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# Semi-definite program for conditional smooth max-entropy

I am aware of a SDP formulation for smooth min-entropy: question link. That program for smooth min-entropy was found in this book by Tomachiel: page 91. However, I am yet to come across a semi-definite formulation for smooth max-entropy. There is however, a formulation for the non-smoothed version of max-entropy, found in this link: page 4, lemma 8. Here is the detailed program for a bipartite density operator $$\rho_{AB}$$, $$2^{H_{\text{max}}(A|B)_\rho}$$ =
$$\text{minimize }\lambda \\ \text{subject to} \\ Z_{AB} \otimes \mathbb{I} \ge \rho_{ABC} \\ \lambda \mathbb{I}_B \ge \text{tr}_A [Z_{AB}] \\ Z_{AB} \ge 0 \\ \lambda \ge 0$$

Where $$Z_{AB}$$ runs over all positive semi-definite operators in $$\mathcal{H}_{AB}$$, $$\lambda$$ is a real number. The smooth max-entropy is then: $$H^{\epsilon}_{\text{max}}(A|B)ρ := \underset{\rho'_{AB} \in \mathcal{B}^\epsilon (\rho_{AB})}{\min}H_{\text{max}}(A|B)_{\rho'}$$ i.e., just the minimum over all bipartite operators which are at most $$\epsilon$$ distance away from $$\rho_{AB}$$. But the primal or dual SDP formulation for the smooth version of max-entropy was not found anywhere. Is there one? How could I transform it into a smooth version? TIA.

In particular they say that the smoothing constraint $$\tilde{\rho}_{AB} \in B^\epsilon(\rho_{AB})$$ can be reformulated as the triple of constraints $$\mathrm{Tr}[\tilde\rho_{ABC} \rho_{ABC}] \geq 1- \epsilon^2~\\ \mathrm{Tr}[\tilde\rho_{ABC}] \leq 1 ~\\ \tilde \rho_{ABC} \geq 0$$ where $$\rho_{ABC}$$ is any purification of $$\rho_{AB}$$.
Now we can incorporate these extra constraints with the SDP formulation of $$H_{\max}(A|B)$$. In particular \begin{aligned} 2^{H^\epsilon_{\max}(A|B)_\rho} &= \min_{\tilde{\rho}_{AB} \in B^\epsilon(\rho_{AB})} 2^{H_{\max}(A|B)_{\tilde\rho}} \\ &= \min_{\tilde{\rho}_{AB} \in B^\epsilon(\rho_{AB})}\min \lambda \\ &\qquad\mathrm{s.t.} \quad Z_{AB} \otimes I_C \geq \tilde\rho_{ABC} \\ &\qquad \qquad \lambda I_B \geq \mathrm{Tr}_A[Z_{AB}] \\ &\qquad\qquad Z_{AB} \geq 0, \quad \lambda \geq 0 \\ &= \,\,\,\min \quad\lambda \\ &\qquad\mathrm{s.t.} \quad Z_{AB} \otimes I_C \geq \tilde\rho_{ABC} \\ &\qquad \qquad \lambda I_B \geq \mathrm{Tr}_A[Z_{AB}] \\ &\qquad \qquad \mathrm{Tr}[\tilde\rho_{ABC} \rho_{ABC}] \geq 1- \epsilon^2~\\ &\qquad \qquad \mathrm{Tr}[\tilde\rho_{ABC}] \leq 1 ~\\ &\qquad\qquad \rho_{ABC} \geq 0, \quad Z_{AB} \geq 0, \quad \lambda \geq 0 \\ \end{aligned}


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# Kelly criterion

In probability theory, the Kelly criterion (or Kelly strategy or Kelly bet), is a formula for sizing a bet. The Kelly bet size is found by maximizing the expected value of the logarithm of wealth, which is equivalent to maximizing the expected geometric growth rate. It assumes that the expected returns are known and is optimal for a bettor who values their wealth logarithmically. J. L. Kelly Jr, a researcher at Bell Labs, described the criterion in 1956.[1] Because the Kelly Criterion leads to higher wealth than any other strategy in the long run (i.e., the theoretical maximum return as the number of bets goes to infinity), it is a scientific gambling method.

Example of the optimal Kelly betting fraction, versus expected return of other fractional bets.

The practical use of the formula has been demonstrated for gambling[2][3] and the same idea was used to explain diversification in investment management.[4] In the 2000s, Kelly-style analysis became a part of mainstream investment theory[5] and the claim has been made that well-known successful investors including Warren Buffett[6] and Bill Gross[7] use Kelly methods. [8] Also see Intertemporal portfolio choice.

## Optimal betting example

In a study, each participant was given $25 and asked to place even-money bets on a coin that would land heads 60% of the time. Participants had 30 minutes to play, so could place about 300 bets, and the prizes were capped at$250. But the behavior of the test subjects was far from optimal:

Remarkably, 28% of the participants went bust, and the average payout was just $91. Only 21% of the participants reached the maximum. 18 of the 61 participants bet everything on one toss, while two-thirds gambled on tails at some stage in the experiment.[9][10] Using the Kelly criterion and based on the odds in the experiment (ignoring the cap of$250 and the finite duration of the test), the right approach would be to bet 20% of one's bankroll on each toss of the coin, which works out to a 2.034% average gain each round. This is a geometric mean, not the arithmetic rate of 4% (${\displaystyle r=(1+0.2\cdot 1.0)^{0.6}\cdot (1-0.2\cdot 1.0)^{0.4}}$ ). The theoretical expected wealth after 300 rounds works out to $10,505 (${\displaystyle =25\cdot (1.02034)^{300}}$ ) if it were not capped. In this particular game, because of the cap, a strategy of betting only 12% of the pot on each toss would have even better results (a 95% probability of reaching the cap and an average payout of$242.03).

## Gambling formula

Where losing the bet involves losing the entire wager, the Kelly bet is:

${\displaystyle f^{*}=p-{\frac {q}{b}}=p-{\frac {1-p}{b}}}$

where:

• ${\displaystyle f^{*}}$  is the fraction of the current bankroll to wager.
• ${\displaystyle p}$  is the probability of a win.
• ${\displaystyle q}$  is the probability of a loss (${\displaystyle q=1-p}$ ).
• ${\displaystyle b}$  is the proportion of the bet gained with a win. E.g. If betting $10 on a 2-to-1 odds bet, (upon win you are returned$30, winning you $20), then ${\displaystyle b=\20/\10=2.0}$ . The figure plots the amount gained with a win on the x-axis against the fraction of portfolio to bet on the y-axis. This figure assumes p=0.5 (that the probability of both a win and a loss is 50%). If the amount gained with a win is 1, then the Kelly betting amount is$0, which makes sense in a fair bet with no expected gain.

As an example, if a gamble has a 60% chance of winning (${\displaystyle p=0.6}$ , ${\displaystyle q=0.4}$ ), and the gambler receives 1-to-1 odds on a winning bet (${\displaystyle b=1}$ ), then to maximize the long-run growth rate of the bankroll, the gambler should bet 20% of the bankroll at each opportunity (${\textstyle f^{*}=0.6-{\frac {0.4}{1}}=0.2}$ ).

The figure plots the amount gained with a win on the x-axis against the fraction of portfolio to bet on the y-axis. This figure assumes p=0.6 (that the probability of both a win is 60%).

3D figure representing the optimal Kelly bet size (vertical axis) as a function of win probability and amount gained with win.

If the gambler has zero edge, i.e. if ${\displaystyle b=q/p}$ , then the criterion recommends for the gambler to bet nothing.

If the edge is negative (${\displaystyle b ) the formula gives a negative result, indicating that the gambler should take the other side of the bet. For example, in American roulette, the bettor is offered an even money payoff (${\displaystyle b=1}$ ) on red, when there are 18 red numbers and 20 non-red numbers on the wheel (${\displaystyle p=18/38}$ ). The Kelly bet is ${\displaystyle -1/19}$ , meaning the gambler should bet one-nineteenth of their bankroll that red will not come up. There is no explicit anti-red bet offered with comparable odds in roulette, so the best a Kelly gambler can do is bet nothing.

## Investment formula

A more general form of the Kelly formula allows for partial losses, which is relevant for investments:

${\displaystyle f^{*}={\frac {p}{a}}-{\frac {q}{b}}}$

where:

• ${\displaystyle f^{*}}$  is the fraction of the assets to apply to the security.
• ${\displaystyle p}$  is the probability that the investment increases in value.
• ${\displaystyle q}$  is the probability that the investment decreases in value (${\displaystyle q=1-p}$ ).
• ${\displaystyle a}$  is the fraction that is lost in a negative outcome. If the security price falls 10%, then ${\displaystyle a=0.1}$
• ${\displaystyle b}$  is the fraction that is gained in a positive outcome. If the security price rises 10%, then ${\displaystyle b=0.1}$ .

Note that the Kelly Criterion is valid only for known outcome probabilities, which is not the case with investments. Risk averse investors should not invest the full Kelly fraction.

This formula can result in Kelly fractions higher than 1. In this case, it is theoretically advantageous to use leverage to purchase additional securities on margin.

## Proof

Heuristic proofs of the Kelly criterion are straightforward.[11] The Kelly criterion maximizes the expected value of the logarithm of wealth (the expectation value of a function is given by the sum, over all possible outcomes, of the probability of each particular outcome multiplied by the value of the function in the event of that outcome). We start with 1 unit of wealth and bet a fraction ${\displaystyle f}$  of that wealth on an outcome that occurs with probability ${\displaystyle p}$  and offers odds of ${\displaystyle b}$ . The probability of winning is ${\displaystyle p}$ , and in that case the resulting wealth is equal to ${\displaystyle 1+fb}$ . The probability of losing is ${\displaystyle q=1-p}$ , and in that case the resulting wealth is equal to ${\displaystyle 1-fa}$ . Therefore, the expected geometric growth rate ${\displaystyle r}$  is:

${\displaystyle r=(1+fb)^{p}\cdot (1-fa)^{q}}$

We want to find the maximum r of this curve (as a function of f), which involves finding the derivative of the equation. This is more easily accomplished by taking the logarithm of each side first. The resulting equation is:

${\displaystyle E=\log(r)=p\log(1+fb)+q\log(1-fa)}$

with ${\displaystyle E}$  denoting logarithmic wealth growth. To find the value of ${\displaystyle f}$  for which the growth rate is maximized, denoted as ${\displaystyle f^{*}}$ , we differentiate the above expression and set this equal to zero. This gives:

${\displaystyle \left.{\frac {dE}{df}}\right|_{f=f^{*}}={\frac {pb}{1+f^{*}b}}+{\frac {-qa}{1-f^{*}a}}=0}$

Rearranging this equation to solve for the value of ${\displaystyle f^{*}}$  gives the Kelly criterion:

${\displaystyle f^{*}={\frac {p}{a}}-{\frac {q}{b}}}$

Notice that this expression reduces to the simple gambling formula when ${\displaystyle a=1=100\%}$ , when a loss results in full loss of the wager.

## Bernoulli

In a 1738 article, Daniel Bernoulli suggested that, when one has a choice of bets or investments, one should choose that with the highest geometric mean of outcomes. This is mathematically equivalent to the Kelly criterion, although the motivation is different (Bernoulli wanted to resolve the St. Petersburg paradox).

An English-language translation of the Bernoulli article was not published until 1954,[12] but the work was well known among mathematicians and economists.

## Application to the stock market

In mathematical finance, if security weights maximize the expected geometric growth rate (which is equivalent to maximizing log wealth), then a portfolio is growth optimal.

Computations of growth optimal portfolios can suffer tremendous garbage in, garbage out problems. For example, the cases below take as given the expected return and covariance structure of assets, but these parameters are at best estimates or models that have significant uncertainty. If portfolio weights are largely a function of estimation errors, then Ex-post performance of a growth-optimal portfolio may differ fantastically from the ex-ante prediction. Parameter uncertainty and estimation errors are a large topic in portfolio theory. An approach to counteract the unknown risk is to invest less than the Kelly criterion (e.g., half).

## Criticism

Although the Kelly strategy's promise of doing better than any other strategy in the long run seems compelling, some economists have argued strenuously against it, mainly because an individual's specific investing constraints may override the desire for optimal growth rate.[8] The conventional alternative is expected utility theory which says bets should be sized to maximize the expected utility of the outcome (to an individual with logarithmic utility, the Kelly bet maximizes expected utility, so there is no conflict; moreover, Kelly's original paper clearly states the need for a utility function in the case of gambling games which are played finitely many times[1]). Even Kelly supporters usually argue for fractional Kelly (betting a fixed fraction of the amount recommended by Kelly) for a variety of practical reasons, such as wishing to reduce volatility, or protecting against non-deterministic errors in their advantage (edge) calculations.[13]

## Advanced mathematics

For a rigorous and general proof, see Kelly's original paper[1] or some of the other references listed below. Some corrections have been published.[14] We give the following non-rigorous argument for the case with ${\displaystyle b=1}$  (a 50:50 "even money" bet) to show the general idea and provide some insights.[1] When ${\displaystyle b=1}$ , a Kelly bettor bets ${\displaystyle 2p-1}$  times their initial wealth ${\displaystyle W}$ , as shown above. If they win, they have ${\displaystyle 2pW}$  after one bet. If they lose, they have ${\displaystyle 2(1-p)W}$ . Suppose they make ${\displaystyle N}$  bets like this, and win ${\displaystyle K}$  times out of this series of ${\displaystyle N}$  bets. The resulting wealth will be:

${\displaystyle 2^{N}p^{K}(1-p)^{N-K}W\!.}$

Note that the ordering of the wins and losses does not affect the resulting wealth. Suppose another bettor bets a different amount, ${\displaystyle (2p-1+\Delta )W}$  for some value of ${\displaystyle \Delta }$  (where ${\displaystyle \Delta }$  may be positive or negative). They will have ${\displaystyle (2p+\Delta )W}$  after a win and ${\displaystyle [2(1-p)-\Delta ]W}$  after a loss. After the same series of wins and losses as the Kelly bettor, they will have:

${\displaystyle (2p+\Delta )^{K}[2(1-p)-\Delta ]^{N-K}W}$

Take the derivative of this with respect to ${\displaystyle \Delta }$  and get:

${\displaystyle K(2p+\Delta )^{K-1}[2(1-p)-\Delta ]^{N-K}W-(N-K)(2p+\Delta )^{K}[2(1-p)-\Delta ]^{N-K-1}W}$

The function is maximized when this derivative is equal to zero, which occurs at:

${\displaystyle K[2(1-p)-\Delta ]=(N-K)(2p+\Delta )}$

which implies that

${\displaystyle \Delta =2\left({\frac {K}{N}}-p\right)}$

but the proportion of winning bets will eventually converge to:

${\displaystyle \lim _{N\to +\infty }{\frac {K}{N}}=p}$

according to the weak law of large numbers. So in the long run, final wealth is maximized by setting ${\displaystyle \Delta }$  to zero, which means following the Kelly strategy. This illustrates that Kelly has both a deterministic and a stochastic component. If one knows K and N and wishes to pick a constant fraction of wealth to bet each time (otherwise one could cheat and, for example, bet zero after the Kth win knowing that the rest of the bets will lose), one will end up with the most money if one bets:

${\displaystyle \left(2{\frac {K}{N}}-1\right)W}$

each time. This is true whether ${\displaystyle N}$  is small or large. The "long run" part of Kelly is necessary because K is not known in advance, just that as ${\displaystyle N}$  gets large, ${\displaystyle K}$  will approach ${\displaystyle pN}$ . Someone who bets more than Kelly can do better if ${\displaystyle K>pN}$  for a stretch; someone who bets less than Kelly can do better if ${\displaystyle K  for a stretch, but in the long run, Kelly always wins. The heuristic proof for the general case proceeds as follows.[citation needed] In a single trial, if you invest the fraction ${\displaystyle f}$  of your capital, if your strategy succeeds, your capital at the end of the trial increases by the factor ${\displaystyle 1-f+f(1+b)=1+fb}$ , and, likewise, if the strategy fails, you end up having your capital decreased by the factor ${\displaystyle 1-fa}$ . Thus at the end of ${\displaystyle N}$  trials (with ${\displaystyle pN}$  successes and ${\displaystyle qN}$  failures), the starting capital of \$1 yields

${\displaystyle C_{N}=(1+fb)^{pN}(1-fa)^{qN}.}$

Maximizing ${\displaystyle \log(C_{N})/N}$ , and consequently ${\displaystyle C_{N}}$ , with respect to ${\displaystyle f}$  leads to the desired result

${\displaystyle f^{*}=p/a-q/b.}$

Edward O. Thorp provided a more detailed discussion of this formula for the general case.[15] There, it can be seen that the substitution of ${\displaystyle p}$  for the ratio of the number of "successes" to the number of trials implies that the number of trials must be very large, since ${\displaystyle p}$  is defined as the limit of this ratio as the number of trials goes to infinity. In brief, betting ${\displaystyle f^{*}}$  each time will likely maximize the wealth growth rate only in the case where the number of trials is very large, and ${\displaystyle p}$  and ${\displaystyle b}$  are the same for each trial. In practice, this is a matter of playing the same game over and over, where the probability of winning and the payoff odds are always the same. In the heuristic proof above, ${\displaystyle pN}$  successes and ${\displaystyle qN}$  failures are highly likely only for very large ${\displaystyle N}$ .

### Multiple outcomes

Kelly's criterion may be generalized[16] on gambling on many mutually exclusive outcomes, such as in horse races. Suppose there are several mutually exclusive outcomes. The probability that the ${\displaystyle k}$ -th horse wins the race is ${\displaystyle p_{k}}$ , the total amount of bets placed on ${\displaystyle k}$ -th horse is ${\displaystyle B_{k}}$ , and

${\displaystyle \beta _{k}={\frac {B_{k}}{\sum _{i}B_{i}}}={\frac {D}{1+Q_{k}}},}$

where ${\displaystyle Q_{k}}$  are the pay-off odds. ${\displaystyle D=1-tt}$ , is the dividend rate where ${\displaystyle tt}$  is the track take or tax, ${\displaystyle {\frac {D}{\beta _{k}}}}$  is the revenue rate after deduction of the track take when ${\displaystyle k}$ -th horse wins. The fraction of the bettor's funds to bet on ${\displaystyle k}$ -th horse is ${\displaystyle f_{k}}$ . Kelly's criterion for gambling with multiple mutually exclusive outcomes gives an algorithm for finding the optimal set ${\displaystyle S^{o}}$  of outcomes on which it is reasonable to bet and it gives explicit formula for finding the optimal fractions ${\displaystyle f_{k}^{o}}$  of bettor's wealth to be bet on the outcomes included in the optimal set ${\displaystyle S^{o}}$ . The algorithm for the optimal set of outcomes consists of four steps:[16]

1. Calculate the expected revenue rate for all possible (or only for several of the most promising) outcomes:
${\displaystyle er_{i}={\frac {Dp_{i}}{\beta _{i}}}=p_{i}(Q_{i}+1)}$

2. Reorder the outcomes so that the new sequence ${\displaystyle er_{k}}$  is non-increasing. Thus ${\displaystyle er_{1}}$  will be the best bet.
3. Set ${\displaystyle S=\varnothing }$  (the empty set), ${\displaystyle k=1}$ , ${\displaystyle R(S)=1}$ . Thus the best bet ${\displaystyle er_{k}=er_{1}}$  will be considered first.
4. Repeat:
If ${\displaystyle er_{k}={\frac {D}{\beta _{k}}}p_{k}>R(S)}$  then insert ${\displaystyle k}$ -th outcome into the set: ${\displaystyle S=S\cup \{k\}}$ , recalculate ${\displaystyle R(S)}$  according to the formula:
${\displaystyle R(S)={\frac {D\sum _{k\notin S}p_{k}}{D-\sum _{k\in S}\beta _{k}}}}$

and then set ${\displaystyle k=k+1}$ ,
Otherwise, set ${\displaystyle S^{o}=S}$  and stop the repetition.

If the optimal set ${\displaystyle S^{o}}$  is empty then do not bet at all. If the set ${\displaystyle S^{o}}$  of optimal outcomes is not empty, then the optimal fraction ${\displaystyle f_{k}^{o}}$  to bet on ${\displaystyle k}$ -th outcome may be calculated from this formula:

${\displaystyle f_{i}=p_{i}-\beta _{i}{\frac {\sum _{k\notin S}p_{k}}{\left(D-\sum _{k\in S}\beta _{k}\right)}}.}$

One may prove[16] that

${\displaystyle R(S^{o})=1-\sum _{i\in S^{o}}{f_{i}^{o}}}$

where the right hand-side is the reserve rate[clarification needed]. Therefore, the requirement ${\displaystyle er_{k}={\frac {D}{\beta _{k}}}p_{k}>R(S)}$  may be interpreted[16] as follows: ${\displaystyle k}$ -th outcome is included in the set ${\displaystyle S^{o}}$  of optimal outcomes if and only if its expected revenue rate is greater than the reserve rate. The formula for the optimal fraction ${\displaystyle f_{k}^{o}}$  may be interpreted as the excess of the expected revenue rate of ${\displaystyle k}$ -th horse over the reserve rate divided by the revenue after deduction of the track take when ${\displaystyle k}$ -th horse wins or as the excess of the probability of ${\displaystyle k}$ -th horse winning over the reserve rate divided by revenue after deduction of the track take when ${\displaystyle k}$ -th horse wins. The binary growth exponent is

${\displaystyle G^{o}=\sum _{i\in S}{p_{i}\log _{2}(er_{i})}+\left(1-\sum _{i\in S}{p_{i}}\right)\log _{2}(R(S^{o})),}$

and the doubling time is

${\displaystyle T_{d}={\frac {1}{G^{o}}}.}$

This method of selection of optimal bets may be applied also when probabilities ${\displaystyle p_{k}}$  are known only for several most promising outcomes, while the remaining outcomes have no chance to win. In this case it must be that

${\displaystyle \sum _{i}{p_{i}}<1,}$  and
${\displaystyle \sum _{i}{\beta _{i}}<1}$ .

### Stock investments

The second-order Taylor polynomial can be used as a good approximation of the main criterion. Primarily, it is useful for stock investment, where the fraction devoted to investment is based on simple characteristics that can be easily estimated from existing historical data – expected value and variance. This approximation leads to results that are robust and offer similar results as the original criterion.[17]

For single assets(stock, index fund, etc.), and a risk-free rate, it is easy to obtain the optimal fraction to invest through geometric Brownian motion. The stochastic differential equation governing the evolution of a lognormally distributed asset ${\displaystyle S}$  at time ${\displaystyle t}$  (${\displaystyle S_{t}}$ ) is

${\displaystyle dS_{t}/S_{t}=\mu dt+\sigma dW_{t}}$

whose solution is

${\displaystyle S_{t}=S_{0}\exp \left(\left(\mu -{\frac {\sigma ^{2}}{2}}\right)t+\sigma W_{t}\right)}$

where ${\displaystyle W_{t}}$  is a Wiener process, and ${\displaystyle \mu }$  (percentage drift) and ${\displaystyle \sigma }$  (the percentage volatility) are constants. Taking expectations of the logarithm:

${\displaystyle \mathbb {E} \log(S_{t})=\log(S_{0})+\left(\mu -{\frac {\sigma ^{2}}{2}}\right)t.}$

Then the expected log return ${\displaystyle R_{s}}$  is

${\displaystyle R_{s}=\left(\mu -{\frac {\sigma ^{2}}{2}}\,\right)s.}$

Consider a portfolio made of an asset ${\displaystyle S}$  and a bond paying risk-free rate ${\displaystyle r}$ , with fraction ${\displaystyle f}$  invested in ${\displaystyle S}$  and ${\displaystyle (1-f)}$  in the bond. The aforementioned equation for ${\displaystyle dS_{t}}$  must be modified by this fraction, ie ${\displaystyle dS_{t}'=fdS_{t}}$ , with associated solution

${\displaystyle S'_{t}=S'_{0}\exp \left(\left(f\mu -{\frac {(f\sigma )^{2}}{2}}\right)t+f\sigma W_{t}\right)}$

the expected one-period return is given by

${\displaystyle \mathbb {E} {\left(\left[{\frac {S'_{1}}{S'_{0}}}-1\right]+(1-f)r\right)}=\mathbb {E} {\left(\left[\exp \left(\left(f\mu -{\frac {(f\sigma )^{2}}{2}}\right)+f\sigma W_{1}\right)-1\right]\right)}+(1-f)r}$

For small ${\displaystyle \mu }$ , ${\displaystyle \sigma }$ , and ${\displaystyle W_{t}}$ , the solution can be expanded to first order to yield an approximate increase in wealth

${\displaystyle G(f)=f\mu -{\frac {(f\sigma )^{2}}{2}}+((1-f)\ r).}$

Solving ${\displaystyle \max(G(f))}$  we obtain

${\displaystyle f^{*}={\frac {\mu -r}{\sigma ^{2}}}.}$

${\displaystyle f^{*}}$  is the fraction that maximizes the expected logarithmic return, and so, is the Kelly fraction. Thorp[15] arrived at the same result but through a different derivation. Remember that ${\displaystyle \mu }$  is different from the asset log return ${\displaystyle R_{s}}$ . Confusing this is a common mistake made by websites and articles talking about the Kelly Criterion.

For multiple assets, consider a market with ${\displaystyle n}$  correlated stocks ${\displaystyle S_{k}}$  with stochastic returns ${\displaystyle r_{k}}$ , ${\displaystyle k=1,\dots ,n,}$  and a riskless bond with return ${\displaystyle r}$ . An investor puts a fraction ${\displaystyle u_{k}}$  of their capital in ${\displaystyle S_{k}}$  and the rest is invested in the bond. Without loss of generality, assume that investor's starting capital is equal to 1. According to the Kelly criterion one should maximize

${\displaystyle \mathbb {E} \left[\ln \left((1+r)+\sum \limits _{k=1}^{n}u_{k}(r_{k}-r)\right)\right].}$

Expanding this with a Taylor series around ${\displaystyle {\vec {u_{0}}}=(0,\ldots ,0)}$  we obtain

${\displaystyle \mathbb {E} \left[\ln(1+r)+\sum \limits _{k=1}^{n}{\frac {u_{k}(r_{k}-r)}{1+r}}-{\frac {1}{2}}\sum \limits _{k=1}^{n}\sum \limits _{j=1}^{n}u_{k}u_{j}{\frac {(r_{k}-r)(r_{j}-r)}{(1+r)^{2}}}\right].}$

Thus we reduce the optimization problem to quadratic programming and the unconstrained solution is

${\displaystyle {\vec {u^{\star }}}=(1+r)({\widehat {\Sigma }})^{-1}({\widehat {\vec {r}}}-r)}$

where ${\displaystyle {\widehat {\vec {r}}}}$  and ${\displaystyle {\widehat {\Sigma }}}$  are the vector of means and the matrix of second mixed noncentral moments of the excess returns. There is also a numerical algorithm for the fractional Kelly strategies and for the optimal solution under no leverage and no short selling constraints.[18]

## References

1. ^ a b c d Kelly, J. L. (1956). "A New Interpretation of Information Rate" (PDF). Bell System Technical Journal. 35 (4): 917–926. doi:10.1002/j.1538-7305.1956.tb03809.x.
2. ^ Thorp, E. O. (January 1961), "Fortune's Formula: The Game of Blackjack", American Mathematical Society
3. ^ Thorp, E. O. (1962), Beat the dealer: a winning strategy for the game of twenty-one. A scientific analysis of the world-wide game known variously as blackjack, twenty-one, vingt-et-un, pontoon or Van John, Blaisdell Pub. Co
4. ^ Thorp, Edward O.; Kassouf, Sheen T. (1967), Beat the Market: A Scientific Stock Market System (PDF), Random House, ISBN 0-394-42439-5, archived from the original (PDF) on 2009-10-07, page 184f.
5. ^ Zenios, S. A.; Ziemba, W. T. (2006), Handbook of Asset and Liability Management, North Holland, ISBN 978-0-444-50875-1
6. ^ Pabrai, Mohnish (2007), The Dhandho Investor: The Low-Risk Value Method to High Returns, Wiley, ISBN 978-0-470-04389-9
7. ^ Thorp, E. O. (September 2008), "The Kelly Criterion: Part II", Wilmott Magazine
8. ^ a b Poundstone, William (2005), Fortune's Formula: The Untold Story of the Scientific Betting System That Beat the Casinos and Wall Street, New York: Hill and Wang, ISBN 0-8090-4637-7
9. ^ Haghani, Victor; Dewey, Richard (19 October 2016). "Rational Decision-Making under Uncertainty: Observed Betting Patterns on a Biased Coin". SSRN 2856963. arXiv:1701.01427
10. ^ "Buttonwood", "Irrational tossers", The Economist Newspaper Limited 2016, Nov 1st 2016.
11. ^ Press, W. H.; Teukolsky, S. A.; Vetterling, W. T.; Flannery, B. P. (2007), "Section 14.7 (Example 2.)", Numerical Recipes: The Art of Scientific Computing (3rd ed.), New York: Cambridge University Press, ISBN 978-0-521-88068-8
12. ^ Bernoulli, Daniel (1954) [1738]. "Exposition of a New Theory on the Measurement of Risk". Econometrica. The Econometric Society. 22 (1): 22–36. doi:10.2307/1909829. JSTOR 1909829.
13. ^ Thorp, E. O. (May 2008), "The Kelly Criterion: Part I", Wilmott Magazine
14. ^ Thorp, E. O. (1969). "Optimal Gambling Systems for Favorable Games". Revue de l'Institut International de Statistique / Review of the International Statistical Institute. International Statistical Institute (ISI). 37 (3): 273–293. doi:10.2307/1402118. JSTOR 1402118. MR 0135630.
15. ^ a b Thorp, Edward O. (June 1997). "The Kelly criterion in blackjack, sports betting, and the stock market" (PDF). 10th International Conference on Gambling and Risk Taking. Montreal. Archived from the original (PDF) on 2009-03-20. Retrieved 2009-03-20.
16. ^ a b c d Smoczynski, Peter; Tomkins, Dave (2010) "An explicit solution to the problem of optimizing the allocations of a bettor’s wealth when wagering on horse races", Mathematical Scientist", 35 (1), 10-17
17. ^ Marek, Patrice; Ťoupal, Tomáš; Vávra, František (2016). "Efficient Distribution of Investment Capital". 34th International Conference Mathematical Methods in Economics, MME2016, Conference Proceedings: 540–545. Retrieved 24 January 2018.
18. ^ Nekrasov, Vasily (2013). "Kelly Criterion for Multivariate Portfolios: A Model-Free Approach". SSRN 2259133.

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SNV - SN EN ISO 4545-1

Metallic materials - Knoop hardness test - Part 1: Test method

active, Most Current
 Organization: SNV Publication Date: 1 August 2018 Status: active Page Count: 36 ICS Code (Mechanical testing of metals): 77.040.10
scope:

ISO 4545-1:2017 specifies the Knoop hardness test method for metallic materials for test forces from 0,009 807 N to 19,613 N. The Knoop hardness test is specified in this document for lengths of indentation diagonals =0,020 mm. Using this method to determine Knoop hardness from smaller indentations is outside the scope of this document as results would suffer from large uncertainties due to the limitations of optical measurement and imperfections in tip geometry. ISO 14577-1 allows the determination of hardness from smaller indentations. A periodic verification method is specified for routine checking of the testing machine in service by the user. Special considerations for Knoop testing of metallic coatings can be found in ISO 4516.

Document History

SN EN ISO 4545-1
August 1, 2018
Metallic materials - Knoop hardness test - Part 1: Test method
ISO 4545-1:2017 specifies the Knoop hardness test method for metallic materials for test forces from 0,009 807 N to 19,613 N. The Knoop hardness test is specified in this document for lengths of...
February 1, 2006
Metallic materials - Knoop hardness test - Part 1: Test method (ISO 4545-1:2005)
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text:
# SyncTeX points to a wrong location in source or preview document

I have the following pretty simple document:

\documentclass[11pt]{article}
\usepackage{xltxtra}
\usepackage{fontspec}
\usepackage{amsmath}
\usepackage{polyglossia}
\setmainlanguage{russian}
\setotherlanguage{english}
\setmainfont[Mapping=tex-text]{Linux Libertine}

\begin{document}

\section{Выражения}

\subsection{Числовые выражения}

Решим задачу: Туристы в течение двух часов ехали на велосипеде по шоссе
со скоростью 16~км/ч, а затем шли лесом еще 7~км. Какова длина всего
маршрута?''

\end{document}

I use TeXworks. When I compile the document with XeLaTeX and then Ctrl + click on either the source or preview it highlights the corresponding line incorrectly. What can be wrong here?

Also I tried the same in Texmaker and it also synchronizes preview with source incorrectly.

-
I am not able to compile this as it seems that the font Linux Libertine is not part of the standard TeXLive2011 distribution. –  Peter Grill Nov 17 '11 at 16:20
@PeterGrill: I'd say that Linux Libertine is indeed a part of TL2011. But the font name should be Linux Libertine O. –  Leo Liu Nov 17 '11 at 17:27
@Maxim: Well, everything is OK on my computer. (with TeXworks) –  Leo Liu Nov 17 '11 at 17:30
@LeoLiu It's strange. You said that Linux Libertine is part of TexLive installation, but when I try to use Linux Libertine O it throws "The font "Linux Libertine O" cannot be found.". I installed Linux Libertine from their site and it is accessible as Linux Libertine (without O at the end). So probably something wrong with my fonts or something like that? I am just a noob in the TeX world... –  Maxim Nov 17 '11 at 18:11
@Maxim: Please read the document of TeX Live 2011, "3.4 Post-install actions" -> "3.4.4 Font configuration for XeTeX" –  Leo Liu Nov 17 '11 at 18:14

Note (added 2012-07-17). The bug has been fixed in TeX Live 2012

I can only confirm the bad behavior, at least on Mac OS X. It seems that the synchronization with XeTeX is off by one paragraph. I've checked with other documents and different pairs of editor/previewer.

The bug seems to have been introduced with the 2011 TeX Live, as synchronization works when the document is compiled with the 2010 binaries.

A minimal document exhibiting the problem is

\documentclass{article}
\begin{document}
1 par

2 par
\end{document}

Clicking on 1par in the source jumps to the left corner of the (absent) headline, while 2par jumps to the right corner. Clicking either on "1 par" or "2 par" in the PDF jumps after \end{document} in the source.

-
Is it font-specific? Can a minimal example be made with plain XeTeX (to make sure it isn't a bug cause by the various packages Maxim loads)? –  Bruno Le Floch Jan 8 '12 at 5:21
This looks OK in TL2012: could you check? –  Joseph Wright Jul 17 '12 at 15:09
I didn't realize I was running an old version of TeX Live. I'll check it out. If this does prove to be the solution, @JosephWright, you should make your comment an answer - I don't know if you get the bounty credit for community wiki –  keflavich Jul 23 '12 at 15:49


text:
## Monday, August 14, 2006

### Time magazine: physics is a sin

I was alerted that the "be very afraid" Time magazine has joined the broad family of people and institutions that I simply can't have any respect for because of their complete lack of intellectual integrity. On
where they even don't know how to spell the name of John Schwarz correctly, they use not only the words of the physicists whom serious physicists almost never take seriously - for good reasons - as well as science critics who have not even tried to do science but also the words of Sean Carroll to achieve their main objective, namely to completely mislead the readers about the status of string theory. The role of Sean Carroll who adds a dumb general comment to this dirty, immoral, dishonest, and anti-scientific enterprise is absolutely no surprise for me.

The very same crap is also printed in New Scientist. This medium has always been dominated by crackpots anyway.


text:
Browse Questions

# Accumulation of excess fluid in tissue spaces is

$\begin{array}{1 1}(A)\;\text{Oedema}\\(B)\;\text{Cirrhosis}\\(C)\;\text{Parkinson's disease}\\(D)\;\text{Hodgkin's disease}\end{array}$

Can you answer this question?

Accumulation of excess fluid in tissue spaces is oedema
Hence (A) is the correct answer.
answered Jul 10, 2014


text:
Others

# Incompressible Navier Stokes Equations

View times: 627
Submitted by: Terry Moschandreou

## Abstract

A closely related problem to The Clay Math Institute "Navier-Stokes, breakdown of smooth solutions here on an arbitrary cube subset of three dimensional space with periodic boundary conditions is examined. The incompressible Navier-Stokes Equations are presented in a new and conventionally different way here, by naturally reducing them to an operator form which is then further analyzed. It is shown that a reduction to a general 2D N-S system decoupled from a 1D non-linear partial differential equation is possible to obtain. This is executed using integration over n-dimensional compact intervals which allows decoupling. Here we extract the measure-zero points in the domain where singularities may occur and are left with a pde that exhibits finite time singularity. The operator form is considered in a physical geometric vorticity case, and a more general case. In the general case, the solution is revealed to have smooth solutions which exhibit finite-time blowup on a fine measure zero set and using the Gagliardo-Nirenberg inequalities it is shown that for any non zero measure set in the form of cube subset of 3D there is finite time blowup.

## 1. Introduction

The question of whether the solutions to the 3-D- Incompressible N-S equations are globally regular or demonstrate finite time blowup has been a long going debate in mathematics and in general the scientific communities. The Millennium problem posed by the Clay Institute [1] is asking for a proof of one of the above conjectures. Seminal papers conducted by Jean Leray [2][3][4] proved that there exists a global (in time) weak solution and a local strong solution of the initial value problem when the domain is all of , that is solutions up to some finite T* on an interval [0, T*]. While specific cases have approached answers in unique cases, the question of whether there is a unique solution for all instants of time, (ie. a global unique solution) is presently open. It has been shown that there exists a unique global solution for the 2D plane-parallel N-S equations [5][6]. While at first glance, the NSEs appear as a compact set of PDEs, the fascination with these Partial Differential Equations is only increased by the fact that the nonlinearity of the ensuing expanded equations, appear to be connected with notions of highly chaotic turbulence and vorticity [7][8][9][10]. Since the announcement of the Millennium Problem, several results have attempted to comment on the existence and uniqueness of the NSEs. One particularly recent result by Kyritsis noted that there existed indications towards establishing a regularity of solutions regarding the Euler Equations and NSEs more generally; this utilized the conservation of particles [11].In the present work, to the best of the authors’ knowledge, the procedure revealed here has not been previously observed in the literature on the question of Incompressible N-S 3-D existence of unique global solutions, except for compressible flows in [12][13][14]. First, a cube in  with boundary conditions that generate a vortex is considered , and an attempt has been made to naturally reduce the 3-D incompressible NSEs to a one component decoupled velocity field solution under scale invariant transformations, with a separate 2-component velocity field solution. For the variable z- component, in particular, a form of solution is extracted in the analysis presented using the divergence form of Green’s identity, Ostogradsky’s theorem. The decomposition method using Geometric Algebra is used in conjunction with a result that a bound for  implies bounds for all derivatives.[10]

## 2. Model

The 3D incompressible unsteady Navier-Stokes Equations (NSEs) in Cartesian coordinates may be listed below in compactified form for the velocity field :

(1)

where ρ is constant density, μ is dynamic viscosity ,, are the body forces on the fluid. In some cases, it may be elected to reparametrize the components of the velocity vector, and pressure to ,coordinates xi and time t according to the following form utilizing the non-dimensional quantity d(assumed negative):

(2)

The continuity equation in Cartesian co-ordinates, is

(3)

### 2.1. Data

Eq.(1), together with Eq.(3) and using the initial condition of  such that encompass the NSEs along with an incompressible initial condition. Ensuring periodic boundary 44 conditions specified in [1] defined on a cube domain Ω in is referred to as the periodic BVP for the NSEs in . See Fig. 1 below for geometry of flow problem.

Figure 1. Vortex generation in a 2-D projection of cube.

## 3. Application

Using Eq.2 above, multiplying the first two components of scale invariant Eq.1 by Cartesian unit vectors respectively and adding modified equations within the set Eq.1 give the following equations, for the resulting composite vector

(4)

Multiplying Eq. 4 by δ3 and by and, the component of Eq.1 by δ3 and by ,(again using Eq.2), the addition of the resulting equations [12][13][14] recalling the product rule, produces a form as displayed below in Eq.5, where . The nonlinear inertial term when added to and factoring out gives, . Here is a dyadic.

(5)

Multiplying Eq.5 by and adding to it gives:

(6)

At this point the z component of the external force, Fz, is assumed to vanish, In this paper Eq.6 is solved, instead of Eq.5. For the ε–periodic solution, it is proposed that integration of divergence or curl of Eq.6 over an arbitrary small volume is equivalent to integration of divergence or curl of Eq.5 for the same volume. That is the extra term’s divergence or curl in Eq.6 when integrated is negligible on set of measure zero. A necessary condition that the form of Eq. 5, call it L1 = 0, is that both the divergence and curl of L1 be zero and upon integrating over a volume U, we introduce a function of t which we call Cs(t).The same is true for the non ε–periodic case where the justification of using Eq 6 instead of 5 will be based on the periodicity of the flow on an interval in . It can be seen that for the general non zero measure set, upon separately taking the divergence and curl of Eq. 6 and integrating over the volume U and using first Ostogradsky’s theorem and the fact that the velocities are periodic on the interval in , we have that the extra term has an integral of it’s divergence equal to zero. Proof is straightforward upon taking the divergence and integrating. Next for the curl of the same term and integrating over the same volume we use the fact that This contribution is also zero due to periodicity of velocities on interval.

Theorem 3.0.1. Reduced structure form of 3-D Navier-Stokes Equations

The 3-D Incompressible Navier Stokes equations can be reduced to a simple form as,

where is the nonlinear partial differential operator given by Eq.13, Q is the two dimensional Navier-Stokes operator acting on the vector defined in the brackets of Eq.17 and . The terms Ωi contained in are defined after Eq.10 and are part of the proof of present Theorem 3.0.1.

Proof. Taking the geometric product with the inertial vector term in the previous equation Eq.6 given by , it can be shown that in the context of Geometric Algebra [12][14], the following scalar and vector grade equations arise:

（7）

（8）

Taking the divergence of the vector equation Eq.8, recalling the product rule, and defining the new term (After taking divergence multiply new equation by H), results in an expression which may be combined with the usage of the scalar equation Eq.7 to produce:

（9）

Continuing with the previous paragraph we use the common term appearing in Eq.7,and in the new equation where we took divergence of Eq.8 and multiplied by H. This term in Eq.8 is . Upon a division of the preceding equation 9 by , it can be seen to result in the general form:

（10）

where for brevity, the following symbols have been defined:

It can be seen that the expression beginning with parentheses may be abbreviated into a nonlinear
vector operator Q and so Eq.10 can be written compactly as:

（11）

where is the non-linear operator associated with remaining part of Eq.10. Utilizing the continuity
Eq.3, it can be seen that the operator is given by the following expression:

（12）

The nonlinear operator form of the NSEs presented is:

（13）

This completes the proof.

An important observation is that in Eq.12,

（14）

Equation 13 displays a general form which may be expanded and analyzed by allowing a geometric assumption to be undergone, or the general case may also be considered.

## 4. Two Cases

### 4.1. The Geometric Case

As a special case, one may consider the case where:

（15）

This condition means that the Lie Product of the velocity inertia is entirely perpendicular to the Force terms, and thus refers to a vortex fluid scenario. This condition automatically implies = 0 and so,

（16）

### 4.2. The General Case

For Q, the expressions with and it’s derivative with respect to t factor out completely from the expression and the operator acting on the vector (that is Eq.4), remains which is identically zero, This means that may be assumed to be:

（17）

The expression in brackets in Eq.17, for, is the 2-D "plane-parallel" Navier-Stokes Equations and it is well known that if all data of problem are independent of one of x, y, z, then the BVP in Eqs.1, Eqs.3, has a unique solution for all instants of time with no restrictions on smallness of , or the domain Ω.[6].

Using Green’s identity, for divergence (see Appendix B), also known as Ostogradsky’s Theorem, it is known that for an arbitrary vector field and scalar field ψ, the following identity holds:

（18）

Eq.16 can be expressed due to integration on a general rectangular volume U as:

（19）

Note that the surface integral in Eq.18 is zero since it is taken over six faces of a general rectangular volume, and is a periodic vector field with normals pointing in opposite directions on opposite sides. Here which is assumed to be equal on the four vertical faces of cube and not equal between the top and bottom face. Also Ω1 is zero using Eq.18 and continuity Eq.3, Ω3 vanishes using Eq.18 as well due to periodic boundary conditions on the cube’s surface. The pressure is expressed using the divergence theorem as a surface integral over the surface. In addition Ω4 and Ω7 vanish using the divergence theorem and the fact that is periodic on the surface of U.

Lemma 1 (Maximum Principle For Eq.19). In this lemma we prove a maximum principle for Eq.19 in the general case where there are no restrictions on Φ and the most general Gagliardo [15]-Nirenberg [16] inequality (see also [17]) is used in this lemma and can be stated as follows, Let and , and either

then there exists a constant C independent of u such that ,

（20）

Proof. Using the z momentum equation again and multiplying by , for the case,

we obtain for j = 1, k = 2,

（21）

where in the last two lines of 21, the Gagliardo-Nirenberg [15],[16] inequality, Eq 20, has been used, Eq.2 has been used to express in terms of using the negative term δ, and for the equality (A)  is used and for . We abbreviate to the following expression involving the pressure and velocity terms,

Next it is easy to show using 14, 19 and 21 that,

（22）

（23）

(Note: General case of constant in space but not in time velocity at centre of cube is considered; Also the Poincare inequality has been used. We have used a result which states if the infinity norm is bounded then all the derivatives are bounded.)

The gradient of pressure is negative inorder to drive the flow, so that -. The determining factor is the integration function in t term , which is positive and leads to a blowup. To show this the remaining steps are to use the condition,

（24）

in Eq. 19. Hence it is shown that,

(25)

where B1 is the infimum of F(x, y, z) in the cube which is non-zero since there is a maximum negative flow in the z direction at the centre ( There exists a negative pressure gradient in z driving the flow downwards in cube with vortex in x, y directions ) and . Solving for in previous equation and differentiating with respect to z the form of the equation results in the following two ordinary differential equations,

（26）

（27）

where λ is a separation constant.

## References

1. C.L. Fefferman Existence and Smoothness of the Navier-Stokes Equation, Millennium Prize Problems, Clay.Math.Inst., Cambridge (MA), (2006) ,57-67.
2. J. Leray Sur le mouvement d’un liquide visqueux emplissant l’espace, J. Math. Pures Appl., 12, (1933), 1-82.
3. J. Leray Sur le mouvement d’un liquide visqueux emplissant l’espace, Acta Math., 63, (1934),193-248.
4. J. Leray Essai sur les mouvements plans d’un fluide visqueux que limitent des parois., J. Math. Pures Appl., 13, (1934), 331-418.
5. A.A. Kiselev and O.A. Ladyzhenskaya, On the existence and the uniqueness of the solution of the non-stationary problem for an incompressible visous fluid, Izv.Akad.Nauk.USSR, 21, (1957), 655-680.
6. O.A. Ladyzhenskaya, The Mathematical Theory of Viscous Incompressible Flow Gordon and Breach, New York, (1969)
7. P. Constantin, C. Foias, O.Manley and R. Temam, Connection between the mathematical theory of the Navier-Stokes equations and the conventional theory of turbulence, C. R. Acad. Sc. Paris, 297, (1983), 599-603.
8. P. Constantin, C. Foias, O. Manley and R. Temam, Determining modes and fractal dimension of turbulent flows, J. Fluid Mech, 150, (1985), 427 - 440.
9. P. Constantin, P. Lax and A. Majda, A simple one-dimensional model for the three-dimensional vorticity equation, Comm. Pure Appl. Math., 38, (1985), 715 - 724.
10. Jens Lorenz and Paulo R. Zingano, Properties at potential blow-up times for the incompressible Navier-Stokes equations, Bol. Soc. Paran. Mat., (3s.) v. 35 2 (2017): 127-158, doi:10.5269/bspm.v35i2.27508.
11. C. E. Kyritsis, On the 4th Clay millennium problem: Proof of the regularity of the solutions of the Euler and Navier-Stokes equations, based on the conservation of particles, J. Sci. Res. St., 4, (2017), 304-317.
12. T.E. Moschandreou, A Method of Solving Compressible Navier Stokes Equations in Cylindrical Coordinates Using Geometric Algebra,Mathematics (2019), 7(2), 126; https://doi.org/10.3390/math7020126 - 27 Jan 2019.
13. T.E. Moschandreou, K.C. Afas Compressible Navier-Stokes Equations in Cylindrical Passages and General Dynamics of Surfaces—(I)-Flow Structures and (II)-Analyzing Biomembranes under Static and Dynamic ConditionsMathematics (2019), 7(11), 1060; https://doi.org/10.3390/math7111060 - 05 Nov 2019.
14. T.E. Moschandreou A New Analytical Procedure to Solve Two Phase Flow in Tubes, Math. Comput. Appl. (2018), 23(2), 26; https://doi.org/10.3390/mca23020026 - 23 May 2018
15. E. Gagliardo Ulteriori proprietà di alcune classi di funzioni in più variabili,Ricerche Mat., (1959), 8:24–51.
16. L.Nirenberg On elliptic partial differential equations. Ann. Scuola Norm. Sup. Pisa (3), (1959), 13:115–162
17. A. Fiorenza,M.R. Formica, T. Roskovec and F. Soudský Detailed proof of classical Gagliardo-Nirenberg Interpolation inequality with historical remarks, (2018), arXiv:1812.04281v1 [math. FA]
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text:
In a box, there are 20 cards,
Question:

In a box, there are 20 cards, out of which 10 are labelled as $\mathrm{A}$ and the remaining 10 are labelled as $\mathrm{B}$. Cards are drawn at random, one after the other and with replacement, till a second A-card is obtained. The probability that the second A-card appears before the third B-card is :

1. (1) $\frac{9}{16}$

2. (2) $\frac{11}{16}$

3. (3) $\frac{13}{16}$

4. (4) $\frac{15}{16}$

Correct Option: , 2

Solution:

$P($ second $A$ – card appears before the third $B$ – card)

$=P(A A)+P(A B A)+P(B A A)+P(A B B A)+P(B B A A)$

$+P(B A B A)$

$=\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{11}{16}$


text:
# Only our own ruts

ENTER THE STATEMENT (70 CHARACTERS):
IS IT A DECLARATIVE SENTENCE (Y/N)?
N
ARE THERE ANY 'UNASSIGNED VALUES' (Y/N)?
N
A PREPOSITION IS A DECLARATIVE SENTENCE WITH NO UNASSIGNED VALUES.
YOU ENTERED:
WAS IT A DECLARATIVE SENTENCE? YOU ANSWERED N
DID IT HAVE ANY UNASSIGNED VALUES? YOU ANSWERED N
DO YOU THINK IT IS A PROPOSITION (Y/N)?
N
IT IS UP TO YOU TO DECIDE!

These proofs get me confused because they circle back and shoot down your assumption – which was the point, but it feels like a mental “frame shift” to return to it. One example from Rosen:

Show that at least four days of 22 days fall on the same day.

With proof by contradiction, you assume the opposite and follow that line of reasoning to a true implication. If you can show the conclusion is a contradiction, then it is false. For the implication to be true, !p must also be false. If !p is false, then p must be true.

Rosen uses the term “at most three” to express the negation of “at least four,” but I like “less than four.” (At least being greater than or equal to, and the negation being less than.)

After negating the premise, you show that it could be true for both the given condition and its opposite: so you can fit three of the same day within twenty-two days, but you can also fit them in twenty-one (three weeks’ worth) days. The not-p is true with both circumstances, even though !(twenty-two days) could be twenty-one days.

Then you have

~p -> F

For this implication to be true, !p must also be false. If !p is false, then p must be true.

I can parrot the explanation, but whether I will be able to apply it is another matter.

It’s not enough to just solve problems. Each problem is a litmus test of your personal framework, your process. Each successful answer validates the invention; each wrong answer is opportunity for feedback. It’s not explicit, anywhere, that the student’s thinking paradigm is the innovation produced out of procedure; it is simply accepted. If you do not have a framework, there is no sink to which labor strengthens; it is futile.

I almost want to say that moving forward is a failing if a framework hasn’t been forged through the crucible of getting things wrong: you’ll forget how to solve those problems, because your context will have staled. Then it’s a trip back to reference it, a dozen Google searches, and even then you’re not certain. I would think, “Man, I wish I had started a program back then.”

Honestly, who has time for that? Only priests focus on one book. (Well, maybe I am a priest.) The other alternative is to apply what I’ve learned, and that leads directly into waterfall design – imagine writing predicates and quantifiers for each part of a requirements spec!

I do have all the problems worked on looseleaf. That becomes my reference.

# Semantic pedantics

No one is perfect.
$\forall x \neg P(x)$
$\equiv \neg P(x_1) \wedge \neg P(x_2) \wedge \dots \wedge \neg P(x_n)$
$\equiv \neg \exists x P(x)$

Not everyone is perfect.
$\neg \forall x P(x)$
$\equiv \neg (P(x_1) \wedge P(x_2) \wedge \dots \wedge P(x_n))$

After DeMorgan’s Law
$\neg P(x_1) \vee \neg P(x_2) \vee \dots \vee \neg P(x_n) \equiv \exists x \neg P(x)$

And the answer key also has this:

Not everybody is your friend or someone is not perfect.
$(\forall x \neg F(x)) \vee (\exists x \neg P(x))$

So, “everybody” refers to “one” and “someone” is the same as “everyone.”

# The fear of insufficiency

I began the next section of discrete math. It covers predicates and quantifiers. These are the loops of the logic world, stretched blankets over exact universes. One quantification, the universal, implies a long conjunction: for every  , apply AND to each of them. It takes one counterexample to show a universal quantification false.

I met the propositional function P(x), which encapsulates the subject and predicate of a statement. Specifying the domain lets us assign values to the variable. Since we’re in logic land, each such assignment evaluates to a truth value, True or False.

Multiple variables create instances of tuples. Databases also use the term tuple. Could the row of a database be an instance of a long function P(a, b, c, …, z)? The binding of one or more of those variables specifying uniqueness, and each row mapped to an array with one less dimension: tables as 2D, mappings as collapse to 1-dimensional, linear collections.

But I speak of fear. One example had math: . Rosen broke it down to . Then he goes to say it’s true if and only if  or . How did he get there? I became worried that my math journey would abruptly cease at section three, chapter one.

Wolfram Alpha gave hints, but ultimately I made a table:

Universe: real numbers
x    x*(x-1)  x*(x-1)>=0
-2.0 6        T
-1.5 3.75     T
-1.0 2        T
-0.5 2.25     T
0    0        T
0.5  -0.25    F
1.0  0        T
1.5  0.25     T
2.0  2        T

Somewhere between 0 and 1, the values we assign to the x variable do not result in a true inequality. Another way to say “x <= 0 or x >=1” is . Can we believe that, though? What if there’s some number 0.000029482 that satisfies greater than or equal to zero? We can think through some assumptions:

x*(x-1) >= 0
A. if x = 0, x*(x-1) = 0
B. if x < 0,
x * (x - 1) >= 0
(x - 1) <= 0     | divide by x (-), switches sign
x <= 1           | add 1 to both sides
x >= 1           | multiply by -1 on both sides (x is -)

The inequality is true when x <= 0.

x*(x-1) >= 0
A. if x = 1, x*(x-1) = 0
B. if x > 1,
x * (x - 1) >= 0
(+) * (+) >= 0   | positive times positive is >= 0

The inequality is also true when x >= 1.

x*(x-1) >= 0
if 0 < x < 1,
x * (x - 1) >= 0
(+) * (-) >= 0 ? | x (less than 1) minus 1 is negative (so, False)

When x is between 0 and 1, the inequality is false. Going back to the universal quantifier, for every x, P(x), with the domain of all real numbers, is false.

I generalized the process thus:

• Gather “data”: make a table. play with numbers. fiddle around. You are the compiler.
• Discover the “rules”: 0 < x < 1
• Check yourself: if x <= 0, if x >= 1, if 0 < x < 1
• Conclude: agree with the author or disagree; move on

# Short dip into negative space

A compound proposition is made up of multiple propositions. If we can assign a truth value to each proposition that makes the compound proposition true, then the compound proposition is satisfiable. If every assignment leads to the compound proposition being true, then the compound proposition is a tautology. Tautologies are true “no matter what.”

p | p v ~p
T     T
F     T

The compound proposition is always true regardless of the assignment of truth values to p. The compound proposition p OR NOT p is a tautology.

An algorithm can be written that determines if a compound proposition is satisfiable. Such an algorithm could be used in a loop. For each assignment, continue until an assignment of propositions is unsatisfiable. If so, exit: that compound proposition is not a tautology. You would have to test every assignment.

Rosen hints at an alternative: try the negation of the compound proposition. If a compound proposition is a tautology, it will always be True, True, …, True for any assignment of Ts and Fs. Then its negation would always be False, False, …, False. However, with the negated version of the compound proposition, we only need to test one instance of satisfiability: if the NOT version is satisfiable once, the original compound proposition is not a tautology.

You can show a compound proposition is true every time, or that its negation is true once. If the negated version is True with that assignment, then the original version will evaluate to False. One False means the compound proposition is not always true, and you need “always True” to get a tautology.

When you are proving something, you could assume the opposite – then you only need to show one example. That means the original given can’t always be true. Would you prefer to demonstrate one counterexample or an exhaustive* listing?

* “Define exhaustive.”

# Speculative generator

A story being an aggregate of chunked words to bond various explanations, then it could show us the progression of a thought process where each compound proposition is satisfied by one combination of truth values.

We plug into this “world engine” certain facts [1], and it creates random compound propositions that are all true for some combination of True and False for each assignment of p, q, r, s, etc.

Our work consists of kicking out each false in some way, changing our preconceptions in each chapter, assuming a different role, until things begin to resolve toward the One Path in which every compound proposition is True.

p q r s C1 C2 C3 C4 C5 ...
T T T T F
T T T F T  F
T T F T T  T  F
T T F F T  T  T  F
T F T T T  T  T  T  F
T F T F T  T  T  T  T
...

It’s a harebrained idea, attempting to mechanize what intuition and skill melds in a writer to weave magic. Even if the method is systematic, all the reader cares about is the “user experience.” For books, that’s great writing and consistent verisimilitude. People can tell artificiality without training.

But as a check against our gut in the middle of a complex narrative, it may have some use. If I bang one out before NaNoWriMo, that would be a nice laboratory for its efficacy.

Inspired by exercise 53 of Rosen’s 5e “Discrete Mathematics and Its Applications.” In the picture below, we see several ways that every compound proposition is satisfied. Ideally, there would only be one set of assignments to p, q, r, and s that made every compound proposition True.

The direct solution is p AND NOT q AND NOT r AND NOT s, where p is True and q, r, and s are all False. We call that Author’s Intent.

[1] Cultural perspective or a moral, objective one – the viewpoint of the Main Character?

# The not-not is

Given an implication, we have its converse, contrapositive and inverse:

p -> q
q -> p
~q -> ~p (contrapositive)
~p -> ~q (inverse)

The original implication is logically equivalent to its contrapositive. The converse and inverse are logically equivalent to each other.

p -> q
~p v q, because p -> q = ~p v q
q v ~p, because p v q = q v p
~(~q) v ~p, because ~(~p) = p
~q -> ~p, because p -> q = ~p v q

So we see that an implication and its contrapositive are logically equivalent.

Given p -> q,
q -> p
~q v p, because p -> q = ~p v q
p v ~q, because p v q = q v p
~(~p) v ~q, because ~(~p) = p
~p -> ~q, because p -> q = ~p v q

And this shows the converse and inverse of an implication are logically equivalent.

I ordered a raft of books, old books, costing about a cent each, whose value outweighs the ship rate of which I paid the majority of the total. Books from shops into the hands of a collector, a private math where communion competes with pragmatics.

# Truth table combinatorics

For every n propositions, you have  possibilities. That adds up: given three characters and two propositions for each, you get six propositions – and 2, 4, 8, 16, 32, 64 combinations. Span your cast to seven or eight simple, archetypal characters as per Dramatica, and it’s 128 combinations!

That’s only theoretic: some combinations are too outlandish to work, and some progressions don’t make sense. At the climax, a character may flip from T to F or from F to T, but flip-flopping would be harder to track. Few readers scribble out a truth table for a one-off, I think.

Using the sequences of a truth table to thread your work can give you a roadmap for developing each character. How does the reader perceive this character to be “True” or “False” in regards to a proposition? For example, “The prince is honorable.” It can appear true in the beginning, but his actions may sway the reader to believe otherwise. Redemption is ripe after a steady string of False, only to turn True at the end.

You’re asking yourself, “How do I show this? How do I show that? How would this show that? Is that good enough? Would I believe it?”

Things can be further complicated by combining logic with constraints: societal constraints, traditions, law and culture. Take your reader on a journey into a rich world – a world that is not only punchy, but internally consistent.

This was also inspired by Bioshock. A friend said Spector’s genius lay in leading the gamer by the nose – getting him to think one thing until the very end, when everything completely shifts. Perception. Prejudice. Preservation.

# All the truth in the world

From Discrete Mathematics and Its Applications, 5e, Rosen:

A detective has interviewed four witnesses to a crime. [Propositions follow.]

• B: The butler is telling the truth.
• C: The cook is telling the truth.
• G: The gardener is telling the truth.
• H: The handyman is telling the truth.

From the problem statement, I get the following compound propositions:

B -> C
C + G
~(~G & ~H)
H -> ~C

Here’s the truth table:

B  C  G  H   B->C  C+G  H->~C  ~(~G&~H)
-  -  -  -    -     -     -        -
T  T  T  T    T     F
T  T  T  F    T     F
T  T  F  T    T     T     F
T  T  F  F    T     T     T        F
T  F  T  T    F
T  F  T  F    F
T  F  F  T    F
T  F  F  F    F
F  T  T  T    T     F
F  T  T  F    T     F
F  T  F  T    T     T     F
F  T  F  F    T     T     T        F
F  F  T  T    T     T     T        T
F  F  T  F    T     T     T        T
F  F  F  T    T     F
F  F  F  F    T     F

There are two cases where all propositions together cause consistency: when G is true or when G and H are true. Assuming there was only one criminal, we could point out the gardener and close the case. But the question asks

For each of the four witnesses, can the detective determine whether that person is telling the truth or lying?

What? I looked in the back for the answer. The answer is that we can determine that the butler and cook are lying, but we cannot determine if the gardener or the handyman are telling the truth.

That makes sense: a criminal wouldn’t implicate himself, and presumably the truth-tellers wouldn’t take the fall. Somewhere in here are the seeds of plot and human nature, and why stories can still be the taut battleground of drama.