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# What is the distribution of the 'achieved' $R^2$?

I am interested in the distribution of/performing inference on the 'achieved' $R^2$ coefficient in multiple linear regression. Suppose that $y\sim x\beta + \epsilon$ with $\epsilon \sim \mathcal{N}\left(0,\sigma^2\right)$, where $x$ is a $p$-vector. You observe $x_1,x_2,\ldots,x_n$ and corresponding $y_1,y_2,\ldots,y_n$, with independent errors. You perform linear regression to get $\hat{\beta}$, and then compute the sample $R^2$ in the usual way: $R^2 = 1 - \frac{SS_\mbox{err}}{SS_{\mbox{tot}}}$. Nothing new here.

There are known methods for hypothesis testing the population analogue of $R^2$, as well as computing confidence intervals on it. (I am thinking of Lee's paper, and followups by Algina, O'Brien, inter alios). The population analogue is defined in terms of the (unknown) true regression coefficients $\beta$ and the true variance $\sigma^2$.

I am interested in the achieved $R^2$ which is the amount of variance explained by $\hat{\beta}$. Conditional on $x$ and on $\hat{\beta}$, I would define it as $$R^2_{\mbox{ach}} = 1 - \frac{\mbox{E}\left[\left(y - \hat{\beta}x\right)^2\right]}{\mbox{E}\left[\left(y-\bar{y}\right)^2\right]}.$$ Clearly this is less than the population $R^2$, because the true $\beta$ minimizes $\mbox{E}\left[\left(y - \beta x\right)^2\right]$. I would guess it is less than the sample $R^2$ because that (usually) overestimates the population $R^2$.

I am having problems thinking about this quantity because it is both unobserved and random ($\hat{\beta}$ is a random variable.) I am guesing that, up to some transforms, I can represent this as a non-central chi-square, abusing $\hat{\beta}\sim\mathcal{N}\left(\beta,\sigma^2\left(X^{\top}X\right)^{-1}\right)$, and noting that $\mbox{Var}\left[y - \hat{\beta}x\right] = \sigma^2 + \mbox{Var}\left[(\hat{\beta} - \beta)x\right]$, but I'm being dense and lazy. (Also, treating $x$ as fixed will later be generalized to $x$ being random.)

Is this is a well-known problem? For predicting future performance of a linear model, it would seem to be a more important quantity than the population $R^2$, for example.

• Re "clearly this is less than": this statement appears to be equivalent to asserting that the OLS estimate of $\sigma^2$ is biased! Am I missing something here? – whuber Jun 12 '12 at 19:04
• @whuber, hmm. the achieved $R^2$ is computed in terms of the population parameters, so I'm not sure how the OLS estimate of $\sigma^2$ enters into it. – shabbychef Jun 12 '12 at 19:19
• ahh, I see the confusion; the $\mbox{Var}$ is conditional on $x$ and $\hat{\beta}$. That is, I have run the regression, computed $\hat{\beta}$ and then will use the regression model 'out of sample'. I can see how this is confusing. edit coming. – shabbychef Jun 12 '12 at 19:22
• the problem is that 'conditional on $x$' forces $y-\hat{\beta}x$ to be just a shift of $y$ and so $R^2_{\mbox{ach}}$ is just identically zero as defined. As noted, I will later have to consider $x$ as being normally distributed, in which case, I believe, the achieved $R^2$ as defined is no longer forced to be zero. bleah. – shabbychef Jun 12 '12 at 19:29


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# Explain about right linear regular grammars in TOC

Data Structure AlgorithmsComputer ScienceComputers

Regular grammar describes a regular language. It consists of four components, which are as follows −

G = (N, E, P, S)

Where,

• N − finite set of non-terminal symbols,

• E − a finite set of terminal symbols,

• P − a set of production rules, each of one is in the forms

• S → aB

• S → a

• S → ∈,

• S ∈ N is the start symbol.

The above grammar can be of two forms −

• Right Linear Regular Grammar

• Left Linear Regular Grammar

## Linear Grammar

When the right side of the Grammar part has only one terminal then it's linear else non linear.

Let’s discuss about right linear grammar −

## Right linear grammar

Right linear grammar means that the non-terminal symbol will be at the right side of the production.

It is a formal grammar (N, Σ, P, S) such that all the production rules in P are of one of the following forms −

L → a, { L is a non-terminal and a is a terminal in Σ}
L → aM, {L and M are non-terminals in N and a is in Σ}
L → ∈.

### Example

Consider a language L= {bnabma | n>=2, m>=2}

The production rules or grammar for the given language L= {bnabma | n>=2, m>=2} is −

S→bbB    ⇒for first 2 b’s
B→bB|aC  ⇒ any number of b’s followed by a
C→bbD    ⇒ 2b’s
D→ bD|a  ⇒ any number of b’s followed by a
Updated on 14-Jun-2021 14:46:58

format: plain


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# Inducing maps between Boolean completions of posets

Given a separative poset $P$, we can form it's Boolean completion $B(P)$. This is a Boolean algebra whose elements are regular cuts on $P$, as defined here. Also, $P$ embeds densely into $B(P)$ by mapping $p$ to the principal ideal $U_p$.

Let's now take a poset $Q$ and a $P\subseteq Q$, both of which are separative. The inclusion $P\to Q$ is monotone. What I want to know is, can we somehow induce a complete embedding $f\colon B(P)\to B(Q)$ from this that makes the obvious diagram commute?

It seems to me there is only one thing that should work. Since $P$ is densely embedded in $B(P)$ and we know $f$ on $P$, the only reasonable choice seems to be $$f(U)=\bigvee_{p\in U}U_p$$

But this doesn't even seem to be a Boolean algebra homomorphism. Let's look at the meets: $$f(U\cap V)=\bigvee_{p\in U\cap V} U_p \stackrel{?}{=} \bigvee_{p\in U} U_p\cap \bigvee_{q\in V} U_q= f(U)\cap f(V)$$

Obviously, the left hand set is contained in the right hand set, but do we have equality? I just don't see it.

Have I missed something obvious? Is this even the right way of defining the map $f$?

Added: Apostolos dealt with the meets in the comments, just using distributivity. But I'm having trouble with joins now. We have $$f(U\vee V)=\bigvee_{p\in U\vee V} U_p^Q \stackrel{?}{=} \bigvee_{p\in U\cup V} U_p^Q= f(U)\vee f(V)$$ (I'll be using $U_p^P$ and $U_p^Q$ to distinguish principal ideals in $P$ and $Q$ respectively)

To focus on the nontrivial inclusion, take $x\in f(U\vee V)$ and $y\leq x$. We wish to find a $p_0\in U\cup V$ such that $U_y^Q\cap U_{p_0}^Q\neq\emptyset$. According to the definition of the join, there exists a $p\in U\vee V$ such that $U_y^Q\cap U_p^Q\neq\emptyset$. Taking an element $q$ of this set, we get $q\leq y$ and $q\leq p$. Now, if we could find an element of $P$ below this $q$, we would be done (since $p\in U\vee V$), but I don't see any reason for such an element to exist. In particular, in the application I have in mind, $P$ is very far from being dense in $Q$. How do I proceed here?

-
For every complete Boolean algebra you have $(\bigvee_{i\in I}a_i)\land(\bigvee_{j\in J}b_j)=\bigvee_{(i,j)\in I\times J}(a_i\land b_j)$. Thus your right hand side is $\bigvee_{(p,q)\in U\times V}U_p\cap U_q$. So if $a$ is in your right hand side it is in some $U_p\cap U_q$ for $p\in U$ and $q\in V$. Then $a\in U$ and $a\in V$ because they are downwards closed (and they contain $p$ and $q$ and $a$ is below them). Hence $a\in f(U\cap V)$. – Apostolos Jun 6 '12 at 19:41
@Apostolos You need to be a bit careful, since the join in $B(P)$ isn't just the union but the "regular closure" of the union. But you're basically right. I was missing distributivity. If you see an elegant way to show this is a complete embedding, please write something up. Otherwise, I'll think about this more in the morning. – Miha Habič Jun 6 '12 at 21:03
If you want the diagram induced by the inclusion map to commute there is really only one way to define your map. Whether or not it is complete, I'm not sure. Note that $P\subseteq Q$ means that $B(P)$ is realized inside $B(Q)$ as the regular cuts defined by the elements of $P$. I suppose it might be possible to create a nice counterexample of two separative posets $P\subseteq Q$ such that $B(P)$ is not a complete subalgebra of $B(Q)$ by using an example of $B'\subseteq B$ which are both complete BA, but $B'$ is not a complete sub-alg. of $B$, then taking $B^+$ and $B'^+$ as your posets. – Asaf Karagila Jun 13 '12 at 17:24
@AsafKaragila : what is it that you denote by $B^{+}$ ? – Ewan Delanoy Jun 14 '12 at 4:55
@EwanDelanoy I expect $B^+$ is the poset of positive elements of $B$. – Miha Habič Jun 14 '12 at 5:21

Take $Q$ to be the power set of $\mathbb R$ and $P$ the power set of $\mathbb N$, both without the zero element, i.e. only non-empty sets. The order on both is inclusion.
Trivially $P$ and $Q$ are separative and $P\subseteq Q$. However since both are the positive elements of a complete Boolean algebra, and the completion is unique [up to isomorphism], the completion is both is simply adding the zero element.
However an embedding of a Boolean algebra into another must respect the $0,1$ elements, but $1_{B(P)}=\mathbb N\neq\mathbb R=1_{B(Q)}$. So there is no obvious way of closing the diagram.
Wow, I didn't expect this to fail this badly. Thanks for the counterexample Asaf! Although this doesn't quite apply in the particular case I'm considering (there, $Q$ is a "lower extension" of $P$, i.e. if an element of $Q$ is above an element of $P$, it was already in $P$), it's enough for me to start to be very careful. – Miha Habič Jul 13 '12 at 6:33


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1. ## Help with Radian Measure

A belt connects pulleys of diameter 2r and 2R with centres x apart. If r=20mm, R=120mm, and x=100*21/2mm, find the length L of the belt and the area of the region it encloses. Find the general expression for L in terms of r, R, x and Ø

3. ## Re: Help with Radian Measure

Yes, from that diagram I was able to work out that:
cosØ=100/100sqrt2
i.e.Ø=pi/4

But from that how do you work out the length of the belt?

4. ## Re: Help with Radian Measure

Originally Posted by KiaraG16
Yes, from that diagram I was able to work out that:
cosØ=100/100sqrt2
i.e.Ø=pi/4???? But from that how do you work out the length of the belt?
The non-blue part of the pulley's path is $\frac{3}{4}$ of the length circumference of the large circle.

The length of the linear blue part is $20+20=40$.
The length along the small circle is $\frac{1}{4}$ of its circumference.

In both cases the central angle is a right angle.

ADD these three lengths to get a total.


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# Measure where every set is Measurable

For the Lebesgue measure the Vitali set is a famous construction of a nonmeasurable set.

I was wondering if we could define some non-trivial outer measure on $2^{\mathbb{R}}$ such that every set is measurable with respect to this measure, or if there always exist nonmeasurable sets with respect to nontrivial measures. (of course we can define the outer measure that is zero on all subsets of $\mathbb{R}$ but i was wondering if we could say something more interesting. If such a measure exists does it have any interesting properties, i.e. does there exist some measure with this property that is Borel-regular? or even Radon? (this seems highly unlikely to me but I'm interested in) I would also like to assume the axiom of choice here.

• What about taking one point and assigning measure one to the sets that contain it, and measure zero to those which don't? – Emilio Oct 15 '16 at 19:19
• I think one can show that if you require translation invariance and countable additivity then non-measurable sets exist. – 3-in-441 Oct 15 '16 at 19:19
• @3-in-441 If you reject the axiom of choice, it is consistent that every subset of $\mathbb{R}$ is measurable. Rejecting choice makes really really weird things possible, however, so please don't. – Daniel Fischer Oct 15 '16 at 19:49
• What counts as nontrivial? We can take a countable set $D = \{ x_n : n \in \mathbb{N}\}$ and set $\mu(A) = \sum_{x_n \in A} 2^{-n}$. Then all sets are $\mu$-measurable. $\mu$ is outer and inner regular. If $D$ is dense, every nonempty open set has positive measure. – Daniel Fischer Oct 15 '16 at 19:55


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## Trigonometry (11th Edition) Clone

The propeller rotates through an angle of $2\pi$ radians 5000 times each minute. We can find the angular speed of the propeller: $\omega = \frac{\theta}{t}$ $\omega = \frac{(2\pi~rad)(5000)}{60~s}$ $\omega = 523.6~rad/s$ The angular speed of the propeller is 523.6 rad/s


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## Gluons are not all the story: An update

30/05/2009

In a recent post of mine (see here) I have pointed out to you a beautiful paper by Dan Pirjol and Carlos Schat. This paper is now appeared on Physical Review Letters (see here) but you can find also the preprint on arxiv (see here). I think its content is really important as it gives a serious clue toward our understanding of low-energy QCD. Dan agreed to publish here a contribution about his work and I am glad to post it.

We find that, within experimental errors on the hadron masses, the
so-called gluon-exchange model (OGE) is disfavored by data. This is probably not very  surprising, since at low energies the real degrees of freedom of QCD should include, in addition to the  gluons, also pions (the Goldstone bosons of the spontaneously broken  chiral symmetry). The OGE model does not include the pion exchange effects; an alternative to the OGE model  which includes their effects is the so-called GBE (Goldstone boson exchange) model.

There has been a long-standing debate about the most appropriate model
of quark forces in the framework of the constituent quark model, in  particular about their spin-flavor dependence. The main candidates are
the OGE and GBE models (see e.g. the second paper in Ref.[3] for a
discussion of this controversy). Our letter attempts to resolve this controversy using only minimal assumptions about the orbital dependence
of the hadronic wave functions. More precisely, we assume only isospin
symmetry, but no other assumption is made about the form of the wave
functions. The novel mathematical tool which makes our analysis possible is the application of the permutation group $S_3$, which allows one to study the implications of the most general spin-flavor structure of the quark forces.

## Sigma resonance again

29/05/2009

José Pelaéz and Guillermo Rìos published today a paper on arxiv (see here). The argument is an understanding of the nature of $\sigma$ and $\kappa$ resonances. The technique they use is Chiral Perturbation Theory (ChPT) but the idea is to see the behavior of the amplitudes at increasing number of colors. They get again a confirmation that the very nature of $\sigma$ is not a typical $\bar qq$ state. Rather, a subdominant $\bar qq$ component is seen at larger energies with larger values of the number of colors. This conclusion  agrees with our theorem proved here.

The current situation forces the authors to prudence. They do not draw any conclusion about the real nature of $\sigma$ and $\kappa$ but their results still appear impressive. These authors have a long file of very good works about the quest for an understanding of the lower part of QCD spectrum and they have given the mass and the width of $\sigma$  with really increased precision. They belong to a group headed by Paco Yndurain. You can find a tribute to Paco by Stephan Narison here.

From my view you can see this as another confirmation to the idea that $\sigma$ is a glueball and the lowest state of a pure Yang-Mills theory. This evidence is becoming overwhelming but other interpretations are not ruled out yet. The fact that $\kappa$ or else f0(980) are glueballs would give further strong support to this as I expect a glueball state at this value of energy.

## A review on proton spin

29/05/2009

As you may know, there is a lot of experimental activity to understand as the spin of the proton can arise from its components, i.e. quarks and gluons. The great difficulties we have to manage low-energy QCD makes this problem fundamental toward an improved comprehension of this limit. In arxiv today an interesting review by Steven Bass is appeared (see here and here). Bass gives a brief  overview of the current situation mostly from the experimental side. As reader from this blog may know (see here), glue contribution to spin is about zero and the proton spin appears mostly due to valence quarks and their interplay with vacuum. Indeed this is Bass’ conclusion and we fully agree with it.

The emerging scenario is really striking. It appears that QCD behavior in a non-perturbative regime goes completely off known tracks. This implies that there is a lot of problems to be solved in the future for us working in this field.

## NIF at start

26/05/2009

National Ignition Facility started this year at Lawrence Livermore Laboratories. This project has the ambition to give an answer to our quest for nuclear fusion as a reliable source of energy.Funded by US government, it has implied delays and cost increase during its realization. But the aims are so relevant that this is worthwhile spent money. The approach for this experiment is that of inertial fusion. This technique uses power lasers to hit a pellet of fusion material. Pressure of light, when applied uniformly on the target, will push nuclei so near to win the effect of the Coulomb barrier and so they start to fuse each other relaxing a large amount of energy. This is our dream of a sun on earth. I think people is also aware of the other way research is pursuing through tokamak where a plasma is heated through different means to achieve the same goal. For this track, ITER is still at very start of its realization.

As usual,  there is a very good article on the New York Times about NIF (see here).  The hope is to solve one of the most difficult problems of humankind. So, we can only wish the best of lucks at NIF.

## Pioneers

20/05/2009

Jeffrey Mandula is a well-knwon theoretical physicist whose main result, Coleman-Mandula theorem, opened the road to the discovery of supersymmetry. But Mandula is also known for his pioneering works on lattice QCD. Looking back to old papers on the question of the Yang-Mills propagators, I have found two beautiful papers by Mandula and Ogilvie (see here and here) published in the eighties, where they arrive to the following conclusion

From the behavior of the gluon propagator reported here, it appears that in pure Yang-Mills theory, a dynamical Higgs phenomenon occurs. Our best estimate of the effective gluon mass, as determined at large distances, is about 600 MeV, with finite size effects, possible scaling violations, and statistical uncertainties of at least $\pm 25\%$. In analogy with the concept of a constituent quark mass, it may be useful to think of the mass in the gluon propagator as a constituent gluon mass. The massiveness of the gluon may be connected to the apparent suppression of many-gluon intermediate states in $J/\psi$ decay, and the relative absence of the mixing between the lowest quark model states and those with gluonic excitations.

They used small lattices due to the computer limitations at that time and ideas about infrared behavior of Yang-Mills theory were just beginning to flourish. But, with our hindsight, we should emphasize the deep intuition that these authors put forward when such analysis were just starting. Besides, $\sigma$ resonance was not yet seen and their estimate of the gluon mass appears really good.

As you may know, after these works, things took a different turn and for a long time since now we have been coping with a different scenario from that devised by Mandula and Ogilvie that took the scenes and not yet left them. This scenario appears today to be in a serious difficulty against lattice computations but people do not generally agree about what the right view should be, making painfully slow truth achievement .

## Cramer-Rao bound and Ricci flow II

18/05/2009

The paper I presented about this matter (see here) has been accepted by EuRad 2009 Conference. This will result in a publication in IEEE Proceedings. IEEE is the most important engineering society. I cannot made public this paper until it will appear in the proceedings. After this date you can read it at IEEE Xplore where you can find another paper of mine about scattering of electromagnetic waves by a rough surface (see here). As you can see, the way publishing is operated by engineers is quite different from that of physicists.

Anyhow, the idea is quite simple and use the fact that for a two-dimensional Riemann manifold one has always a conformal metric. Then, Fischer information matrix can be expressed in a diagonal form with new estimators that are always optimal with respect to Cramer-Rao bound. So, due to the fact that there exists a vast set of probability distributions with two parameters, the application areas of this result are huge. In my paper I make the case of sea clutter for radar applications but what I prove is a theorem in statistics and you can realize by yourself the importance.

The n-parameter case can also be made but here there are two more demanding requests: the existence of a conformal metric and the existence of a potential for a vector field that satisfies Liouville equation. These cannot always be satisfied and so the two-dimensional case appears a rather lucky one.

## A beautiful paper on arxiv

15/05/2009

Keeping on their way of producing sound work, Bogolubsky, Ilgenfritz, Mueller-Preussker and Sternbeck have got their paper (see here) published on Physics Letters B. This is a collaboration between people working in Russia, Germany and Australia. The main aim of this work is the computation on a lattice of the two-point functions and the running coupling of a pure Yang-Mills theory. They carry on lattice computations from $(64)^4$ to $(96)^4$ points entering into a deep enough infrared limit to get a meaningful behavior of the lattice theory in this case. I give below their main results

Gluon propagator

Dressing function of the ghost propagator

Running coupling

These results confirm completely the decoupling solution. The definition of the running coupling is the one proposed by Alkofer and von Smekal and it is my personal conviction that it conveys the right physical behavior of the theory. This is exactly the scenario I have derived in my paper (see here) that has been published on Physics Letters B too and has arisen a lot of rumors around. You will not find this paper cited in this work as these authors have concerns about gauge invariance in my computations. As you may know from my dispute with Terry Tao, gauge invariance is not a problem here. One could ask why a mathematical technique, like a gradient expansion is, should not work for Yang-Mills equations but it does for all other equations of mathematical physics. Anyhow, I am here ready to listen to whoever is able to prove this. With this proof in hand one should also warn all general relativists that use this technique and put it in their handbooks.

The authors conclude their paper by pointing out weaknesses in lattice computations that may bring in discussion their results. Finally, they ask if the other solution, the one with a scaling behavior, can emerge from lattice computations. The understanding of this question is surely of relevant interest. We stay tuned to hear news about.


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Save up to $499! Grab all Python courses for$49 or all online courses we’ve ever launched for only \$169. Only Feb 11-16. Happy Valentine's!
Introduction
Simple OVER()
PARTITION BY
Ranking functions
Window Frames
Analytic Functions
PARTITION BY ORDER BY
28. PARTITION BY ORDER BY – Exercise 3
Order of Evaluation
Finished!

## Instruction

Impressive! Okay, this is going to be the very last exercise in this section.

## Exercise

For each movie, show the following information: Title, Genre, the average user rating for that Movie (column AvgRating) and the movie's Rank within its Genre, based on that average rating. Use descending order so that the highest-ranked movies will be shown first.

### Stuck? Here's a hint!

GROUP BY the movie Title and Genre, In the window function, partition by Genre, and order by the average rating.


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## Coarse-graining discrete cell cycle models

### Introduction

One observation that often goes underappreciated in computational biology discussions is that a computational model is often a model of a model of a model of biology: that is, it’s a numerical approximation (a model) of a mathematical model of an experimental model of a real-life biological system. Thus, there are three big places where a computational investigation can fall flat:

1. The experimental model may be a bad choice for the disease or process (not our fault).
2. Second, the mathematical model of the experimental system may have flawed assumptions (something we have to evaluate).
3. The numerical implementation may have bugs or otherwise be mathematically inconsistent with the mathematical model.

Critically, you can’t use simulations to evaluate the experimental model or the mathematical model until you verify that the numerical implementation is consistent with the mathematical model, and that the numerical solution converges as $$\Delta t$$ and $$\Delta x$$ shrink to zero.

There are numerous ways to accomplish this, but ideally, it boils down to having some analytical solutions to the mathematical model, and comparing numerical solutions to these analytical or theoretical results. In this post, we’re going to walk through the math of analyzing a typical type of discrete cell cycle model.

### Discrete model

Suppose we have a cell cycle model consisting of phases $$P_1, P_2, \ldots P_n$$, where cells in the $$P_i$$ phase progress to the $$P_{i+1}$$ phase after a mean waiting time of $$T_i$$, and cells leaving the $$P_n$$ phase divide into two cells in the $$P_1$$ phase. Assign each cell agent $$k$$ a current phenotypic phase $$S_k(t)$$. Suppose also that each phase $$i$$ has a death rate $$d_i$$, and that cells persist for on average $$T_\mathrm{A}$$ time in the dead state before they are removed from the simulation.

The mean waiting times $$T_i$$ are equivalent to transition rates $$r_i = 1 / T_i$$ (Macklin et al. 2012). Moreover, for any time interval $$[t,t+\Delta t]$$, both are equivalent to a transition probability of
$\mathrm{Prob}\Bigl( S_k(t+\Delta t) = P_{i+1} | S(t) = P_i \Bigr) = 1 – e^{ -r_i \Delta t } \approx r_i \Delta t = \frac{ \Delta t}{ T_i}.$ In many discrete models (especially cellular automaton models) with fixed step sizes $$\Delta t$$, models are stated in terms of transition probabilities $$p_{i,i+1}$$, which we see are equivalent to the work above with $$p_{i,i+1} = r_i \Delta t = \Delta t / T_i$$, allowing us to tie mathematical model forms to biological, measurable parameters. We note that each $$T_i$$ is the average duration of the $$P_i$$ phase.

#### Concrete example: a Ki67 Model

Ki-67 is a nuclear protein that is expressed through much of the cell cycle, including S, G2, M, and part of G1 after division. It is used very commonly in pathology to assess proliferation, particularly in cancer. See the references and discussion in (Macklin et al. 2012). In Macklin et al. (2012), we came up with a discrete cell cycle model to match Ki-67 data (along with cleaved Caspase-3 stains for apoptotic cells). Let’s summarize the key parts here.

Each cell agent $$i$$ has a phase $$S_i(t)$$. Ki67- cells are quiescent (phase $$Q$$, mean duration $$T_\mathrm{Q}$$), and they can enter the Ki67+ $$K_1$$ phase (mean duration $$T_1$$). When $$K_1$$ cells leave their phase, they divide into two Ki67+ daughter cells in the $$K_2$$ phase with mean duration $$T_2$$. When cells exit $$K_2$$, they return to $$Q$$. Cells in any phase can become apoptotic (enter the $$A$$ phase with mean duration $$T_\mathrm{A}$$), with death rate $$r_\mathrm{A}$$.

### Coarse-graining to an ODE model

If each phase $$i$$ has a death rate $$d_i$$, if $$N_i(t)$$ denotes the number of cells in the $$P_i$$ phase at time $$t$$, and if $$A(t)$$ is the number of dead (apoptotic) cells at time $$t$$, then on average, the number of cells in the $$P_i$$ phase at the next time step is given by
$N_i(t+\Delta t) = N_i(t) + N_{i-1}(t) \cdot \left[ \textrm{prob. of } P_{i-1} \rightarrow P_i \textrm{ transition} \right] – N_i(t) \cdot \left[ \textrm{prob. of } P_{i} \rightarrow P_{i+1} \textrm{ transition} \right]$ $– N_i(t) \cdot \left[ \textrm{probability of death} \right]$ By the work above, this is:
$N_i(t+\Delta t) \approx N_i(t) + N_{i-1}(t) r_{i-1} \Delta t – N_i(t) r_i \Delta t – N_i(t) d_i \Delta t ,$ or after shuffling terms and taking the limit as $$\Delta t \downarrow 0$$, $\frac{d}{dt} N_i(t) = r_{i-1} N_{i-1}(t) – \left( r_i + d_i \right) N_i(t).$ Continuing this analysis, we obtain a linear system:
$\frac{d}{dt}{ \vec{N} } = \begin{bmatrix} -(r_1+d_1) & 0 & \cdots & 0 & 2r_n & 0 \\ r_1 & -(r_2+d_2) & 0 & \cdots & 0 & 0 \\ 0 & r_2 & -(r_3+d_3) & 0 & \cdots & 0 \\ & & \ddots & & \\0&\cdots&0 &r_{n-1} & -(r_n+d_n) & 0 \\ d_1 & d_2 & \cdots & d_{n-1} & d_n & -\frac{1}{T_\mathrm{A}} \end{bmatrix}\vec{N} = M \vec{N},$ where $$\vec{N}(t) = [ N_1(t), N_2(t) , \ldots , N_n(t) , A(t) ]$$.

For the Ki67 model above, let $$\vec{N} = [K_1, K_2, Q, A]$$. Then the linear system is
$\frac{d}{dt} \vec{N} = \begin{bmatrix} -\left( \frac{1}{T_1} + r_\mathrm{A} \right) & 0 & \frac{1}{T_\mathrm{Q}} & 0 \\ \frac{2}{T_1} & -\left( \frac{1}{T_2} + r_\mathrm{A} \right) & 0 & 0 \\ 0 & \frac{1}{T_2} & -\left( \frac{1}{T_\mathrm{Q}} + r_\mathrm{A} \right) & 0 \\ r_\mathrm{A} & r_\mathrm{A} & r_\mathrm{A} & -\frac{1}{T_\mathrm{A}} \end{bmatrix} \vec{N} .$
(If we had written $$\vec{N} = [Q, K_1, K_2 , A]$$, then the matrix above would have matched the general form.)

### Some theoretical results

If $$M$$ has eigenvalues $$\lambda_1 , \ldots \lambda_{n+1}$$ and corresponding eigenvectors $$\vec{v}_1, \ldots , \vec{v}_{n+1}$$, then the general solution is given by
$\vec{N}(t) = \sum_{i=1}^{n+1} c_i e^{ \lambda_i t } \vec{v}_i ,$ and if the initial cell counts are given by $$\vec{N}(0)$$ and we write $$\vec{c} = [c_1, \ldots c_{n+1} ]$$, we can obtain the coefficients by solving $\vec{N}(0) = [ \vec{v}_1 | \cdots | \vec{v}_{n+1} ]\vec{c} .$ In many cases, it turns out that all but one of the eigenvalues (say $$\lambda$$ with corresponding eigenvector $$\vec{v}$$) are negative. In this case, all the other components of the solution decay away, and for long times, we have $\vec{N}(t) \approx c e^{ \lambda t } \vec{v} .$ This is incredibly useful, because it says that over long times, the fraction of cells in the $$i^\textrm{th}$$ phase is given by $v_{i} / \sum_{j=1}^{n+1} v_{j}.$

#### Matlab implementation (with the Ki67 model)

First, let’s set some parameters, to make this a little easier and reusable.

parameters.dt = 0.1; % 6 min = 0.1 hours
parameters.time_units = 'hour';
parameters.t_max = 3*24; % 3 days

parameters.K1.duration =  13;
parameters.K1.death_rate = 1.05e-3;
parameters.K1.initial = 0;

parameters.K2.duration = 2.5;
parameters.K2.death_rate = 1.05e-3;
parameters.K2.initial = 0;

parameters.Q.duration = 74.35 ;
parameters.Q.death_rate = 1.05e-3;
parameters.Q.initial = 1000;

parameters.A.duration = 8.6;
parameters.A.initial = 0;

Next, we write a function to read in the parameter values, construct the matrix (and all the data structures), find eigenvalues and eigenvectors, and create the theoretical solution. It also finds the positive eigenvalue to determine the long-time values.

function solution = Ki67_exact( parameters )

% allocate memory for the main outputs

solution.T = 0:parameters.dt:parameters.t_max;
solution.K1 = zeros( 1 , length(solution.T));
solution.K2 = zeros( 1 , length(solution.T));
solution.K = zeros( 1 , length(solution.T));
solution.Q = zeros( 1 , length(solution.T));
solution.A = zeros( 1 , length(solution.T));
solution.Live = zeros( 1 , length(solution.T));
solution.Total = zeros( 1 , length(solution.T));

% allocate memory for cell fractions

solution.AI = zeros(1,length(solution.T));
solution.KI1 = zeros(1,length(solution.T));
solution.KI2 = zeros(1,length(solution.T));
solution.KI = zeros(1,length(solution.T));

% get the main parameters

T1 = parameters.K1.duration;
r1A = parameters.K1.death_rate;

T2 = parameters.K2.duration;
r2A = parameters.K2.death_rate;

TQ = parameters.Q.duration;
rQA = parameters.Q.death_rate;

TA = parameters.A.duration;

% write out the mathematical model:
% d[Populations]/dt = Operator*[Populations]

Operator = [ -(1/T1 +r1A) , 0 , 1/TQ , 0; ...
2/T1 , -(1/T2 + r2A) ,0 , 0; ...
0 , 1/T2 , -(1/TQ + rQA) , 0; ...
r1A , r2A, rQA , -1/TA ];

% eigenvectors and eigenvalues

[V,D] = eig(Operator);
eigenvalues = diag(D);

% save the eigenvectors and eigenvalues in case you want them.

solution.V = V;
solution.D = D;
solution.eigenvalues = eigenvalues;

% initial condition

VecNow = [ parameters.K1.initial ; parameters.K2.initial ; ...
parameters.Q.initial ; parameters.A.initial ] ;
solution.K1(1) = VecNow(1);
solution.K2(1) = VecNow(2);
solution.Q(1) = VecNow(3);
solution.A(1) = VecNow(4);
solution.K(1) = solution.K1(1) + solution.K2(1);
solution.Live(1) = sum( VecNow(1:3) );
solution.Total(1) = sum( VecNow(1:4) );

solution.AI(1) = solution.A(1) / solution.Total(1);
solution.KI1(1) = solution.K1(1) / solution.Total(1);
solution.KI2(1) = solution.K2(1) / solution.Total(1);
solution.KI(1) = solution.KI1(1) + solution.KI2(1);

% now, get the coefficients to write the analytic solution
% [Populations] = c1*V(:,1)*exp( d(1,1)*t) + c2*V(:,2)*exp( d(2,2)*t ) +
%                 c3*V(:,3)*exp( d(3,3)*t) + c4*V(:,4)*exp( d(4,4)*t );

coeff = linsolve( V , VecNow );

% find the (hopefully one) positive eigenvalue.
% eigensolutions with negative eigenvalues decay,
% leaving this as the long-time behavior.

eigenvalues = diag(D);
n = find( real( eigenvalues ) &gt; 0 )
solution.long_time.KI1 = V(1,n) / sum( V(:,n) );
solution.long_time.KI2 = V(2,n) / sum( V(:,n) );
solution.long_time.QI = V(3,n) / sum( V(:,n) );
solution.long_time.AI = V(4,n) / sum( V(:,n) ) ;
solution.long_time.KI = solution.long_time.KI1 + solution.long_time.KI2;

% now, write out the solution at all the times
for i=2:length( solution.T )
% compact way to write the solution
VecExact = real( V*( coeff .* exp( eigenvalues*solution.T(i) ) ) );

solution.K1(i) = VecExact(1);
solution.K2(i) = VecExact(2);
solution.Q(i) = VecExact(3);
solution.A(i) = VecExact(4);
solution.K(i) = solution.K1(i) + solution.K2(i);
solution.Live(i) = sum( VecExact(1:3) );
solution.Total(i) = sum( VecExact(1:4) );

solution.AI(i) = solution.A(i) / solution.Total(i);
solution.KI1(i) = solution.K1(i) / solution.Total(i);
solution.KI2(i) = solution.K2(i) / solution.Total(i);
solution.KI(i) = solution.KI1(i) + solution.KI2(i);
end

return;

Now, let’s run it and see what this thing looks like:

Next, we plot KI1, KI2, and AI versus time (solid curves), along with the theoretical long-time behavior (dashed curves). Notice how well it matches–it’s neat when theory works! 🙂

Some readers may recognize the long-time fractions: KI1 + KI2 = KI = 0.1743, and AI = 0.00833, very close to the DCIS patient values from our simulation study in Macklin et al. (2012) and improved calibration work in Hyun and Macklin (2013).

### Comparing simulations and theory

I wrote a small Matlab program to implement the discrete model: start with 1000 cells in the $$Q$$ phase, and in each time interval $$[t,t+\Delta t]$$, each cell “decides” whether to advance to the next phase, stay in the same phase, or apoptose. If we compare a single run against the theoretical curves, we see hints of a match:

If we average 10 simulations and compare, the match is better:

And lastly, if we average 100 simulations and compare, the curves are very difficult to tell apart:

Even in logarithmic space, it’s tough to tell these apart:

### Code

The following matlab files (available here) can be used to reproduce this post:

Ki67_exact.m
The function defined above to create the exact solution using the eigenvalue/eignvector approach.
Ki67_stochastic.m
Runs a single stochastic simulation, using the supplied parameters.
script.m
Runs the theoretical solution first, creates plots, and then runs the stochastic model 100 times for comparison.

To make it all work, simply run “script” at the command prompt. Please note that it will generate some png files in its directory.

### Closing thoughts

In this post, we showed a nice way to check a discrete model against theoretical behavior–both in short-term dynamics and long-time behavior. The same work should apply to validating many discrete models. However, when you add spatial effects (e.g., a cellular automaton model that won’t proliferate without an empty neighbor site), I wouldn’t expect a match. (But simulating cells that initially have a “salt and pepper”, random distribution should match this for early times.)

Moreover, models with deterministic phase durations (e.g., K1, K2, and A have fixed durations) aren’t consistent with the ODE model above, unless the cells they are each initialized with a random amount of “progress” in their initial phases. (Otherwise, the cells in each phase will run synchronized, and there will be fixed delays before cells transition to other phases.) Delay differential equations better describe such models. However, for long simulation times, the slopes of the sub-populations and the cell fractions should start to better and better match the ODE models.

Now that we have verified that the discrete model is performing as expected, we can have greater confidence in its predictions, and start using those predictions to assess the underlying models. In ODE and PDE models, you often validate the code on simpler problems where you have an analytical solution, and then move on to making simulation predictions in cases where you can’t solve analytically. Similarly, we can now move on to variants of the discrete model where we can’t as easily match ODE theory (e.g., time-varying rate parameters, spatial effects), but with the confidence that the phase transitions are working as they should.

## Paul Macklin featured in Biotechniques article

Paul Macklin’s work on mathematical modeling of breast cancer and BioFVM was recently featured in Kelly Rae Chi’s article in Biotechniques on virtual cell cultures. It also included great work by James Glazier (CompuCell3D) and Kristin Swanson (glioblastoma modeling).

Read the article: http://www.biotechniques.com/news/Mighty-Modelers-The-Art-of-Virtual-Cell-Culture/biotechniques-364893.html (July 20, 2016)

## Paul Macklin calls for common standards in cancer modeling

At a recent NCI-organized mini-symposium on big data in cancer, Paul Macklin called for new data standards in Multicellular data in simulations, experiments, and clinical science. USC featured the talk (abstract here) and the work at news.usc.edu.

Read the article: http://news.usc.edu/59091/usc-researcher-calls-for-common-standards-in-cancer-modeling/ (Feb. 21, 2014)

## Paul Macklin interviewed at 2013 PSOC Annual Meeting

Paul Macklin gave a plenary talk at the 2013 NIH Physical Sciences in Oncology Annual Meeting. After the talk, he gave an interview to the Pauline Davies at the NIH on the need for data standards and model compatibility in computational and mathematical modeling of cancer. Of particular interest:

Pauline Davies: How would you ever get this standardization? Who would be responsible for saying we want it all reported in this particular way?

Paul Macklin: That’s a good question. It’s a bit of the chicken and the egg problem. Who’s going to come and give you data in your standard if you don’t have a standard? How do you plan a standard without any data? And so it’s a bit interesting. I just think someone needs to step forward and show leadership and try to get a small working group together, and at the end of the day, perfect is the enemy of the good. I think you start small and give it a go, and you add more to your standard as you need it. So maybe version one is, let’s say, how quickly the cells divide, how often they do it, how quickly they die, and what their oxygen level is, and maybe their positions. And that can be version one of this standard and a few of us try it out and see what we can do. I think it really comes down to a starting group of people and a simple starting point, and you grow it as you need it.

Shortly after, the MultiCellDS project was born (using just this strategy above!), with the generous assistance of the Breast Cancer Research Foundation.

Read / Listen to the interview: http://physics.cancer.gov/report/2013report/PaulMacklin.aspx (2013)

## Macklin Lab featured on March 2013 cover of Notices of the American Mathematical Society

I’m very excited to be featured on this month’s cover of the Notices of the American Mathematical Society. The cover shows a series of images from a multiscale simulation of a tumor growing in the brain, made with John Lowengrub while I was a Ph.D. student at UC Irvine. (See Frieboes et al. 2007, Macklin et al. 2009, and Macklin and Lowengrub 2008.) The “about the cover” write-up (Page 325) gives more detail.

The inside has a short interview on our more current work, particularly 3-D agent-based modeling. You should also read Rick Durrett‘s perspective piece on cancer modeling (Page 304)—it’s a great read! (And yup, Figure 3 is from our patient-calibrated breast cancer modeling in Macklin et al. 2012. 😉 )

The entire March 2013 issue can be accessed for free at the AMS Notices website:

http://www.ams.org/notices/201303/

I want to thank Bill Casselman and Rick Durrett for making this possible. I had a lot of fun in the process, and I’m grateful for the opportunity to trade ideas!

## DCIS modeling paper accepted

Recently, I wrote about a major work we submitted to the Journal of Theoretical Biology: “Patient-calibrated agent-based modelling of ductal carcinoma in situ (DCIS): From microscopic measurements to macroscopic predictions of clinical progression.”

I am pleased to report that our paper has now been accepted.  You can download the accepted preprint here. We also have a lot of supplementary material, including simulation movies, simulation datasets (for 0, 15, 30, adn 45 days of growth), and open source C++ code for postprocessing and visualization.

I discussed the results in detail here, but here’s the short version:

1. We use a mechanistic, agent-based model of individual cancer cells growing in a duct. Cells are moved by adhesive and repulsive forces exchanged with other cells and the basement membrane.  Cell phenotype is controlled by stochastic processes.
2. We constrained all parameter expected to be relatively independent of patients by a careful analysis of the experimental biological and clinical literature.
3. We developed the very first patient-specific calibration method, using clinically-accessible pathology.  This is a key point in future patient-tailored predictions and surgical/therapeutic planning.
4. The model made numerous quantitative predictions, such as:
1. The tumor grows at a constant rate, between 7 to 10 mm/year. This is right in the middle of the range reported in the clinic.
2. The tumor’s size in mammgraphy is linearly correlated with the post-surgical pathology size.  When we linearly extrapolate our correlation across two orders of magnitude, it goes right through the middle of a cluster of 87 clinical data points.
3. The tumor necrotic core has an age structuring: with oldest, calcified material in the center, and newest, most intact necrotic cells at the outer edge.
4. The appearance of a “typical” DCIS duct cross-section varies with distance from the leading edge; all types of cross-sections predicted by our model are observed in patient pathology.
5. The model also gave new insight on the underlying biology of breast cancer, such as:
1. The split between the viable rim and necrotic core (observed almost universally in pathology) is not just an artifact, but an actual biomechanical effect from fast necrotic cell lysis.
2. The constant rate of tumor growth arises from the biomechanical stress relief provided by lysing necrotic cells. This points to the critical role of intracellular and intra-tumoral water transport in determining the qualitative and quantitative behavior of tumors.
3. Pyknosis (nuclear degradation in necrotic cells), must occur at a time scale between that of cell lysis (on the order of hours) and cell calcification (on the order of weeks).
4. The current model cannot explain the full spectrum of calcification types; other biophysics, such as degradation over a long, 1-2 month time scale, must be at play.
I hope you enjoy this article and find it useful. It is our hope that it will help drive our field from qualitative theory towards quantitative, patient-tailored predictions.
Direct link to the preprint: http://www.mathcancer.org/Publications.php#macklin12_jtb
I want to express my greatest thanks to my co-authors, colleagues, and the editorial staff at the Journal of Theoretical Biology.

## PSOC Short Course on Multidisciplinary Cancer Modeling

Next Monday (October 17, 2011), the USC-led Physical Sciences Oncology Center / CAMM will host a short course on multidisciplinary cancer modeling, combining the expertise of biologists, oncologists, and physical scientists. I’ll attach a PDF flyer of the schedule below.  I am giving a talk during “Session II – The Physicist Perspective on Cancer.”  I will focus on tailoring mathematical models from the ground up to clinical data from individual patients, with an emphasis on using computational models to make testable clinical predictions, and using these models a platforms to generate hypotheses on cancer biology.

The response to our short course has been overwhelming (in a good way), with around 200 registrants! So, registration is unfortunately closed at this time.  However, the talks will be broadcast live via a webcast.  The link and login details are in the PDF below. I hope to see you there! — Paul

Agenda
7:00 am – 8:25 am : Registration, breakfast, and opening comments, etc.
David B. Agus, M.D. (Director of USC CAMM)
W. Daniel Hillis, Ph.D. (PI of USC PSOC, Applied Minds)
Larry A. Nagahara (NCI PSOC Program Director)

8:30 am – 10:15 am : Session I – Cancer Biology and the Cancer Genome
Paul Mischel, UCLA – The Biology of Cancer from Cell to Patient, Oncogenesis to Therapeutic Response
Matteo Pellegrini, UCLA – Evolution in Cancer
Mitchelll Gross, USC – Historical Perspective on Cancer Diagnosis and Treatment

10:30 am – 12:15 pm : Session II – The Physicist Perspective on Cancer
Dan Ruderman, USC – Cancer as a Multi-scale Problem
Paul Macklin, USC – Computational Models of Cancer Growth
Tom Tombrello, Cal-Tech – Perspective: Big Problems in Physics vs. Cancer

1:45 pm. – 3:10 pm : Session III – Novel Measurement Platforms & Data Management & Integration
Michelle Povinelli, USC – The Role of Novel Microdevices in Dissecting Cellular Phenomena
Carl Kesselman, USC – Data Management & Integration Challenges in Interdisciplinary Studies

3:10 pm – 3:40 pm : Session IV – Creativity in Research at the Interface between the Life and Physical Sciences
‘Fireside Chat’ David Agus and Danny Hillis, USC

4:00 pm – 5:00 pm : Capstone – Keynote Speaker
Tim Walsh, Game Inventor, Keynote Speaker

5:30 pm – 8:00 pm : Poster Session and Reception

Click here for the poster, including webcast information: http://mathcancer.org/publications/PSOC_short_course_2011.pdf

## DCIS paper resubmitted; lots of clinical predictions, lots of validation

After a lot of revision, I have merged my two papers on ductal carcinoma in situ (DCIS) in to a single manuscript and resubmitted to the Journal of Theoretical Biology for final review.  A preprint is posted at my website, along with considerable supplementary material (data sets and source code, and animations).  Thanks to my co-authors and other friends for all the help in the revisions. Thanks also the reviewers for insightful comments. I think this revised manuscript is all the better for it.

DCIS is a precursor to invasive breast cancer, and it is generally detected by annual mammographic screening. More advanced DCIS (with greater risk) tends to have comedonecrosis–a type of cell death that leaves calcium phosphate deposits in the centers of the ducts. In fact, this is generally what’s detected in mammograms. DCIS is usually surgically removed by cutting out a small ball of tissue around what’s found in the mammograms (breast-conserving surgery, or lumpectomy).  But current planning isn’t so great. Even with the state-of-the-art in patient imaging and surgical planning, about 20%-50% of women need to get a second surgery because the first one didn’t get the entire tumor.

So, there’s great need to understand calcifications, and how what you see in mammography relates to the actual tumor size and shape.  And if you do have a model to do this, there’s great need to calibrate it to patient pathology data (the stuff you get from your biopsies) so that the models say something meaningful about individual patients.  And there has been no method to do that. Until now.

(As far as I know), this paper is the first to calibrate to individual patient immunohistochemistry and histopathology.  This, along with some parameter estimates to the theoretical and experimental biology literature, allows us to fully constrain the model. No free parameters to play with until it looks right. Any results are fully emergent from a mechanistic model and realistic parameter estimates rooted in the biology.

This model also includes the most detailed description of necrosis–the type of cell death that results in the comedonecrosis seen in mammograms. We include cell swelling, cell bursting, gradual loss of fluid, and the very first model of calcification.

Clinical predictions, with lots of validation:
All said and done, the model gives some big (and validated!) predictions:

• The model predicts that a tumor grows through the duct at a constant rate.  This is consistent with what’s actually seen in mammography.
• The model gives a new explanation for the known trend: when necrotic cells burst and lose fluid, it makes it more mechanically favorable for proliferating cells to push into the center of the duct, rather than along the duct.  For this reason the model predicts faster growth in smaller ducts, and slower growth in larger ducts.
• The model predicts growth rates between 7.5 and 10 mm per year.  This is quantitatively consistent with published values in the clinical literature.
• The model predicts the difference between the size in a mammogram and the actual size (as measured by a pathologist after surgery) grows in time. This unfortunately means that it’s unlikely that there is some “fixed” safe distance to cut around the mammographic findings.
• On the other hand, the model predicts that there is a linear correlation between the size in a mammogram and the actual (pathology) tumor size. This bodes well for future surgical planning.
• Better still, the linear correlation we found quantitatively fits through 87 published patients, spanning two orders of magnitude.
New insights on DCIS biology:
The model also makes several key predictions on the smaller-scale biology:

• The model predicts that fast swelling of necrotic cells (on the order of 6 hours) is responsible for the tear between the viable rim and necrotic core seen in just about every pathology image of DCIS.
• The model predicts that the necrotic core is “age structured”, with newly necrotic cells (with relatively intact nuclei) on the outer edge, and interior band of mostly degraded but noncalcified cells, and a central core of oldest, calcified material.  This compares well with patient histopathology.
• Comparing the model-predicted age structuring to histopathology predicts a sharper estimate on the various necrosis time scales: swelling and lysis (~6 h) < slow fluid loss (~days to a week) < pyknosis (~10+ days) < calcification (~2 weeks).
• Because the model only predicts linear / casting-type calcifications (long “plugs” of calcification), other biophysics must be responsible for the variety of calcification types seen in mammography.
• Among other mechanisms, we postulate a very long-timescale (1-2 months) process of degradation of the phospholipid “backbone” of the calcifications, resulting in degradation of the calcifications. The cracks seen in the central portions of calcifications (in histopathology) supports this view.
This last point is interesting: only 30-50% of solid-type DCIS has linear calcifications. This could provide an explanation for that, and may help improve the accuracy of diagnostic mammography. Furthermore, it may explain spontaneous resolution: where calcifications sometimes disappear from mammograms, while the underlying tumor is still present.

Long-term outlook, and next steps:
In this work, we have taken a step towards moving mathematical models from the blackboard to the clinic. We actually can calibrate models to individual patient data. We actually can make testable predictions on things like growth rates and mammography sizes.

The next step is to start validating the predictions in individual patients, rather than by the clinical literature. Our team has started doing just that.  Our pathologists are getting histopathology measurements from several patients.  Our mammographer is giving us calcification sizes and other data from 2 time points for each patient.  This will allow us to validate the predicted growth rate in each patient!

In the longer term, I’d like to use the model to develop a spatial mapping between the calcification appearance in the mammogram and its actual shape in the breast, as an improved surgical planning tool.  I’d like to study the impact of inadequate surgical margins in our simulated tumors.  And I’d like to expand the model to the next natural (and significant!) step of microinvasion, and progression to full invasive ductal carcinoma.

We’ve taken some nice baby steps towards making an impact in the clinic.  And that’s what this modeling is all about.


text:
# Invariance of $ds^2$ from invariance of all null intervals

Is this linear algebra statement true?

Let $$\eta= \begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}$$.

If $$x^T (\Lambda^T\eta\Lambda) x$$=0 for all $$x$$ such that $$x^T \eta x=0$$, then $$\Lambda^T\eta\Lambda=a\eta$$ for some $$a \in \mathbb{R}$$.

If so how does one prove it? Is a stronger statement true (e.g. $$a>0$$)?

Motivation: I can't figure out why this statement from the Wikipedia article is true using the above mathematical language:

Since if $$ds^{2}=0$$, then the interval will be null in any other system (second postulate), and since $$ds^{2}$$ and $$ds'^{2}$$ are infinitesimals of the same order, they must be proportional to each other,

$$ds^{2}=ads'^{2}$$.

The translation of the above is that $$x^T (\Lambda^T\eta\Lambda) x$$ must be proportional to $$x^T \eta x$$ for all $$x$$. Why?

• For the $\Lambda$ you gave $\Lambda^T \eta \Lambda=\begin{pmatrix}0&-2\\-2&0\end{pmatrix}$ and $x^T\Lambda^T \eta \Lambda x$ doesn't equal zero for any null vector (i.e. an $x$ where $x^T \eta x=0$) so I don't see how this is a counterexample. Also, what is your definition of "a set of Minkowski coordinates"? The mathematical condition I gave is equivalent to saying that the new coordinates are a linear transformation of inertial coordinates that preserves the speed of light. Is that not a suitable definition of Minkowski coordinates? – Alex Mar 12 at 15:53
• I've converted my comment into an answer, and replied to your comment as part of the answer. I've also given a simpler counterexample than the one you were replying to above. – Ben Crowell Mar 12 at 19:08
• FWIW, the statement is taken pretty much verbatim from Landau & Lifshitz, who don't provide a proof either. – Javier Mar 12 at 20:15

# Prelude: 1+2D

In 1+2D we have a matrix $$\Lambda = \begin{bmatrix}a&b&c\\ d&e&f\\ g&h&i \end{bmatrix}$$ which is being used to generate a symmetric matrix $$\eta' = \Lambda^T \eta \Lambda$$. We can rotate the two spatial dimensions into each other to find a family of null vectors $$v(\theta) = [1, \cos\theta,\sin\theta]^T$$ such that $$v^T \eta v = 0$$ and you want to consider only the $$\Lambda$$ such that $$v^T \eta' v =0$$ too, for all $$\theta$$. This would mean that $$(a + b\cos\theta + c\sin\theta)^2 = (d+e\cos\theta + f \sin\theta)^2 + (g + h\cos\theta + i\sin\theta)^2.$$ So we have 6 degrees of freedom (symmetric 3x3 matrix $$\eta'$$) but presumably we have 5 equations here: terms in $$\theta$$ proportional to $$1,$$ $$\cos\theta,$$ $$\sin\theta,$$ $$\cos(2\theta),$$ and $$\sin(2\theta):$$ \begin{align} 2a^2 + b^2 + c^2 &= 2d^2 + e^2 + f^2 + 2g^2 + h^2 + i^2\\ ab &= de + gh\\ ac &= df + gi\\ b^2 - c^2 &= e^2 - f^2 + h^2 - i^2\\ bc &= ef + hi \end{align}The three "small" equations above set all of the off-diagonal elements to be 0 in the resulting matrix. The first "big" equation can be reduced to $$a^2 + b^2 = d^2 + e^2 + g^2 + h^2$$ in light of the second, thus we would have \begin{align}a^2 - d^2 - g^2 &= - b^2 + e^2 + h^2 &:= K\\ b^2 - e^2 - h^2 &= c^2 - f^2 - i^2 &= -K \end{align} proving the form $$\begin{bmatrix}K&0&0\\0&-K&0\\0&0&-K\end{bmatrix}$$ for some $$K$$ which likewise does not have to be positive.

# Extending to 1+N dimensions

Now let's just do the same thing as before, but probe $$\eta'$$ in 1+N dimensions with some basic null vectors that comprise the unit vector in the time direction $$\hat w$$ and the unit vector in some arbitrary space dimension $$\hat x$$, e.g. $$(\hat w \pm \hat x)^T\eta'(\hat w \pm \hat x) = 0.$$Since $$\eta'$$ is symmetric one gets results like $$\hat w^T \eta' \hat w ~\pm~ 2 \hat w^T\eta'\hat x ~+~ \hat x^T\eta'\hat x ~=~ 0$$ and this then argues that these off-diagonal elements $$\hat w^T\eta'\hat x = 0$$ directly.

The above rotational argument from 1+2D gives the same for the $$\hat x^T\eta'\hat y$$ terms if we just do a rotation from any spatial coordinate into any other, call them $$\hat x$$ and $$\hat y$$: we have even that $$(\hat w + \hat x \cos\theta + \hat y \sin\theta)^T \eta' (\hat w + \hat x \cos\theta + \hat y \sin\theta) = 0$$ and the $$\sin(2\theta)$$ component of that equation comes exclusively from $$2 \hat x^T \eta' \hat y \cos\theta \sin\theta$$ and this can only be zero if $$\hat x^T \eta' \hat y = 0.$$

So we've proven that all off-diagonal elements must be zero and then we can just probe with those first null vectors again, so if we use $$\hat w + \hat x$$ then we determine that the $$(w, w)$$ diagonal element must be the negative of the $$(x, x)$$ diagonal element, but since we chose $$\hat x$$ arbitrarily this must apply to all of the other diagonal elements: it must have the form $$\operatorname{diag}(K, -K, -K, \dots)$$.

Proposition. In a space $$V\cong \mathbb{R}^n$$ of dimension $$n=p+q$$ let there be given an indefinite metric tensor $$\eta~=~\begin{pmatrix} \mathbf{1}_{p\times p}& \mathbf{0}_{p\times q} \cr \mathbf{0}_{q\times p} & -\mathbf{1}_{q\times q} \end{pmatrix}_{n\times n}~=~{\rm diag}(\underbrace{+1,\ldots,+1}_{p\text{ times}},\underbrace{-1,\ldots,-1}_{q\text{ times}})\tag{1}$$ of signature $$(p,q)$$, and a (possibly degenerate & indefinite) metric tensor $$g$$. Assume that all null-vectors for $$\eta$$ are also null-vectors for $$g$$: $$\forall v\in V :~~v^t\eta v~=~0~~\Rightarrow~~v^tg v~=~0.\tag{2}$$ Then $$g$$ is proportional to $$\eta$$: $$\exists \lambda\in \mathbb{R}:~~g~=~\lambda \eta.\tag{3}$$

Sketched proof of proposition:

1. Let $$e_i=(0,\ldots, 0,1,0,\ldots, 0)^t$$ be the $$i$$th unit-vector (of $$\eta$$-length-square $$\pm 1$$). Write the metric tensor $$g~=~ \begin{pmatrix} a& b^t \cr b & c \end{pmatrix} \tag{4}$$ in terms of a symmetric $$p\times p$$ matrix $$a$$, a symmetric $$q\times q$$ matrix $$c$$, and a rectangular $$q\times p$$ matrix $$b$$.

2. Use the following "polarization trick" to show that the $$b$$-block vanishes: $$b~=~0.\tag{5}$$ If $$g_{ij}=e_i^tge_j$$ corresponds to a matrix element in the $$b$$-block, then $$v_{\pm}:=e_i\pm e_j$$ are null-vectors, so $$4g_{ij}~=~4e_i^tge_j~=~(v_++v_-)^tg(v_+-v_-)~=~v_+^tgv_+ -v_-^tgv_-~\stackrel{(2)}{=}~0+0~=~0. \tag{6}$$

3. We can diagonalize the symmetric $$a$$ and $$c$$ blocks by orthogonal matrices while keeping $$\eta$$ invariant. In other words, we may assume w.l.o.g. that $$g\text{ is diagonal}.\tag{7}$$

4. Finally, by considering null-vectors of the form $$v:=e_i+ e_j$$, it becomes clear that $$g_{ii}+g_{jj}~=~e_i^tge_i+e_j^tge_j~\stackrel{(7)}{=}~v^tgv~\stackrel{(2)}{=}~0.\tag{8}$$ This implies that both $$a$$ and $$c$$ are proportional to an identity matrix. The sought-for eq. (3) follows. $$\Box$$

[converted from a comment into an answer]

The WP article isn't claiming that it's true on mathematical grounds, it's just saying that it needs to be true on physical grounds. It also isn't claiming that it holds for any Λ, but only for a Λ that represents a change of coordinates to a new set of Minkowski coordinates.

I don't think your claim, without any conditions on Λ, holds. Let $$\Lambda=\operatorname{diag}(2,2)$$. This is just rescaling the coordinates by a factor of 2. For this $$\Lambda$$, we have $$\Lambda^T\eta\Lambda=4\eta$$. This is a counterexample to your conjecture that "If $$x^T (\Lambda^T\eta\Lambda) x$$=0 for all $$x$$ such that $$x^T \eta x=0$$, then $$\Lambda^T\eta\Lambda=\eta$$. "

Also, what is your definition of "a set of Minkowski coordinates"?

Minkowski coordinates are coordinates in which the metric has the form $$\operatorname{diag}(1,-1,-1,-1)$$.

So I think what you were interpreting in the WP article as a mathematical theorem is actually a combination of a physical argument (second postulate) with a definition (defining Minkowski coordinates as above).

A straightforward way to see that your conjecture shouldn't be expected to hold without any condtions on $$\Lambda$$ is that it's written in a way that assumes that we can do a certain coordinate transformation and the corresponding inverse transformation using the matrix $$\Lambda$$ and its transpose. This is not generally true for coordinate transformations. It's a mor special property that happens to hold for Lorentz boosts from one set of Minkowski coordinates to another.

• (1) Your counterexample is good and leads me to think that, in accordance with the wikipedia argument (originating with Landau & Lifshiftz?), that my statement needs to be changed to $\Lambda^T\eta\Lambda=\mathrm{const}\eta$. I'll edit the question. (2) I think the definition you give of Minkowski coordinates puts the cart before the horse in this case. I'm trying to show that the transformation $\Lambda$ preserves the Minkowski metric $\eta$, not start from that point. (3) There is no assumption that $\Lambda^T$ is the inverse transformation. $(\Lambda x)^T$ is a forward transformation of $x$ – Alex Mar 12 at 20:01
• Your counterexample is good and leads me to think that, in accordance with the wikipedia argument (originating with Landau & Lifshiftz?), that my statement needs to be changed to ΛTηΛ=constη No, that doesn't work. This counterexample is just a special case of the fact that all of these ideas only work for certain special transformations, not for all changes of coordinates. – Ben Crowell Mar 12 at 21:58
• I think the definition you give of Minkowski coordinates puts the cart before the horse in this case. I'm trying to show that the transformation Λ preserves the Minkowski metric η, not start from that point. Then you have to decide what premise you're going to start from. Your basic problem here is that you haven't clearly identified any particular set of assumptions that you want to start from, so you can't prove anything. There is more than one way to set up the logic here, but it won't work if you don't start with any set of assumptions at all. – Ben Crowell Mar 12 at 22:00
• Assumptions should be clear: 1. transformation is linear (i.e. no preferred origin of space and time). 2. something traveling at the speed of light in the first frame also travels at the speed of light in the other frame. Goal: prove that such a transformation preserves $\eta$ (maybe up to a scalar multiple). Can you explain why you are so sure this statement is false / give a counterexample? – Alex Mar 12 at 23:26


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# 16th KVJ Solutions8

Here you find the solutions of  16th KVJ Solutions8 in detail explanation

8Q) (a) Two parallel chords in a circle have lengths 10cm an 14cm and distance between them is 6 cm. If a chord parallel to these chords and midway between them is length √a, find the value of a.

Solution :

Let O be the center  and r be the radius of the circle.

Let AB and CD are two Parallel chords and distance between them is 6 cm .

Let AB =14 and CD =10 cm .

Let perpendicular from O to AB is Q and from O to CD be P, then PQ=6cm.

AQ =QB =7cm

CP=PD =5 cm

OA= OC =r cm

Let MN is a chord  of length $$\sqrt{a}$$, midway between AB and CD and intersect PQ at E.

MN = $$\sqrt{a}$$ , Then ME= EN =$$\frac{\sqrt{a}}{2}$$

PE = EQ =3 cm.

Let OQ = x cm then OP =6-x      (as PQ= 6cm).

From Triangle OAQ

$$7^{2} +x^{2} =r^{2}$$

x = $$\sqrt{r^{2} - 49}$$ ----------(1)

From Triangle OCP ,

$$OC^{2} =OP^{2} + PC^{2}$$

$$r^{2} = (6-x)^{2} + 5^{2}$$

$$r^{2} = (6 - \sqrt{r^{2}-49})^2 +25$$  (From (1))

$$r^{2} = 36 + r^{2} - 49 - 12\sqrt{r^{2}-49} +25$$

$$12 -12\sqrt{r^{2}-49} =0$$

$$r^{2} -49 = 1$$

$$r^{2} = 50$$

$$r = 5\sqrt{2}$$

$$x=\sqrt{r^2-49}$$

x=1

To find MN length.

MN is midway between AB and CD.

PE= EQ =3

EQ =x+ OE

3 = 1+OE

OE =2

From triangle OME , we get,

$$OM^{2} =OE^{2} +EM^{2}$$

$$r^{2} =2^{2} + (\frac{\sqrt{a}}{2})^{2}$$

50 = 4 + $$\frac{a}{4}$$

46 = $$\frac{a}{4}$$

a = 46 * 4 = 184

Therefore the value of a is 184.

(b) The line joining two points A(2, 0) and B(3, 1) is rotated about point A in anticlockwise direction through an angle of $$15^{\circ}$$. Find the equation of the line in the new position. IF B goes to C in new position, find the coordinates of C.

For Part (b) solution of  16th KVJ Solutions8 , Click on NEXT .