 0°. Remember that there is no such thing as an “absolute” angle of phase shift for a voltage or current, since its always a quantity relative to another waveform.

Phase angles for impedance, however (like those of the resistor, inductor, and capacitor), are known absolutely, because the phase relationships between voltage and current at each component are absolutely defined.

Notice that I’m assuming a perfectly reactive inductor and capacitor, with impedance phase angles of exactly +90 and -90°, respectively.

Although real components won’t be perfect in this regard, they should be fairly close. For simplicity, I’ll assume perfectly reactive inductors and capacitors from now on in my example calculations except where noted otherwise.

Since the above example circuit is a series circuit, we know that the total circuit impedance is equal to the sum of the individuals, so:

Inserting this figure for total impedance into our table:

We can now apply Ohm’s Law (I=E/R) vertically in the “Total” column to find total current for this series circuit:

Being a series circuit, the current must be equal through all components. Thus, we can take the figure obtained for total current and distribute it to each of the other columns:

Now we’re prepared to apply Ohm’s Law (E=IZ) to each of the individual component columns in the table, to determine voltage drops:

Notice something strange here: although our supply voltage is only 120 volts, the voltage across the capacitor is 137.46 volts! How can this be? The answer lies in the interaction between the inductive and capacitive reactances.

Expressed as impedances, we can see that the inductor opposes current in a manner precisely opposite that of the capacitor. Expressed in rectangular form, the inductor’s impedance has a positive imaginary term and the capacitor has a negative imaginary term.

When these two contrary impedances are added (in series), they tend to cancel each other out! Although they’re still added together to produce a sum, that sum is actually less than either of the individual (capacitive or inductive) impedances alone.

It is analogous to adding together a positive and a negative (scalar) number: the sum is a quantity less than either one’s individual absolute value.

If the total impedance in a series circuit with both inductive and capacitive elements is less than the impedance of either element separately, then the total current in that circuit must be greater than what it would be with only the inductive or only the capacitive elements there.

With this abnormally high current through each of the components, voltages greater than the source voltage may be obtained across some of the individual components! Further consequences of inductors’ and capacitors’ opposite reactances in the same circuit will be explored in the next chapter.

Once you’ve mastered the technique of reducing all component values to impedances (Z), analyzing any AC circuit is only about as difficult as analyzing any DC circuit, except that the quantities dealt with are vector instead of a scalar.

With the exception of equations dealing with power (P), equations in AC circuits are the same as those in DC circuits, using impedances (Z) instead of resistances (R). Ohm’s Law (E=IZ) still holds true, and so do Kirchhoff’s Voltage and Current Laws.

To demonstrate Kirchhoff’s Voltage Law in an AC circuit, we can look at the answers we derived for component voltage drops in the last circuit. KVL tells us that the algebraic sum of the voltage drops across the resistor, inductor, and capacitor should equal the applied voltage from the source.

Even though this may not look like it is true at first sight, a bit of complex number addition proves otherwise:

Aside from a bit of rounding error, the sum of these voltage drops does equal 120 volts. Performed on a calculator (preserving all digits), the answer you will receive should be exactly 120 + j0 volts.

We can also use SPICE to verify our figures for this circuit:

Example series R, L, and C SPICE circuit.

r1 1 2 250
l1 2 3 650m
c1 3 0 1.5u
.ac lin 1 60 60
.print ac v(1,2) v(2,3) v(3,0) i(v1)
.print ac vp(1,2) vp(2,3) vp(3,0) ip(v1)
.end
freq            v(1,2)          v(2,3)          v(3)            i(v1)
6.000E+01       1.943E+01       1.905E+01       1.375E+02       7.773E-02

freq            vp(1,2)         vp(2,3)         vp(3)           ip(v1)
6.000E+01       8.068E+01       1.707E+02       -9.320E+00       -9.932E+01

The SPICE simulation shows our hand-calculated results to be accurate.

As you can see, there is little difference between AC circuit analysis and DC circuit analysis, except that all quantities of voltage, current, and resistance (actually, impedance) must be handled in complex rather than scalar form so as to account for phase angle.

This is good since it means all you’ve learned about DC electric circuits applies to what you’re learning here. The only exception to this consistency is the calculation of power, which is so unique that it deserves a chapter devoted to that subject alone.

REVIEW:

• Impedances of any kind add in series: ZTotal = Z1 + Z2 + . . . Zn
• Although impedances add in series, the total impedance for a circuit containing both inductance and capacitance may be less than one or more of the individual impedances, because series inductive and capacitive impedances tend to cancel each other out. This may lead to voltage drops across components exceeding the supply voltage!
• All rules and laws of DC circuits apply to AC circuits, so long as values are expressed in complex form rather than a scalar. The only exception to this principle is the calculation of power, which is very different for AC.

RELATED WORKSHEET:

• S
soncu August 09, 2020

Correction: Rule of series circuit: Itotal = IR = IL = IC

Like.
• blahhhhh April 28, 2021

how did it became 1.5437 kohms?

Like.
• alpacas May 07, 2022
So X = XR + XL + XC since series circuit which = R + j * ( 2*pi*60*L - 1/(2*pi*60*C) ) (I'm taking out the j from the equations) (also note -j for the C equation since phase is opposite (-90 deg)) After that you plug in values for L,R,C and you get your number.
Like.

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# If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)

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Joined: 02 Sep 2009
Posts: 62298
Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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20 Jun 2017, 21:01
uvee wrote:
Bunuel wrote:
BANON wrote:
If x, y, and k are positive numbers such that x/(x+y)*10 + y/(x+y)*20 = k and if x < y, which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30

$$10*\frac{x}{x+y}+20*\frac{y}{x+y}=k$$

$$10*\frac{x+2y}{x+y}=k$$

$$10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k$$

Finally we get: $$10*(1+\frac{y}{x+y})=k$$

We know that $$x<y$$

Hence $$\frac{y}{x+y}$$ is more than $$0.5$$ and less than $$1$$

$$0.5<\frac{y}{x+y}<1$$

So, $$15<10*(1+\frac{y}{x+y})<20$$

Only answer between $$15$$ and $$20$$ is $$18$$.

There can be another approach:

We have: $$\frac{10x+20y}{x+y}=k$$, if you look at this equation you'll notice that it's a weighted average.

There are $$x$$ red boxes and $$y$$ blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes?

k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition.

Hi, Bunuel! I have a silly question. Can you please explain how you arrived at the following?
We know that $$x<y$$

Hence $$\frac{y}{x+y}$$ is more than $$0.5$$ and less than $$1$$

$$\frac{y}{x+y}$$

Since both x and y are positive, then the denominator is greater than numerator. So, $$\frac{y}{x+y}<1$$.

Next, if x were equal to y, then we'd have y/(2y) = 1/2 but x < y, so denominator is less than 2y, which makes the fraction greater than 1/2.

Hope it's clear.
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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19 Jan 2018, 07:31
Could you guys please explain if the answer holds true when x<0 and y>0 ? Thanks !
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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19 Jan 2018, 19:26
Hi Phlaryu,

The prompt tells us that X, Y and K are all POSITIVE, so you shouldn't be wasting time on the outcome if X < 0.

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Re: If x, y, and k are positive numbers such that 10*x/(x+y)+ 20  [#permalink]

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03 Dec 2018, 23:58
1
1
This one's all about pattern recognition.

Notice that the equation is giving us a weighted average. Because y is bigger than x, the average must be weighted towards y. Visually:

x---------avg---y
10--------avg---20

Thus, the answer must be between 15 and 20, exclusive.

Among the answer choices, only 18 fits the bill.

Choose D.
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Re: If x, y, and k are positive numbers such that 10*x/(x+y)+ 20  [#permalink]

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03 Dec 2018, 23:59
2
pzazz12 wrote:
If x, y, and k are positive numbers such that (x/x+y)(10)+(y/x+y)(20)=k and if x < y, which of the following could be the value of k?

A. 10
B. 12
C. 15
D. 18
E. 30

I suspect that many test-takers would not recognize that this is a weighted average problem. For those test-takers, a straightforward and efficient approach would be to plug in the answer choices, which represent the value of k.

(10x + 20y)/(x+y) = 15
10x + 20y = 15x + 15y
5y = 5x
y = x
Doesn't work because the problem states that x<y.

We need y to be larger, so let's try answer choice D: k=18
10x + 20y = 18x + 18y
2y = 8x
y/x = 8/2
Success! x<y.

The process would be very quick if you recognized right away that since k = one of the answer choices, you could simply plug them in for k.
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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01 Sep 2019, 08:08
BANON wrote:
If x, y, and k are positive numbers such that $$\frac{x}{x + y}*10 + \frac{y}{x + y}*20 = k$$ and if x < y, which of the following could be the value of k?

A. 10
B. 12
C. 15
D. 18
E. 30

Given:
1. x, y, and k are positive numbers such that $$\frac{x}{x + y}*10 + \frac{y}{x + y}*20 = k$$
2. x < y

Asked: Which of the following could be the value of k?

$$\frac{x}{x + y}*10 + \frac{y}{x + y}*20 = k$$
$$\frac{x+y}{x + y}*10 + \frac{y}{x + y}*10 = k$$
$$10 + \frac{y}{x + y}*10 = k$$
SInce y>x
$$y> \frac{(x+y)}{2}$$
$$\frac{y}{x+y}>\frac{1}{2}$$
$$k > 10 + \frac{1}{2}*10$$
$$k > 15$$
$$10 + \frac{y}{x + y}*10 = k$$
Since $$\frac{y}{x+y}< 1$$
$$k< 10 + 1*10$$
k<20
15<k<20
Only option D 18 satisfies these conditions

IMO D
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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01 Sep 2019, 20:58
10x/ (X+Y) + 20Y/(X+Y) = K

10*{ (X+2Y)/ (X+Y) } = K

(X+2Y) / (X+Y) = K/10 and we know that X<Y, so here we plug in values.

(X+2Y) / (X+Y) = 10/10......Can't be true as Y will have to be 0 for this.
(X+2Y)/ (X+Y) = 12/10.... Gives us 2 equations. X+ 2Y = 12 and X+Y = 10...solving this we get X= 8 and Y =2, so incorrect.
(X+2Y) / (X+Y) = 15/10...similar to above, we will get two equations and by solving them, we get X= 5 and Y= 5, so incorrect.
(X+2Y) / (X+Y) = 18/10.......X+ 2Y= 18 and X+Y = 10........solving them we get, X= 2 and Y = 8...This is the correct solution.
(X+2Y) / (X+Y) = 30/10......we get -ve value of X. Hence the answer is (D)
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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15 Nov 2019, 23:24
I encountered this in a GMATFocus test.

Here's the GMATFocus solution.
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)   [#permalink] 15 Nov 2019, 23:24

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# LuaLaTeX error when compiling visual counter [closed]

I asked a question about visual counters in beamer, for which I got this answer. Now I need to compile it by LuaLaTeX in order to use the package luapersian(below code), but it returns some errors.

\documentclass{beamer}
\usepackage{lmodern}
\usepackage{tikz}
\usetikzlibrary{calc}

\pgfkeys{/visual counter/.cd,
thickness/.store in=\thickness,
thickness=0.4ex,
segment distance/.store in=\segdist,
segment distance=8,
color current frame/.store in=\colcurrframe,
color current frame=orange,
color old frame/.store in=\cololdframe,
color old frame=blue,
color next frame/.store in=\colnextframe,
color next frame=gray!30,
color page number/.store in=\colpagenum,
color page number=white,
current value/.store in=\currentv,
current value=1,
total value/.store in=\totalv,
total value=2,
circled page number/.code={
\begin{tikzpicture}[fill color/.style={}]
\pgfkeys{/visual counter/.cd,
}
\pgfmathtruncatemacro\current{\currentv+1}
\def\tot{\totalv}

\foreach \s in {1,...,\tot}
{
\ifnum\s>\current%
\tikzset{fill color/.append style={\colnextframe}}%
\fi%
\ifnum\s=\current%
\tikzset{fill color/.append style={\colcurrframe}}%
\fi%
\ifnum\s<\current%
\tikzset{fill color/.append style={\cololdframe}}%
\fi%
\fill[fill color]
({90-360/\tot * (\s - 1)-\segdist}:\radiusout) arc
({90-360/\tot * (\s - 1)-\segdist}:{90-360/\tot * (\s)+\segdist}:\radiusout) --
({90-360/\tot * (\s)+\segdist}:{90-360/\tot * (\s - 1)-\segdist}:\radiusin);
}
\end{tikzpicture}
},
}

% new footline with
\setbeamertemplate{footline}{
\begin{beamercolorbox}[wd=0.95\textwidth, ht=2ex,dp=1ex,sep=1ex]{footline}
\hfill%
\tikz\node[/visual counter/.cd,
segment distance=-2pt,
color old frame=orange!50,
color current frame=cyan!80!gray!50,
color next frame=cyan!80!gray!50,
circled page number,
]{};
\end{beamercolorbox}
}

\usepackage{luapersian}

\begin{document}

\begin{frame}

\end{frame}

\end{document}

## closed as unclear what you're asking by user36296, Stefan Pinnow, Jesse, Martin Schröder, Au101Jun 12 '17 at 15:15

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Please give us some code. I don't use Persian and don't want to find out how to do that in LaTeX just in order to help here with some compilation issue. And what are the errors? – LaRiFaRi Apr 15 '14 at 15:21
• Yes, please, edit your post and include all the relevant TeX codes, otherwise we are wildly guessing what might go wrong. – Malipivo Apr 15 '14 at 16:20
• Welcome to TeX.SX! Please help us to help you and add a minimal working example (MWE) that illustrates your problem. It will be much easier for us to reproduce your situation and find out what the issue is when we see compilable code, starting with \documentclass{...} and ending with \end{document}. – Martin Schröder Apr 16 '14 at 11:58
• I edited the post, please see again. – user49875 Apr 16 '14 at 13:25
• Your example document fails to compile (even without luapersian); something seems to be wrong. Does it compile for you as is (without loading luapersian)? – Gonzalo Medina Apr 16 '14 at 13:36

Caveat emptor I don't have \usepackage{luapersian}, so I only tried to compile with lualatex

Your visible code looks fine, but when copying from your question above there where tons of invisible characters in front of after the visible text. After removing all of them, you example compiles fine with lualatex

\documentclass{beamer}
\usepackage{lmodern}
\usepackage{tikz}
\usetikzlibrary{calc}

\pgfkeys{/visual counter/.cd,
thickness/.store in=\thickness,
thickness=0.4ex,
segment distance/.store in=\segdist,
segment distance=8,
color current frame/.store in=\colcurrframe,
color current frame=orange,
color old frame/.store in=\cololdframe,
color old frame=blue,
color next frame/.store in=\colnextframe,
color next frame=gray!30,
color page number/.store in=\colpagenum,
color page number=white,
current value/.store in=\currentv,
current value=1,
total value/.store in=\totalv,
total value=2,
circled page number/.code={
\begin{tikzpicture}[fill color/.style={}]
\pgfkeys{/visual counter/.cd,
}
\pgfmathtruncatemacro\current{\currentv+1}
\def\tot{\totalv}

\foreach \s in {1,...,\tot}
{
\ifnum\s>\current%
\tikzset{fill color/.append style={\colnextframe}}%
\fi%
\ifnum\s=\current%
\tikzset{fill color/.append style={\colcurrframe}}%
\fi%
\ifnum\s<\current%
\tikzset{fill color/.append style={\cololdframe}}%
\fi%
\fill[fill color]
({90-360/\tot * (\s - 1)-\segdist}:\radiusout) arc
({90-360/\tot * (\s - 1)-\segdist}:{90-360/\tot * (\s)+\segdist}:\radiusout) --
({90-360/\tot * (\s)+\segdist}:{90-360/\tot * (\s - 1)-\segdist}:\radiusin);
}
\end{tikzpicture}
},
}

% new footline with
\setbeamertemplate{footline}{
\begin{beamercolorbox}[wd=0.95\textwidth, ht=2ex,dp=1ex,sep=1ex]{footline}
\hfill%
\tikz\node[/visual counter/.cd,
segment distance=-2pt,
color old frame=orange!50,
color current frame=cyan!80!gray!50,
color next frame=cyan!80!gray!50,
circled page number,
]{};
\end{beamercolorbox}
}

%\usepackage{luapersian}

\begin{document}

\begin{frame}
abc
\end{frame}

\begin{frame}
abc
\end{frame}

\begin{frame}
abc
\end{frame}

\end{document}


text:
# 4.6: Generating Functions

As usual, our starting point is a random experiment modeled by a probability sace $$(\Omega, \mathscr F, \P)$$. A generating function of a real-valued random variable is an expected value of a certain transformation of the random variable involving another (deterministic) variable. Most generating functions share four important properties:

1. Under mild conditions, the generating function completely determines the distribution of the random variable.
2. The generating function of a sum of independent variables is the product of the generating functions
3. The moments of the random variable can be obtained from the derivatives of the generating function.
4. Ordinary (pointwise) convergence of a sequence of generating functions corresponds to the special convergence of the corresponding distributions.

Property 1 is perhaps the most important. Often a random variable is shown to have a certain distribution by showing that the generating function has a certain form. The process of recovering the distribution from the generating function is known as inversion. Property 2 is frequently used to determine the distribution of a sum of independent variables. By contrast, recall that the probability density function of a sum of independent variables is the convolution of the individual density functions, a much more complicated operation. Property 3 is useful because often computing moments from the generating function is easier than computing the moments directly from the probability density function. The last property is known as the continuity theorem. Often it is easer to show the convergence of the generating functions than to prove convergence of the distributions directly.

The numerical value of the generating function at a particular value of the free variable is of no interest, and so generating functions can seem rather unintuitive at first. But the important point is that the generating function as a whole encodes all of the information in the probability distribution in a very useful way. Generating functions are important and valuable tools in probability, as they are in other areas of mathematics, from combinatorics to differential equations.

We will study the three generating functions in the list below, which correspond to increasing levels of generality. The fist is the most restrictive, but also by far the simplest, since the theory reduces to basic facts about power series that you will remember from calculus. The third is the most general and the one for which the theory is most complete and elegant, but it also requires basic knowledge of complex analysis. The one in the middle is perhaps the one most commonly used, and suffices for most distributions in applied probability.

We will also study the characteristic function for multivariate distributions, although analogous results hold for the other two types. In the basic theory below, be sure to try the proofs yourself before reading the ones in the text.

## Basic Theory

### The Probability Generating Function

For our first generating function, assume that $$N$$ is a random variable taking values in $$\N$$.

The probability generating function $$P$$ of $$N$$ is defined by $P(t) = \E\left(t^N\right)$ for all $$t \in \R$$ for which the expected value exists in $$\R$$.

That is, $$P(t)$$ is defined when $$\E\left(|t|^N\right) \lt \infty$$. The probability generating function can be written nicely in terms of the probability density function.

Suppose that $$N$$ has probability density function $$f$$ and probability generating function $$P$$. Then $P(t) = \sum_{n=0}^\infty f(n) t^n, \quad t \in (-r, r)$ where $$r \in [1, \infty]$$ is the radius of convergence of the series.

Proof

The expansion follows from the discrete change of variables theorem for expected value. Note that the series is a power series in $$t$$, and hence by basic calculus, converges absolutely for $$t \in (-r, r)$$ where $$r \in [0, \infty]$$ is the radius of convergence. But since $$\sum_{n=0}^\infty f(n) = 1$$ we must have $$r \ge 1$$, and the series converges absolutely at least for $$t \in [-1, 1]$$.

In the language of combinatorics, $$P$$ is the ordinary generating function of $$f$$. Of course, if $$N$$ just takes a finite set of values in $$\N$$ then $$r = \infty$$. Recall from calculus that a power series can be differentiated term by term, just like a polynomial. Each derivative series has the same radius of convergence as the original series (but may behave differently at the endpoints of the interval of convergence). We denote the derivative of order $$n$$ by $$P^{(n)}$$. Recall also that if $$n \in \N$$ and $$k \in \N$$ with $$k \le n$$, then the number of permutations of size $$k$$ chosen from a population of $$n$$ objects is $n^{(k)} = n (n - 1) \cdots (n - k + 1)$ The following theorem is the inversion result for probability generating functions: the generating function completely determines the distribution.

Suppose again that $$N$$ has probability density function $$f$$ and probability generating function $$P$$. Then $f(k) = \frac{P^{(k)}(0)}{k!}, \quad k \in \N$

Proof

This is a standard result from the theory of power series. Differentiating $$k$$ times gives $$P^{(k)}(t) = \sum_{n=k}^\infty n^{(k)} f(n) t^{n-k}$$ for $$t \in (-r, r)$$. Hence $$P^{(k)}(0) = k^{(k)} f(k) = k! f(k)$$

Our next result is not particularly important, but has a certain curiosity.

$$\P(N \text{ is even}) = \frac{1}{2}\left[1 + P(-1)\right]$$.

Proof

Note that $P(1) + P(-1) = \sum_{n=0}^\infty f(n) + \sum_{n=0}^\infty (-1)^n f(n) = 2 \sum_{k=0}^\infty f(2 k) = 2 \P(N \text{ is even })$ We can combine the two sum since we know that the series converge absolutely at 1 and $$-1$$.

Recall that the factorial moment of $$N$$ of order $$k \in \N$$ is $$\E\left[N^{(k)}\right]$$. The factorial moments can be computed from the derivatives of the probability generating function. The factorial moments, in turn, determine the ordinary moments about 0 (sometimes referred to as raw moments).

Suppose that the radius of convergence $$r \gt 1$$. Then $$P^{(k)}(1) = \E\left[N^{(k)}\right]$$ for $$k \in \N$$. In particular, $$N$$ has finite moments of all orders.

Proof

As before, $$P^{(k)}(t) = \sum_{n=k}^\infty n^{(k)} f(n) t^{n-k}$$ for $$t \in (-r, r)$$. Hence if $$r \gt 1$$ then $$P^{(k)}(1) = \sum_{n=k}^\infty n^{(k)} f(n) = \E\left[N^{(k)}\right]$$

Suppose again that $$r \gt 1$$. Then

1. $$\E(N) = P^\prime(1)$$
2. $$\var(N) = P^{\prime \prime}(1) + P^\prime(1)\left[1 - P^\prime(1)\right]$$
Proof
1. $$\E(N) = \E\left[N^{(1)}\right] = P^\prime(1)$$.
2. $$\E\left(N^2\right) = \E[N (N - 1)] + \E(N) = \E\left[N^{(2)}\right] + \E(N) = P^{\prime\prime}(1) + P^\prime(1)$$. Hence from (a), $$\var(N) = P^{\prime\prime}(1) + P^\prime(1) - \left[P^\prime(1)\right]^2$$.

Suppose that $$N_1$$ and $$N_2$$ are independent random variables taking values in $$\N$$, with probability generating functions $$P_1$$ and $$P_2$$ having radii of convergence $$r_1$$ and $$r_2$$, respectively. Then the probability generating function $$P$$ of $$N_1 + N_2$$ is given by $$P(t) = P_1(t) P_2(t)$$ for $$\left|t\right| \lt r_1 \wedge r_2$$.

Proof

Recall that the expected product of independent variables is the product of the expected values. Hence $P(t) = \E\left(t^{N_1 + N_2}\right) = \E\left(t^{N_1} t^{N_2}\right) = \E\left(t^{N_1}\right) \E\left(t^{N_2}\right) = P_1(t) P_2(t), \quad \left|t\right| \lt r_1 \wedge r_2$

### The Moment Generating Function

Our next generating function is defined more generally, so in this discussion we assume that the random variables are real-valued.

The moment generating function of $$X$$ is the function $$M$$ defined by $M(t) = \E\left(e^{tX}\right), \quad t \in \R$

Note that since $$e^{t X} \ge 0$$ with probability 1, $$M(t)$$ exists, as a real number or $$\infty$$, for any $$t \in \R$$. But as we will see, our interest will be in the domain where $$M(t) \lt \infty$$.

Suppose that $$X$$ has a continuous distribution on $$\R$$ with probability density function $$f$$. Then $M(t) = \int_{-\infty}^\infty e^{t x} f(x) \, dx$

Proof

This follows from the change of variables theorem for expected value.

Thus, the moment generating function of $$X$$ is closely related to the Laplace transform of the probability density function $$f$$. The Laplace transform is named for Pierre Simon Laplace, and is widely used in many areas of applied mathematics, particularly differential equations. The basic inversion theorem for moment generating functions (similar to the inversion theorem for Laplace transforms) states that if $$M(t) \lt \infty$$ for $$t$$ in an open interval about 0, then $$M$$ completely determines the distribution of $$X$$. Thus, if two distributions on $$\R$$ have moment generating functions that are equal (and finite) in an open interval about 0, then the distributions are the same.

Suppose that $$X$$ has moment generating function $$M$$ that is finite in an open interval $$I$$ about 0. Then $$X$$ has moments of all orders and $M(t) = \sum_{n=0}^\infty \frac{\E\left(X^n\right)}{n!} t^n, \quad t \in I$

Proof

Under the hypotheses, the expected value operator can be interchanged with the infinite series for the exponential function: $M(t) = \E\left(e^{t X}\right) = \E\left(\sum_{n=0}^\infty \frac{X^n}{n!} t^n\right) = \sum_{n=0}^\infty \frac{\E(X^n)}{n!} t^n, \quad t \in I$ The interchange is a special case of Fubini's theorem, named for Guido Fubini. For more details see the advanced section on properties of the integral in the chapter on Distributions.

So under the finite assumption in the last theorem, the moment generating function, like the probability generating function, is a power series in $$t$$.

Suppose again that $$X$$ has moment generating function $$M$$ that is finite in an open interval about 0. Then $$M^{(n)}(0) = \E\left(X^n\right)$$ for $$n \in \N$$

Proof

This follows by the same argument as above for the PGF: $$M^{(n)}(0) / n!$$ is the coefficient of order $$n$$ in the power series above, namely $$\E\left(X^n\right) / n!$$. Hence $$M^{(n)}(0) = \E\left(X^n\right)$$.

Thus, the derivatives of the moment generating function at 0 determine the moments of the variable (hence the name). In the language of combinatorics, the moment generating function is the exponential generating function of the sequence of moments. Thus, a random variable that does not have finite moments of all orders cannot have a finite moment generating function. Even when a random variable does have moments of all orders, the moment generating function may not exist. A counterexample is constructed below.

For nonnegative random variables (which are very common in applications), the domain where the moment generating function is finite is easy to understand.

Suppose that $$X$$ takes values in $$[0, \infty)$$ and has moment generating function $$M$$. If $$M(t) \lt \infty$$ for $$t \in \R$$ then $$M(s) \lt \infty$$ for $$s \le t$$.

Proof

Since $$X \ge 0$$, if $$s \le t$$ then $$s X \le t X$$ and hence $$e^{s X} \le e^{t X}$$. Hence $$\E\left(e^{s X}\right) \le \E\left(e^{t X} \right)$$.

So for a nonnegative random variable, either $$M(t) \lt \infty$$ for all $$t \in \R$$ or there exists $$r \in (0, \infty)$$ such that $$M(t) \lt \infty$$ for $$t \lt r$$. Of course, there are complementary results for non-positive random variables, but such variables are much less common. Next we consider what happens to the moment generating function under some simple transformations of the random variables.

Suppose that $$X$$ has moment generating function $$M$$ and that $$a, \, b \in \R$$. The moment generating function $$N$$ of $$Y = a + b X$$ is given by $$N(t) = e^{a t} M(b t)$$ for $$t \in \R$$.

Proof

$$\E\left[e^{t (a + b X)}\right] = \E\left(e^{t a} e^{t b X}\right) = e^{t a} \E\left[e^{(t b) X}\right] = e^{a t} M(b t)$$ for $$t \in \R). Recall that if \( a \in \R$$ and $$b \in (0, \infty)$$ then the transformation $$a + b X$$ is a location-scale transformation on the distribution of $$X$$, with location parameter $$a$$ and scale parameter $$b$$. Location-scale transformations frequently arise when units are changed, such as length changed from inches to centimeters or temperature from degrees Fahrenheit to degrees Celsius.

Suppose that $$X_1$$ and $$X_2$$ are independent random variables with moment generating functions $$M_1$$ and $$M_2$$ respectively. The moment generating function $$M$$ of $$Y = X_1 + X_2$$ is given by $$M(t) = M_1(t) M_2(t)$$ for $$t \in \R$$.

Proof

As with the PGF, the proof for the MGF relies on the law of exponents and the fact that the expected value of a product of independent variables is the product of the expected values: $\E\left[e^{t (X_1 + X_2)}\right] = \E\left(e^{t X_1} e^{t X_2}\right) = \E\left(e^{t X_1}\right) \E\left(e^{t X_2}\right) = M_1(t) M_2(t), \quad t \in \R$

The probability generating function of a variable can easily be converted into the moment generating function of the variable.

Suppose that $$X$$ is a random variable taking values in $$\N$$ with probability generating function $$G$$ having radius of convergence $$r$$. The moment generating function $$M$$ of $$X$$ is given by $$M(t) = G\left(e^t\right)$$ for $$t \lt \ln(r)$$.

Proof

$$M(t) = \E\left(e^{t X}\right) = \E\left[\left(e^t\right)^X\right] = G\left(e^t\right)$$ for $$e^t \lt r$$.

The following theorem gives the Chernoff bounds, named for the mathematician Herman Chernoff. These are upper bounds on the tail events of a random variable.

If $$X$$ has moment generating function $$M$$ then

1. $$\P(X \ge x) \le e^{-t x} M(t)$$ for $$t \gt 0$$
2. $$\P(X \le x) \le e^{-t x} M(t)$$ for $$t \lt 0$$
Proof
1. From Markov's inequality, $$\P(X \ge x) = \P\left(e^{t X} \ge e^{t x}\right) \le \E\left(e^{t X}\right) \big/ e^{t x} = e^{-t x} M(t)$$ if $$t \gt 0$$.
2. Similarly, $$\P(X \le x) = \P\left(e^{t X} \ge e^{t x}\right) \le e^{-t x} M(t)$$ if $$t \lt 0$$.

Naturally, the best Chernoff bound (in either (a) or (b)) is obtained by finding $$t$$ that minimizes $$e^{-t x} M(t)$$.

### The Characteristic Function

Our last generating function is the nicest from a mathematical point of view. Once again, we assume that our random variables are real-valued.

The characteristic function of $$X$$ is the function $$\chi$$ defined by by $\chi(t) = \E\left(e^{i t X}\right) = \E\left[\cos(t X)\right] + i \E\left[\sin(t X)\right], \quad t \in \R$

Note that $$\chi$$ is a complex valued function, and so this subsection requires some basic knowledge of complex analysis. The function $$\chi$$ is defined for all $$t \in \R$$ because the random variable in the expected value is bounded in magnitude. Indeed, $$\left|e^{i t X}\right| = 1$$ for all $$t \in \R$$. Many of the properties of the characteristic function are more elegant than the corresponding properties of the probability or moment generating functions, because the characteristic function always exists.

If $$X$$ has a continuous distribution on $$\R$$ with probability density function $$f$$ and characteristic function $$\chi$$ then $\chi(t) = \int_{-\infty}^{\infty} e^{i t x} f(x) dx, \quad t \in \R$

Proof

This follows from the change of variables theorem for expected value, albeit a complex version.

Thus, the characteristic function of $$X$$ is closely related to the Fourier transform of the probability density function $$f$$. The Fourier transform is named for Joseph Fourier, and is widely used in many areas of applied mathematics.

As with other generating functions, the characteristic function completely determines the distribution. That is, random variables $$X$$ and $$Y$$ have the same distribution if and only if they have the same characteristic function. Indeed, the general inversion formula given next is a formula for computing certain combinations of probabilities from the characteristic function.

Suppose again that $$X$$ has characteristic function $$\chi$$. If $$a, \, b \in \R$$ and $$a \lt b$$ then $\int_{-n}^n \frac{e^{-i a t} - e^{- i b t}}{2 \pi i t} \chi(t) \, dt \to \P(a \lt X \lt b) + \frac{1}{2}\left[\P(X = b) - \P(X = a)\right] \text{ as } n \to \infty$

The probability combinations on the right side completely determine the distribution of $$X$$. A special inversion formula holds for continuous distributions:

Suppose that $$X$$ has a continuous distribution with probability density function $$f$$ and characteristic function $$\chi$$. At every point $$x \in \R$$ where $$f$$ is differentiable, $f(x) = \frac{1}{2 \pi} \int_{-\infty}^\infty e^{-i t x} \chi(t) \, dt$

This formula is essentially the inverse Fourrier transform. As with the other generating functions, the characteristic function can be used to find the moments of $$X$$. Moreover, this can be done even when only some of the moments exist.

Suppose again that $$X$$ has characteristic function $$\chi$$. If $$n \in \N_+$$ and $$\E\left(\left|X^n\right|\right) \lt \infty$$. Then $\chi(t) = \sum_{k=0}^n \frac{\E\left(X^k\right)}{k!} (i t)^k + o(t^n)$ and therefore $$\chi^{(n)}(0) = i^n \E\left(X^n\right)$$.

Details

Recall that the last term is a generic function that satisfies $$o(t^n) / t^n \to 0$$ as $$t \to \infty$$.

Next we consider how the characteristic function is changed under some simple transformations of the variables.

Suppose that $$X$$ has characteristic function $$\chi$$ and that $$a, \, b \in \R$$. The characteristic function $$\psi$$ of $$Y = a + b X$$ is given by $$\psi(t) = e^{i a t} \chi(b t)$$ for $$t \in \R$$.

Proof

The proof is just like the one for the MGF: $$\psi(t) = \E\left[e^{i t (a + b X)}\right] = \E\left(e^{i t a} e^{i t b X}\right) = e^{i t a} \E\left[e^{i (t b) X}\right] = e^{i a t} \chi(b t)$$ for $$t \in \R$$.

Suppose that $$X_1$$ and $$X_2$$ are independent random variables with characteristic functions $$\chi_1$$ and $$\chi_2$$ respectively. The characteristic function $$\chi$$ of $$Y = X_1 + X_2$$ is given by $$\chi(t) = \chi_1(t) \chi_2(t)$$ for $$t \in \R$$.

Proof

Again, the proof is just like the one for the MGF: $\chi(t) = \E\left[e^{i t (X_1 + X_2)}\right] = \E\left(e^{i t X_1} e^{i t X_2}\right) = \E\left(e^{i t X_1}\right) \E\left(e^{i t X_2}\right) = \chi_1(t) \chi_2(t), \quad t \in \R$

The characteristic function of a random variable can be obtained from the moment generating function, under the basic existence condition that we saw earlier.

Suppose that $$X$$ has moment generating function $$M$$ that satisfies $$M(t) \lt \infty$$ for $$t$$ in an open interval $$I$$ about 0. Then the characteristic function $$\chi$$ of $$X$$ satisfies $$\chi(t) = M(i t)$$ for $$t \in I$$.

The final important property of characteristic functions that we will discuss relates to convergence in distribution. Suppose that $$(X_1, X_2, \ldots)$$ is a sequence of real-valued random with characteristic functions $$(\chi_1, \chi_2, \ldots)$$ respectively. Since we are only concerned with distributions, the random variables need not be defined on the same probability space.

The Continuity Theorem

1. If the distribution of $$X_n$$ converges to the distribution of a random variable $$X$$ as $$n \to \infty$$ and $$X$$ has characteristic function $$\chi$$, then $$\chi_n(t) \to \chi(t)$$ as $$n \to \infty$$ for all $$t \in \R$$.
2. Conversely, if $$\chi_n(t)$$ converges to a function $$\chi(t)$$ as $$n \to \infty$$ for $$t$$ in an open interval about 0, and if $$\chi$$ is continuous at 0, then $$\chi$$ is the characteristic function of a random variable $$X$$, and the distribution of $$X_n$$ converges to the distribution of $$X$$ as $$n \to \infty$$.

There are analogous versions of the continuity theorem for probability generating functions and moment generating functions. The continuity theorem can be used to prove the central limit theorem, one of the fundamental theorems of probability. Also, the continuity theorem has a straightforward generalization to distributions on $$\R^n$$.

### The Joint Characteristic Function

All of the generating functions that we have discussed have multivariate extensions. However, we will discuss the extension only for the characteristic function, the most important and versatile of the generating functions. There are analogous results for the other generating functions. So in this discussion, we assume that $$(X, Y)$$ is a random vector for our experiment, taking values in $$\R^2$$.

The (joint) characteristic function $$\chi$$ of $$(X, Y)$$ is defined by $\chi(s, t) = \E\left[\exp(i s X + i t Y)\right], \quad (s, t) \in \R^2$

Once again, the most important fact is that $$\chi$$ completely determines the distribution: two random vectors taking values in $$\R^2$$ have the same characteristic function if and only if they have the same distribution.

The joint moments can be obtained from the derivatives of the characteristic function.

Suppose that $$(X, Y)$$ has characteristic function $$\chi$$. If $$m, \, n \in \N$$ and $$\E\left(\left|X^m Y^n\right|\right) \lt \infty$$ then $\chi^{(m, n)}(0, 0) = e^{i \, (m + n)} \E\left(X^m Y^n\right)$

The marginal characteristic functions and the characteristic function of the sum can be easily obtained from the joint characteristic function:

Suppose again that $$(X, Y)$$ has characteristic function $$\chi$$, and let $$\chi_1$$, $$\chi_2$$, and $$\chi_+$$ denote the characteristic functions of $$X$$, $$Y$$, and $$X + Y$$, respectively. For $$t \in \R$$

1. $$\chi(t, 0) = \chi_1(t)$$
2. $$\chi(0, t) = \chi_2(t)$$
3. $$\chi(t, t) = \chi_+(t)$$
Proof

All three results follow immediately from the definitions.

Suppose again that $$\chi_1$$, $$\chi_2$$, and $$\chi$$ are the characteristic functions of $$X$$, $$Y$$, and $$(X, Y)$$ respectively. Then $$X$$ and $$Y$$ are independent if and only if $$\chi(s, t) = \chi_1(s) \chi_2(t)$$ for all $$(s, t) \in \R^2$$.

Naturally, the results for bivariate characteristic functions have analogies in the general multivariate case. Only the notation is more complicated.

## Examples and Applications

As always, be sure to try the computational problems yourself before expanding the solutions and answers in the text.

### Dice

Recall that an ace-six flat die is a six-sided die for which faces numbered 1 and 6 have probability $$\frac{1}{4}$$ each, while faces numbered 2, 3, 4, and 5 have probability $$\frac{1}{8}$$ each. Similarly, a 3-4 flat die is a six-sided die for which faces numbered 3 and 4 have probability $$\frac{1}{4}$$ each, while faces numbered 1, 2, 5, and 6 have probability $$\frac{1}{8}$$ each.

Suppose that an ace-six flat die and a 3-4 flat die are rolled. Use probability generating functions to find the probability density function of the sum of the scores.

Solution

Let $$X$$ and $$Y$$ denote the score on the ace-six die and 3-4 flat die, respectively. Then $$X$$ and $$Y$$ have PGFs $$P$$ and $$Q$$ given by \begin{align*} P(t) &= \frac{1}{4} t + \frac{1}{8} t^2 + \frac{1}{8} t^3 + \frac{1}{8} t^4 + \frac{1}{8} t^5 + \frac{1}{4} t^6, \quad t \in \R\\ Q(t) &= \frac{1}{8} t + \frac{1}{8} t^2 + \frac{1}{4} t^3 + \frac{1}{4} t^4 + \frac{1}{8} t^5 + \frac{1}{8} t^6, \quad t \in \R \end{align*} Hence $$X + Y$$ has PGF $$P Q$$. Expanding (a computer algebra program helps) gives $P(t) Q(t) = \frac{1}{32} t^2 + \frac{3}{64} t^3 + \frac{3}{32} t^4 + \frac{1}{8} t^5 + \frac{1}{8} t^6 + \frac{5}{32} t^7 + \frac{1}{8} t^8 + \frac{1}{8} t^9 + \frac{3}{32} t^{10} + \frac{3}{64} t^{11} + \frac{1}{32} t^{12}, \quad t \in \R$ Thus the PDF $$f$$ of $$X + Y$$ is given by $$f(2) = f(12) = \frac{1}{32}$$, $$f(3) = f(11) = \frac{3}{64}$$, $$f(4) = f(10) = \frac{3}{32}$$, $$f(5) = f(6) = f(8) = f(9) = \frac{1}{8}$$ and $$f(7) = \frac{5}{32}$$.

Two fair, 6-sided dice are rolled. One has faces numbered $$(0, 1, 2, 3, 4, 5)$$ and the other has faces numbered $$(0, 6, 12, 18, 24, 30)$$. Use probability generating functions to find the probability density function of the sum of the scores, and identify the distribution.

Solution

Let $$X$$ and $$Y$$ denote the score on the first die and the second die described, respectively. Then $$X$$ and $$Y$$ have PGFs $$P$$ and $$Q$$ given by \begin{align*} P(t) &= \frac{1}{6} \sum_{k=0}^5 t^k \quad t \in \R \\ Q(t) &= \frac{1}{6} \sum_{j=0}^5 t^{6 j} \quad t \in \R \end{align*} Hence $$X + Y$$ has PGF $$P Q$$. Simplifying gives $P(t) Q(t) = \frac{1}{36} \sum_{j=0}^5 \sum_{k=0}^5 t^{6 j + k} = \frac{1}{36} \sum_{n=0}^{35} t^n, \quad t \in \R$ Hence $$X + Y$$ is uniformly distributed on $$\{0, 1, 2, \ldots, 35\}$$.

Suppose that random variable $$Y$$ has probability generating function $$P$$ given by $P(t) = \left(\frac{2}{5} t + \frac{3}{10} t^2 + \frac{1}{5} t^3 + \frac{1}{10} t^4\right)^5, \quad t \in \R$

1. Interpret $$Y$$ in terms of rolling dice.
2. Use the probability generating function to find the first two factorial moments of $$Y$$.
3. Use (b) to find the variance of $$Y$$.
1. A four-sided die has faces numbered $$(1, 2, 3, 4)$$ with respective probabilities $$\left(\frac{2}{5}, \frac{3}{10}, \frac{1}{5}, \frac{1}{10}\right)$$. $$Y$$ is the sum of the scores when the die is rolled 5 times.
2. $$\E(Y) = P^\prime(1) = 10$$, $$\E[Y(Y - 1)] = P^{\prime \prime}(1) = 95$$
3. $$\var(Y) = 5$$

### Bernoulli Trials

Suppose $$X$$ is an indicator random variable with $$p = \P(X = 1)$$, where $$p \in [0, 1]$$ is a parameter. Then $$X$$ has probability generating function $$P(t) = 1 - p + p t$$ for $$t \in \R$$.

Proof

$$P(t) = \E\left(t^X\right) = t^0 (1 - p) + t^1 p = 1 - p + p t$$ for $$t \in \R$$.

Recall that a Bernoulli trials process is a sequence $$(X_1, X_2, \ldots)$$ of independent, identically distributed indicator random variables. In the usual language of reliability, $$X_i$$ denotes the outcome of trial $$i$$, where 1 denotes success and 0 denotes failure. The probability of success $$p = \P(X_i = 1)$$ is the basic parameter of the process. The process is named for Jacob Bernoulli. A separate chapter on the Bernoulli Trials explores this process in more detail.

For $$n \in \N_+$$, the number of successes in the first $$n$$ trials is $$Y_n = \sum_{i=1}^n X_i$$. Recall that this random variable has the binomial distribution with parameters $$n$$ and $$p$$, which has probability density function $$f_n$$ given by $f_n(y) = \binom{n}{y} p^y (1 - p)^{n - y}, \quad y \in \{0, 1, \ldots, n\}$

Random variable $$Y_n$$ has probability generating function $$P_n$$ given by $$P_n(t) = (1 - p + p t)^n$$ for $$t \in \R$$.

Proof

This follows immediately from the PGF of an indicator variable and the result for sums of independent variables.

Rando variable $$Y_n$$ has the following parameters:

1. $$\E\left[Y_n^{(k)}\right] = n^{(k)} p^k$$
2. $$\E\left(Y_n\right) = n p$$
3. $$\var\left(Y_n\right) = n p (1 - p)$$
4. $$\P(Y_n \text{ is even}) = \frac{1}{2}\left[1 - (1 - 2 p)^n\right]$$
Proof
1. Repeated differentiation gives $$P^{(k)}(t) = n^{(k)} p^k (1 - p + p t)^{n-k}$$. Hence $$P^{(k)}(1) = n^{(k)} p^k$$, which is $$\E\left[X^{(k)} \right]$$ by the moment result above.
2. This follows from the formula for mean.
3. This follows from the formula for variance.
4. This follows from the even value formula.

Suppose that $$U$$ has the binomial distribution with parameters $$m \in \N_+$$ and $$p \in [0, 1]$$, $$V$$ has the binomial distribution with parameters $$n \in \N_+$$ and $$q \in [0, 1]$$, and that $$U$$ and $$V$$ are independent.

1. If $$p = q$$ then $$U + V$$ has the binomial distribution with parameters $$m + n$$ and $$p$$.
2. If $$p \ne q$$ then $$U + V$$ does not have a binomial distribution.
Proof

From the result for sums of independent variables and the PGF of the binomial distribution, note that the probability generating function of $$U + V$$ is $$P(t) = (1 - p + p t)^m (1 - q + q t)^n$$ for $$t \in \R$$.

1. If $$p = q$$ then $$U + V$$ has PGF $$P(t) = (1 - p + p t)^{m + n}$$, which is the PGF of the binomial distribution with parameters $$m + n$$ and $$p$$.
2. On the other hand, if $$p \ne q$$, the PGF $$P$$ does not have the functional form of a binomial PGF.

Suppose now that $$p \in (0, 1]$$. The trial number $$N$$ of the first success in the sequence of Bernoulli trials has the geometric distribution on $$\N_+$$ with success parameter $$p$$. The probability density function $$h$$ is given by $h(n) = p (1 - p)^{n-1}, \quad n \in \N_+$ The geometric distribution is studied in more detail in the chapter on Bernoulli trials.

Let $$Q$$ denote the probability generating function of $$N$$. Then

1. $$Q(t) = \frac{p t}{1 - (1 - p)t}$$ for $$-\frac{1}{1 - p} \lt t \lt \frac{1}{1 - p}$$
2. $$\E\left[N^{(k)}\right] = k! \frac{(1 - p)^{k-1}}{p^k}$$ for $$k \in \N$$
3. $$\E(N) = \frac{1}{p}$$
4. $$\var(N) = \frac{1 - p}{p^2}$$
5. $$\P(N \text{ is even}) = \frac{1 - p}{2 - p}$$
Proof
1. Using the formula for the sum of a geometric series, $Q(t) = \sum_{n=1}^\infty (1 - p)^{n-1} p t^n = p t \sum_{n=1}^\infty [(1 - p) t]^{n-1} = \frac{p t}{1 - (1 - p) t}, \quad \left|(1 - p) t\right| \lt 1$
2. Repeated differentiation gives $$H^{(k)}(t) = k! p (1 - p)^{k-1} \left[1 - (1 - p) t\right]^{-(k+1)}$$ and then the result follows from the inversion formula.
3. This follows from (b) and the formula for mean.
4. This follows from (b) and the formula for variance.
5. This follows from even value formula.

The probability that $$N$$ is even comes up in the alternating coin tossing game with two players.

### The Poisson Distribution

Recall that the Poisson distribution has probability density function $$f$$ given by $f(n) = e^{-a} \frac{a^n}{n!}, \quad n \in \N$ where $$a \in (0, \infty)$$ is a parameter. The Poisson distribution is named after Simeon Poisson and is widely used to model the number of random points in a region of time or space; the parameter is proportional to the size of the region of time or space. The Poisson distribution is studied in more detail in the chapter on the Poisson Process.

Suppose that $$N$$ has Poisson distribution with parameter $$a \in (0, \infty)$$. Let $$P_a$$ denote the probability generating function of $$N$$. Then

1. $$P_a(t) = e^{a (t - 1)}$$ for $$t \in \R$$
2. $$\E\left[N^{(k)}\right] = a^k$$
3. $$\E(N) = a$$
4. $$\var(N) = a$$
5. $$\P(N \text{ is even}) = \frac{1}{2}\left(1 + e^{-2 a}\right)$$
Proof
1. Using the exponential series, $P_a(t) = \sum_{n=0}^\infty e^{-a} \frac{a^n}{n!} t^n = e^{-a} \sum_{n=0}^\infty \frac{(a t)^n}{n!} = e^{-a} e^{a t}, \quad t \in \R$
2. Repeated differentiation gives $$P_a^{(k)}(t) = e^{a (t - 1)} a^k$$, so the result follows from inversion formula.
3. This follows from (b) and the formula for mean.
4. This follows from (b) and the formula for variance.
5. This follows from even value formula.

The Poisson family of distributions is closed with respect to sums of independent variables, a very important property.

Suppose that $$X, \, Y$$ have Poisson distributions with parameters $$a, \, b \in (0, \infty)$$, respectively, and that $$X$$ and $$Y$$ are independent. Then $$X + Y$$ has the Poisson distribution with parameter $$a + b$$.

Proof

In the notation of the previous result, note that $$P_a P_b = P_{a+b}$$.

The right distribution function of the Poisson distribution does not have a simple, closed-form expression. The following exercise gives an upper bound.

Suppose that $$N$$ has the Poisson distribution with parameter $$a \gt 0$$. Then $\P(N \ge n) \le e^{n - a} \left(\frac{a}{n}\right)^n, \quad n \gt a$

Proof

The PGF of $$N$$ is $$P(t) = e^{a (t - 1)}$$ and hence the MGF is $$P\left(e^t\right) = \exp\left(a e^t - a\right)$$. From the Chernov bounds we have $\P(N \ge n) \le e^{-t n} \exp\left(a e^t - a\right) = \exp\left(a e^t - a - tn\right)$ If $$n \gt a$$ the expression on the right is minimized when $$t = \ln(n / a)$$. Substituting gives the upper bound.

The following theorem gives an important convergence result that is explored in more detail in the chapter on the Poisson process.

Suppose that $$p_n \in (0, 1)$$ for $$n \in \N_+$$ and that $$n p_n \to a \in (0, \infty)$$ as $$n \to \infty$$. Then the binomial distribution with parameters $$n$$ and $$p_n$$ converges to the Poisson distribution with parameter $$a$$ as $$n \to \infty$$.

Proof

Let $$P_n$$ denote the probability generating function of the binomial distribution with parameters $$n$$ and $$p_n$$. From the PGF of the binomial distribution we have $P_n(t) = \left[1 + p_n (t - 1)\right]^n = \left[1 + \frac{n p_n (t - 1)}{n}\right]^n, \quad t \in \R$ Using a famous theorem from calculus, $$P_n(t) \to e^{a (t - 1)}$$ as $$n \to \infty$$. But this is the PGF of the Poisson distribution with parameter $$a$$, so the result follows from the continuity theorem for PGFs.

### The Exponential Distribution

Recall that the exponential distribution is a continuous distribution on $$[0, \infty)$$ with probability density function $$f$$ given by $f(t) = r e^{-r t}, \quad t \in (0, \infty)$ where $$r \in (0, \infty)$$ is the rate parameter. This distribution is widely used to model failure times and other random times, and in particular governs the time between arrivals in the Poisson model. The exponential distribution is studied in more detail in the chapter on the Poisson Process.

Suppose that $$T$$ has the exponential distribution with rate parameter $$r \in (0, \infty)$$ and let $$M$$ denote the moment generating function of $$T$$. Then

1. $$M(s) = \frac{r}{r - s}$$ for $$s \in (-\infty, r)$$.
2. $$\E(T^n) = n! / r^n$$ for $$n \in \N$$
Proof
1. $$M(s) = \E\left(e^{s T}\right) = \int_0^\infty e^{s t} r e^{-r t} \, dt = \int_0^\infty r e^{(s - r) t} \, dt = \frac{r}{r - s}$$ for $$s \lt r$$.
2. $$M^{(n)}(s) = \frac{r n!}{(r - s)^{n+1}}$$ for $$n \in \N$$

Suppose that $$(T_1, T_2, \ldots)$$ is a sequence of independent random variables, each having the exponential distribution with rate parameter $$r \in (0, \infty)$$. For $$n \in \N_+$$, the moment generating function $$M_n$$ of $$U_n = \sum_{i=1}^n T_i$$ is given by $M_n(s) = \left(\frac{r}{r - s}\right)^n, \quad s \in (-\infty, r)$

Proof

This follows from the previous result and the result for sums of independent variables.

Random variable $$U_n$$ has the Erlang distribution with shape parameter $$n$$ and rate parameter $$r$$, named for Agner Erlang. This distribution governs the $$n$$th arrival time in the Poisson model. The Erlang distribution is a special case of the gamma distribution and is studied in more detail in the chapter on the Poisson Process.

### Uniform Distributions

Suppose that $$a, \, b \in \R$$ and $$a \lt b$$. Recall that the continuous uniform distribution on the interval $$[a, b]$$ has probability density function $$f$$ given by $f(x) = \frac{1}{b - a}, \quad x \in [a, b]$ The distribution corresponds to selecting a point at random from the interval. Continuous uniform distributions arise in geometric probability and a variety of other applied problems.

Suppose that $$X$$ is uniformly distributed on the interval $$[a, b]$$ and let $$M$$ denote the moment generating function of $$X$$. Then

1. $$M(t) = \frac{e^{b t} - e^{a t}}{(b - a)t}$$ if $$t \ne 0$$ and $$M(0) = 1$$
2. $$\E\left(X^n\right) = \frac{b^{n+1} - a^{n + 1}}{(n + 1)(b - a)}$$ for $$n \in \N$$
Proof
1. $$M(t) = \int_a^b e^{t x} \frac{1}{b - a} \, dx = \frac{e^{b t} - e^{a t}}{(b - a)t}$$ if $$t \ne 0$$. Trivially $$M(0) = 1$$
2. This is a case where the MGF is not helpful, and it's much easier to compute the moments directly: $$\E\left(X^n\right) = \int_a^b x^n \frac{1}{b - a} \, dx = \frac{b^{n+1} - a^{n + 1}}{(n + 1)(b - a)}$$

Suppose that $$(X, Y)$$ is uniformly distributed on the triangle $$T = \{(x, y) \in \R^2: 0 \le x \le y \le 1\}$$. Compute each of the following:

1. The joint moment generating function of $$(X, Y)$$.
2. The moment generating function of $$X$$.
3. The moment generating function of $$Y$$.
4. The moment generating function of $$X + Y$$.
1. $$M(s, t) = 2 \frac{e^{s+t} - 1}{s (s + t)} - 2 \frac{e^t - 1}{s t}$$ if $$s \ne 0, \; t \ne 0$$. $$M(0, 0) = 1$$
2. $$M_1(s) = 2 \left(\frac{e^2}{s^2} - \frac{1}{s^2} - \frac{1}{s}\right)$$ if $$s \ne 0$$. $$M_1(0) = 1$$
3. $$M_2(t) = 2 \frac{t e^t - e^t + 1}{t^2}$$ if $$t \ne 0$$. $$M_2(0) = 1$$
4. $$M_+(t) = \frac{e^{2 t} - 1}{t^2} - 2 \frac{e^t - 1}{t^2}$$ if $$t \ne 0$$. $$M_+(0) = 1$$

### A Bivariate Distribution

Suppose that $$(X, Y)$$ has probability density function $$f$$ given by $$f(x, y) = x + y$$ for $$(x, y) \in [0, 1]^2$$. Compute each of the following:

1. The joint moment generating function $$(X, Y)$$.
2. The moment generating function of $$X$$.
3. The moment generating function of $$Y$$.
4. The moment generating function of $$X + Y$$.
1. $$M(s, t) = \frac{e^{s+t}(-2 s t + s + t) + e^s(s t - s - t) + s + t}{s^2 t^2}$$ if $$s \ne 0, \, t \ne 0$$. $$M(0, 0) = 1$$
2. $$M_1(s) = \frac{3 s e^2 - 2 e^2 - s + 2}{2 s^2}$$ if $$s \ne 0$$. $$M_1(0) = 1$$
3. $$M_2(t) = \frac{3 t e^t - 2 e^t - t + 2}{2 t^2}$$ if $$t \ne 0$$. $$M_2(0) = 1$$
4. $$M_+(t) = \frac{[e^{2 t}(1 - t) + e^t (t - 2) + 1]}{t^3}$$ if $$t \ne 0$$. $$M_+(0) = 1$$

### The Normal Distribution

Recall that the standard normal distribution is a continuous distribution on $$\R) with probability density function \(\phi$$ given by $\phi(z) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} z^2}, \quad z \in \R$ Normal distributions are widely used to model physical measurements subject to small, random errors and are studied in more detail in the chapter on Special Distributions.

Suppose that $$Z$$ has the standard normal distribution and let $$M$$ denote the moment generating function of $$Z$$. Then

1. $$M(t) = e^{\frac{1}{2} t^2}$$ for $$t \in \R$$
2. $$\E\left(Z^n\right) = 1 \cdot 3 \cdots (n - 1)$$ if $$n$$ is even and $$\E\left(Z^n\right) = 0$$ if $$n$$ is odd.
Proof
1. First, $M(t) = \E\left(e^{t Z}\right) = \int_{-\infty}^\infty e^{t z} \frac{1}{\sqrt{2 \pi}} e^{-z^2 / 2} \, dz = \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi}} \exp\left(-\frac{z^2}{2} + t z\right) \, dz$ Completing the square in $$z$$ gives $$\exp\left(-\frac{z^2}{2} + t z\right) = \exp\left[\frac{1}{2} t^2 - \frac{1}{2}(z - t)^2 \right] = e^{\frac{1}{2} t^2} \exp\left[-\frac{1}{2} (z - t)^2\right]$$. hence $M(t) = e^{\frac{1}{2} t^2} \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi}} \exp\left[-\frac{1}{2} (z - t)^2\right] \, dz = e^{\frac{1}{2} t^2}$ because the function of $$z$$ in the last integral is the probability density function for the normal distribution with mean $$t$$ and variance 1.
2. Note that $$M^\prime(t) = t M(t)$$. Thus, repeated differentiation gives $$M^{(n)}(t) = p_n(t) M(t)$$ for $$n \in \N$$, where $$p_n$$ is a polynomial of degree $$n$$ satisfying $$p_{n+1}^\prime(t) = t p_n(t) + p_n^\prime(t)$$. Since $$p_0 = 1$$, it's easy to see that $$p_n$$ has only even or only odd terms, depending on whether $$n$$ is even or odd, respectively. Thus, $$\E\left(X^n\right) = p_n(0)$$. This is 0 if $$n$$ is odd, and is the constant term $$1 \cdot 3 \cdots (n - 1)$$ if $$n$$ is even. Of course, we can also see that the odd order moments must be 0 by symmetry.

More generally, for $$\mu \in \R$$ and $$\sigma \in (0, \infty)$$, recall that the normal distribution with mean $$\mu$$ and standard deviation $$\sigma$$ is a continuous distribution on $$\R$$ with probability density function $$f$$ given by $f(x) = \frac{1}{\sqrt{2 \pi} \sigma} \exp\left[-\frac{1}{2}\left(\frac{x - \mu}{\sigma}\right)^2\right], \quad x \in \R$ Moreover, if $$Z$$ has the standard normal distribution, then $$X = \mu + \sigma Z$$ has the normal distribution with mean $$\mu$$ and standard deviation $$\sigma$$. Thus, we can easily find the moment generating function of $$X$$:

Suppose that $$X$$ has the normal distribution with mean $$\mu$$ and standard deviation $$\sigma$$. The moment generating function of $$X$$ is $M(t) = \exp\left(\mu t + \frac{1}{2} \sigma^2 t^2\right), \quad t \in \R$

Proof

This follows easily the previous result and the result for linear transformations: $$X = \mu + \sigma Z$$ where $$Z$$ has the standard normal distribution. Hence $M(t) = \E\left(e^{t X}\right) = e^{\mu t} \E\left(e^{\sigma t Z}\right) = e^{\mu t} e^{\frac{1}{2} \sigma^2 t ^2}, \quad t \in \R$

So the normal family of distributions in closed under location-scale transformations. The family is also closed with respect to sums of independent variables:

If $$X$$ and $$Y$$ are independent, normally distributed random variables then $$X + Y$$ has a normal distribution.

Proof

Suppose that $$X$$ has the normal distribution with mean $$\mu \in \R$$ and standard deviation $$\sigma \in (0, \infty)$$, and that $$Y$$ has the normal distribution with mean $$\nu \in \R$$ and standard deviation $$\tau \in (0, \infty)$$. By (14), the MGF of $$X + Y$$ is $M_{X+Y}(t) = M_X(t) M_Y(t) = \exp\left(\mu t + \frac{1}{2} \sigma^2 t^2\right) \exp\left(\nu t + \frac{1}{2} \tau^2 t^2\right) = \exp\left[(\mu + \nu) t + \frac{1}{2}\left(\sigma^2 + \tau^2\right) t^2 \right]$ which we recognize as the MGF of the normal distribution with mean $$\mu + \nu$$ and variance $$\sigma^2 + \tau^2$$. Of course, we already knew that $$\E(X + Y) = \E(X) + \E(Y)$$, and since $$X$$ and $$Y$$ are independent, $$\var(X + Y) = \var(X) + \var(Y)$$, so the new information is that the distribution is also normal.

### The Pareto Distribution

Recall that the Pareto distribution is a continuous distribution on $$[1, \infty) with probability density function \(f$$ given by $f(x) = \frac{a}{x^{a + 1}}, \quad x \in [1, \infty)$ where $$a \in (0, \infty)$$ is the shape parameter. The Pareto distribution is named for Vilfredo Pareto. It is a heavy-tailed distribution that is widely used to model financial variables such as income. The Pareto distribution is studied in more detail in the chapter on Special Distributions.

Suppose that $$X$$ has the Pareto distribution with shape parameter $$a$$, and let $$M$$ denote the moment generating function of $$X$$. Then

1. $$\E\left(X^n\right) = \frac{a}{a - n}$$ if $$n \lt a$$ and $$\E\left(X^n\right) = \infty$$ if $$n \ge a$$
2. $$M(t) = \infty$$ for $$t \gt 0$$
Proof
1. We have seen this computation before. $$\E\left(X^n\right) = \int_1^\infty x^n \frac{a}{x^{a+1}} \, dx = \int_1^\infty x^{n - a - 1} \, dx$$. The integral evaluates to $$\frac{a}{a - n}$$ if $$n \lt a$$ and $$\infty$$ if $$n \ge a$$.
2. This follows from part (a). Since $$X \ge 1$$, $$M(t)$$ is increasing in $$t$$. Thus $$M(t) \le 1$$ if $$t \lt 0$$. If $$M(t) \lt \infty$$ for some $$t \gt 0$$, then $$M(t)$$ would be finite for $$t$$ in an open interval about 0, in which case $$X$$ would have finite moments of all orders. Of course, it's also easy to see directly from the integral that $$M(t) = \infty$$ for $$t \gt 0$$

On the other hand, like all distributions on $$\R$$, the Pareto distribution has a characteristic function. However, the characteristic function of the Pareto distribution does not have a simple, closed form.

### The Cauchy Distribution

Recall that the (standard) Cauchy distribution is a continuous distribution on $$\R$$ with probability density function $$f$$ given by $f(x) = \frac{1}{\pi \left(1 + x^2\right)}, \quad x \in \R$ and is named for Augustin Cauchy. The Cauch distribution is studied in more generality in the chapter on Special Distributions. The graph of $$f$$ is known as the Witch of Agnesi, named for Maria Agnesi.

Suppose that $$X$$ has the standard Cauchy distribution, and let $$M$$ denote the moment generating function of $$X$$. Then

1. $$\E(X)$$ does not exist.
2. $$M(t) = \infty$$ for $$t \ne 0$$.
Proof
1. We have seen this computation before. $$\int_a^\infty \frac{x}{\pi (1 + x^2)} \, dx = \infty$$ and $$\int_{-\infty}^a \frac{x}{\pi (1 + x^2)} \, dx = -\infty$$ for every $$a \in \R$$, so $$\int_{-\infty}^\infty \frac{x}{\pi (1 + x^2)} \, dx$$ does not exist.
2. Note that $$\int_0^\infty \frac{e^{t x}}{\pi (1 + x^2) } \, dx = \infty$$ if $$t \ge 0$$ and $$\int_{-\infty}^0 \frac{e^{t x}}{\pi (1 + x^2)} \, dx = \infty$$ if $$t \le 0$$.

Once again, all distributions on $$\R$$ have characteristic functions, and the standard Cauchy distribution has a particularly simple one.

Let $$\chi$$ denote the characteristic function of $$X$$. Then $$\chi(t) = e^{-\left|t\right|}$$ for $$t \in \R$$.

Proof

The proof of this result requires contour integrals in the complex plane, and is given in the section on the Cauchy distribution in the chapter on special distributions.

### Counterexample

For the Pareto distribution, only some of the moments are finite; so course, the moment generating function cannot be finite in an interval about 0. We will now give an example of a distribution for which all of the moments are finite, yet still the moment generating function is not finite in any interval about 0. Furthermore, we will see two different distributions that have the same moments of all orders.

Suppose that Z has the standard normal distribution and let $$X = e^Z$$. The distribution of $$X$$ is known as the (standard) lognormal distribution. The lognormal distribution is studied in more generality in the chapter on Special Distributions. This distribution has finite moments of all orders, but infinite moment generating function.

$$X$$ has probability density function $$f$$ given by $f(x) = \frac{1}{\sqrt{2 \pi} x} \exp\left(-\frac{1}{2} \ln^2(x)\right), \quad x \gt 0$

1. $$\E\left(X^n\right) = e^{\frac{1}{2}n^2}$$ for $$n \in \N$$.
2. $$\E\left(e^{t X}\right) = \infty$$ for $$t \gt 0$$.
Proof

We use the change of variables theorem. The transformation is $$x = e^z$$ so the inverse transformation is $$z = \ln x$$ for $$x \in (0, \infty)$$ and $$z \in \R$$. Letting $$\phi$$ denote the PDF of $$Z$$, it follows that the PDF of $$X$$ is $$f(x) = \phi(z) \, dz / dx = \phi\left(\ln x\right) \big/ x$$ for $$x \gt 0$$.

1. We use the moment generating function of the standard normal distribution given above: $$\E\left(X^n\right) = \E\left(e^{n Z}\right) = e^{n^2 / 2}$$.
2. Note that $\E\left(e^{t X}\right) = \E\left[\sum_{n=0}^\infty \frac{(t X)^n}{n!}\right] = \sum_{n=0}^\infty \frac{\E(X^n)}{n!} t^n = \sum_{n=0}^\infty \frac{e^{n^2 / 2}}{n!} t^n = \infty, \quad t \gt 0$ The interchange of expected value and sum is justified since $$X$$ is nonnegative. See the advanced section on properties of the integral in the chapter on Distributions for more details.

Next we construct a different distribution with the same moments as $$X$$.

Let $$h$$ be the function defined by $$h(x) = \sin\left(2 \pi \ln x \right)$$ for $$x \gt 0$$ and let $$g$$ be the function defined by $$g(x) = f(x)\left[1 + h(x)\right]$$ for $$x \gt 0$$. Then

1. $$g$$ is a probability density function.
2. If $$Y$$ has probability density function $$g$$ then $$\E\left(Y^n\right) = e^{\frac{1}{2} n^2}$$ for $$n \in \N$$
Proof


text:
## Tuesday, May 01, 2007

In recent weeks, monetary policy in India has faced many problems. Every plausible monetary policy regime ends up stabilising something - inflation, or interest rates, or exchange rates. You can disagree with what is being stabilised, but atleast something is. In recent weeks, India has suffered from huge volatility in all three. These suggest that the monetary policy regime is in distress. (See my blog entry from 7 March.)

In recent months, huge purchases of foreign currency were incompletely sterilised, fueling local inflation. These choices by RBI showed that the reaction function of RBI valued exchange rate policy over controlling inflation, even though the public rhetoric was full of talk about inflation. In mid-March, RBI appears to have stopped trading on the currency market. The INR appreciated sharply.

But there are numerous un-answered questions. Has the currency regime changed? What is the stance of monetary policy? Unfortunately, there are many uncertainties. The actions of RBI are the major risk factor that shapes future uncertainty. But it is not clear what RBI will do. The market does not know the monetary policy regime; it doesn't know the reaction function of RBI.

With this breakdown of the monetary policy framework, the recent credit policy statement would have been an ideal time for an introspective RBI to have displayed thought leadership, and offered a positive message of change. If such thinking is taking place, it is certainly not being communicated. RBI says that nothing was ever wrong and nothing needs to change. As an example:

q: Is there a need for the currency management policy to be changed as far as monetary management is concerned?

a: The exchange rate policy has served us very well in the last so many years and the exchange rate has been commended all around.

It has served the purpose of growth, stability, and external sector balance and there is no need to change any element of the exchange rate policy.

RBI believes that inflation targeting hasn't worked in the UK:

"Experience shows that inflation targeting has not worked in countries which had set formal inflation targets," Reserve Bank Governor YV Reddy said here.

Citing the example of the UK, he said that country had an open capital account, a pure monetary authority and an exchange rate which was virtually free, yet it had not succeeded in inflation targeting.

And, RBI seems to believe that an INR appreciation won't generate a negative impact on local prices:

Reserve Bank of India (RBI) governor Y V Reddy today disputed the notion that there is an inverse relation between the movement of the rupee and the direction of prices. The rupees rise to nine-year highs will not necessarily bring inflation down, he pointed out.

These unfortunate statements generate a loss of credibility and leave the market highly confused. Just in case you felt that RBI's retreat from currency trading in mid-March signified a change in the monetary policy regime, RBI is atleast saying this is not the case. Market manipulation of the currency market has gone down - does this mean we are headed for a more floating rate? But no, RBI recently bumped up the limit on MSS - so are we in for a bit more currency trading? (Isn't it time for MOF to question MSS?) Nobody knows what the next steps of monetary policy will be.

Business Standard had a fascinating edit titled Confusing signals on the subject on 30 April:

Has the Reserve Bank of India (RBI) changed its mind onand its attitude tocapital flows, currency management and how to cope with the problems caused by its established monetary policy? That would certainly seem to be the case, given the sharp rise of the rupee against the dollar in recent weeks. Perhaps the continuing strength of the capital inflow made it impossible for the RBI to continue with the old policy of trying to keep the rupee down by buying dollars and then sterilising the rupee funds that were generated by this action so that inflation did not get out of hand. Inflation was refusing to come under control, the dollar inflow was intensifying and monetary policy was coming unstuckwith higher interest rates causing their own problems. A change of tack may therefore have become inevitable.

From November 2006 to February 2007, the Reserve Bank of India spent Rs 88,000 crore to buy $19.7 billion on the currency market. These purchases are now worth about Rs 80,000 crore today, so the fall of the dollar has meant a substantial loss. An interest cost of 7 per cent, applied on Rs 80,000 crore worth of securities issued to sterilise rupee funds, would mean a further outflow of Rs 5,600 crore a year. Those are big numbers: Rs 14,000 crore would build a national highway from Delhi to Kanyakumari. But the true costs of the RBIs currency policy are even bigger. If dollar assets in November were$100 billion, a 10 per cent currency appreciation imposes a fiscal cost of 1 per cent of GDP. On top of that, there are costs owing to the poor returns on the reserves portfolio. The biggest cost of the policy, however, is not fiscal but political. Only a partial sterilisation of the dollar inflow was possible. As a result, in 2006-07, reserve money grew at 24 per cent, in contrast to 17 per cent in the previous year, resulting in higher inflation, which is politically unacceptable.

A higher external value for the rupee, which results from allowing the dollars to flow in and the RBI choosing not to intervene, is not without costs; the export slowdown that is already visible is one result. That is what has now made proponents of an activist currency policy unhappy, mindful as they are of what an export slowdown can mean to GDP growth and other macro-variables. The question is whether the RBI had a choice. Certainly, Y V Reddy, the RBI governor, seems to be singing a new tune after abandoning the erstwhile currency policy. With rising interest rates, inflation rates and money supply, the problems of monetary policy in India are visible to all. The question is whether a critical switch has been made in recent weeks, from the policies which worked fine in a closed economy, to a different policy matrix more in tune with an economy that is integrating with the rest of the world. The RBI does not make clear whether this is the case, and certainly its policy pronouncements should be different if it has indeed changed tack.

The centrepiece of a monetary policy statement is the central banks forecast of future inflation, which should guide interest rate setting today. Last weeks policy statement desires inflation of 4-4.5 per cent, but does not offer a path to that destination. While the stock market has been ecstatic because interest rates have not been jacked up again, the underlying problems remain. The government will want consumer price inflation to be no more than 3 per cent by 2008-09, so it is possible that the RBI may have to surprise markets again.

On the same day, Abheek Barua said:

Whatever the RBIs decision, it is imperative at this stage to communicate this to the markets if it is to put an end to the flux that currently prevails. If indeed, we are on the cusp of structural change in monetary management, the markets need to prepare for it. Second, if a less interventionist structure is indeed a short-term objective, the RBI has to recognise the imperfections and asymmetries in market structure that currently exist. These imperfections make a section of the banking system more vulnerable to the enhanced volatility that comes with a freer regime. Until the financial system moves to a new equilibrium, the central bank could perhaps provide these banks some succour.

What is to be done? The ideal thing is to have a consistent monetary policy regime. E.g. in the UK or the US, economic agents are not puzzled by the central bank; monetary policy decisions are not a source of risk. Even if a proper monetary policy regime is not in place, Ila Patnaik points out that a lot more transparency will help economic agents decipher what RBI is doing. Intuitively, think that economic agents, when armed with enough information, will run the right regressions and figure out what monetary policy is doing - regardless of what the central bank says.

I find that foreign banks or foreign financial firms often strongly advocate currency management by RBI. A pegged exchange rate, after all, is a gift of free risk management services for them. In addition, there are the great opportunities to make money from one-way bets, either speculative attacks or reverse speculative attacks. I find it fascinating to look back at how the people who saw the one-way bet on the INR a while ago have come out right.

Benn Steil and Robert Litan [book] have been preaching that third world countries lack the institutional capacity to do inflation targeting, and should just dollarise. At times like this, it's something to think of.

1. Great article!!!
but inflation targeting isn't the be all end all of monetary policy...as BoE is finding out
This is especially true in India with headline inflation far from the "true" measure of inflation

http://www.bloomberg.com/apps/news?pid=20601087&sid=azKW.ik2JECg&refer=home

2. moreover, narrow inflation targeting (or dollarization, for the matter) can lead to asset price booms (or busts) that pose systemic risks to the economy

3. Problems of targeting inflation in the UK - any RBI governor would LOVE to have these problems. The problems that we face are n times worse.

4. While there may be merits for inflation targeting in advanced industrial countries (US, UK, EU), in my opinion, India is not ready for narrow and explicitly defined inflation targets for the following reasons.

1. Measurement: The problems with WPI as a measure of inflation in India are well known. Most of the components in the WPI (Food articles, Food products, Fruit & vegetables, Milk, Oilseeds) are influenced by seasonal factors. Holding RBI accountable to a narrowly defined WPI target would lead to seasonal factors dominating underlying inflationary trends while setting inflation rates. Deficient rainfall in certain regions will increase headline inflation numbers without affecting underlying trends, leading the RBI to increase interest rates precisely when counter-cyclical policy may required. On the other hand, abundant rainfall would boost output and lower WPI, leading RBI to decrease interest rates when perhaps they should have been increased. RBI will be forced to follow pro-cyclical monetary policy instead of a counter-cyclical one.
2. It is as good as impossible for the RBI and the CSO to come up with a stripped measure of inflation (like core inflation in the US) that reflects the consumption basket of the average Indian household, since expenditure on foodstuffs constitute a large part of most Indian household budget.
3. The credibility of the RBI is as good as non-existent. It is hard to imagine that the financial markets will believe that the RBI will stick to the inflation targets, especially since it does not have legal independence. If setting explicit inflation targets does not increase central banks credibility, why bother. There may be better ways to do that in the Indian context.
4. The government of the day (especially those who believe in 10% growth as India’s divine right) will engage in expansionary fiscal policy to counter a tight monetary policy. The FRBM Act won’t stop the government to engage in off balance sheet expenditures (such as oil bonds or using foreign exchange reserves for infrastructure) to boost growth, further fueling inflation. If the RBI refuses to monetize the expansionary fiscal policy, the government can force public sector banks or EPFO to step in. Persistent contractionary monetary policy many even lead to the central government going back on some of the fiscal reforms and push implementation of FRBM Act forward.
5. Inflation and inflation expectations in India have been the result of supply shocks (rather than demand shocks). Given the weak mechanisms of transmission of monetary policy, it will be difficult for the RBI to control inflation in the short run without hurting the real sector. With strict inflation targets (and the pay of the RBI governor linked to the inflation targets being achieved), more often than not, the RBI will raise interest rates more than necessary chocking growth.
6. For better or worse, the RBI controls the exchange rate. Floating the rupee, as would be required for inflation targeting, would certainly be desirable in the medium term but not in the short term. A high degree of volatility will certainly not suit India’s growth prospects at this critical juncture.

Yes, inflation targeting should be a medium term goal but the institutional deficits of the RBI, lack of good high frequency data and the presence of a large number of households engaged in primary activities make inflation targeting unsuitable for India.

5. You may also want to look at "Inflation Targeting in India: Issues and Prospects" by R. Jha

http://rspas.anu.edu.au/papers/asarc/WP2005_04.pdf

OR

"How Applicable is the Inflation Targeting Framework for India" by S. Chand and K. Singh

http://folk.uio.no/sheetakc/India%20policy%20forum%20paper.pdf

6. Wasn't the Bretton Woods System of the 50s and 60s essentially an example of a global currency regime? Is a return to the pre 70s regime feasible in the present context of free capital and current accounts?

Plese shed light on this.
TIA

7. to shrikanth,

A return to the Bretton Woods system is not possible as long as we have fiat money. And if, with volatile capital flows and countries with huge current account surpluses and deficits, we implement a pegged currency regime, it will be deflationary in some countries and inflationary in others.

I am not sure we have enough gold reserves to support the kind of capital flows we have currently.

In my opinion, it will be better for countries in the optimal currency areas to have a common currency. This will align business cycles and curtail fiscal profilgacy and the tendency of the Fed and BoJ to print money.

8. yeah foreign banks are biased. They want RBi to intervene ? strange. All their reports from March april from their reseach desk paints pictures of huge RBI intervention. May be they want more of forex reserves to manage from their NY offices.

9. Dollarize? I dare say your friends who make up the "financial interest" in Bombay might have something to say about that. Surely hoping for the steady penetration of Government institutions by this interest, leading to gradual market development, is better than annihilating it and forgoing politically-critical monetary sovereignty altogether?

I realize you're being slightly facetious in this suggestion, but once more you betray that neoliberal tendency for utopian, quick-fix solutions by doing so. When will you lot realize that people will not play musical chairs with their lives with global finance calling the tune? In the immortal words of Princess Leia, "The more you tighten your grip, the more [human] systems will slip through your fingers."

10. I am crystal clear that in a first best world, India has monetary policy autonomy plus an open capital account plus a floating rate. But this subtle institutional dance, that is required to master fiat money, is not easy. It requires a sophisticated institutional capacity and fair knowledge of economics.

We are now in the worst of all worlds. We have the inefficiencies and rent seeking of capital controls + lack of stable interest rates + lack of stable currency + lack of stable and predictable inflation + complete confusion in the minds of the market about what is going on.

Dollarisation is not the best, but it's better than this. If we can't make it to first best, we should seriously consider it. Think about it: You'd have all of (a) an open capital account (b) stable currency (c) a well known and understood monetary policy regime. The only thing it lacks when compared with inflation targeting is: an autonomous monetary policy. But it'd be better than where we are.

11. Dollarization would forcibly align India's business cycle with the US. Just take a look at the Irish property bubble to see why this is neither benefitial nor desirable.

In the unlikely event that India dolarizes, The Feds monetary policy would be too accomodative for India, adding fuel to the inflationary fire.

As it is we have asset price bubbles. Dolarization would only fuel those bubbles. And when the bubble bursts, neither of the three benefits you outline would be of any consequence.

12. "The ideal thing is to have a consistent monetary policy regime. E.g. in the UK or the US, economic agents are not puzzled by the central bank; monetary policy decisions are not a source of risk."

Surely you're joking?

In one kind of view, it's the only source of not-already-priced-in risk...there's a reason why the Fed likes being gnomic, it's just that the RBI can't help but be that way.

So, the real question is, as you surely know, what are the different kinds of risks available here? How do different interests within the RBI work themselves out?

Or, are your neoliberal interests such that dollarization will do for any old entirely contrived reason?

13. Fed is not the epitome of certainty you make it out to be. There is no consensus among "economic actors" about the course of monerary policy a few months down the line. Bear Stears projects federal funds rate to by 50 bps higher while Goldman and UBS project it be to 50 bps lower by the end of this year.

A central bank can only be as certain as the data allows them to be.

A bit of uncertainty may not be that bad afterall. Given the weak short term transmission mechanisms for monetary policy in India, the only way to RBI to have an impact is to pull things out of a hat.

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MRI-based vocal tract representations for the three-dimensional finite element synthesis of diphthongs
KTH, School of Electrical Engineering and Computer Science (EECS), Intelligenta system, Speech, Music and Hearing, TMH, Speech Communication and Technology.ORCID iD: 0000-0003-4532-014X
2019 (English)In: IEEE Transactions on Audio, Speech, and Language Processing, ISSN 1558-7916, E-ISSN 1558-7924, Vol. 27, no 12, p. 2173-2182Article in journal (Refereed) In press
##### Abstract [en]

The synthesis of diphthongs in three-dimensions (3D) involves the simulation of acoustic waves propagating through a complex 3D vocal tract geometry that deforms over time. Accurate 3D vocal tract geometries can be extracted from Magnetic Resonance Imaging (MRI), but due to long acquisition times, only static sounds can be currently studied with an adequate spatial resolution. In this work, 3D dynamic vocal tract representations are built to generate diphthongs, based on a set of cross-sections extracted from MRI-based vocal tract geometries of static vowel sounds. A diphthong can then be easily generated by interpolating the location, orientation and shape of these cross-sections, thus avoiding the interpolation of full 3D geometries. Two options are explored to extract the cross-sections. The first one is based on an adaptive grid (AG), which extracts the cross-sections perpendicular to the vocal tract midline, whereas the second one resorts to a semi-polar grid (SPG) strategy, which fixes the cross-section orientations. The finite element method (FEM) has been used to solve the mixed wave equation and synthesize diphthongs [${\alpha i}$] and [${\alpha u}$] in the dynamic 3D vocal tracts. The outputs from a 1D acoustic model based on the Transfer Matrix Method have also been included for comparison. The results show that the SPG and AG provide very close solutions in 3D, whereas significant differences are observed when using them in 1D. The SPG dynamic vocal tract representation is recommended for 3D simulations because it helps to prevent the collision of adjacent cross-sections.

##### Place, publisher, year, edition, pages
IEEE Press, 2019. Vol. 27, no 12, p. 2173-2182
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Computer Sciences
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ISI: 000492183000001Scopus ID: 2-s2.0-85073632242OAI: oai:DiVA.org:kth-259580DiVA, id: diva2:1352328
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Available from: 2019-09-18 Created: 2019-09-18 Last updated: 2019-11-11Bibliographically approved

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# Path Integral Quantization and Stochastic Quantization Paperback

#### Description

In this book, we discuss the path integral quantization and the stochastic quantization of classical mechanics and classical field theory.

Forthe description ofthe classical theory, we have two methods, one based on the Lagrangian formalism and the other based on the Hamiltonian formal- ism.

The Hamiltonian formalism is derived from the Lagrangian*formalism.

In the standard formalism ofquantum mechanics, we usually make use ofthe Hamiltonian formalism.

This fact originates from the following circumstance which dates back to the birth of quantum mechanics.

The first formalism ofquantum mechanics is Schrodinger's wave mechan- ics.

In this approach, we regard the Hamilton-Jacobi equation of analytical mechanics as the Eikonal equation of "geometrical mechanics".

Based on the optical analogy, we obtain the Schrodinger equation as a result ofthe inverse of the Eikonal approximation to the Hamilton-Jacobi equation, and thus we arrive at "wave mechanics".

The second formalism ofquantum mechanics is Heisenberg's "matrix me- chanics".

In this approach, we arrive at the Heisenberg equation of motion from consideration of the consistency of the Ritz combination principle, the Bohr quantization condition and the Fourier analysis of a physical quantity.

These two formalisms make up the Hamiltonian.formalism of quantum me- chanics.

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