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Background. We can play a game in which we can put one dollar and get out $X$ dollars, where $X$ is 2 dollars with probability $p>1/2$, or zero dollars with probability $1-p$. We also assume that different plays of this game are independent. Note that $E[X] > 1$, meaning that we have an advantage.

We begin with one dollar and start playing this game, such that we play with a fraction $f\in[0,1]$ of our current fortune each time we play.

Thus—and here we let $X,X_1,X_2,\ldots$ denote (independent) random variables having the distribution of our plays as described above—our initial fortune is $F_0 = 1$, our fortune after 1 play is $F_1 = 1-f+fX_1 = 1+f(X_1-1)$, and in general, it is easily calculated that our fortune after $n$ plays is $$F_n = \prod_{i=1}^n(1+f(X_i-1)).$$

Since the plays are independent, our expected fortune after $n$ plays is $$E[F_n] = E\bigg[ \prod_{i=1}^n(1+f(X_i-1)) \bigg] = (1+f(E[X]-1))^n.$$ Since $E[X] > 1$, it is clear that $E[F_n] \to \infty$ no matter what $f$ we choose.

However, for any given $n$ we see that $E[F_n]$ is largest for $f=1$. Does this mean we should choose $f=1$? This is probably a bad idea, because in this case, the probability of going broke after $n$ or fewer plays is $1-p^n$, which approaches one as $n\to\infty$. In other words, if we choose $f=1$, we will eventually go broke for sure. (For a further explanation of this, feel free to refer to my previous question, Resolving a paradox concerning an expected value.)

So what strategy should we take to maximize our long-term performance? Some authors have suggested the approach of maximizing the “exponential rate of growth.” Let me explain.

Imagine that $F_n = e^{nG_n}$. Here, $G_n$ is the “exponential rate of growth” of $F_n$. We can write $$G_n = \frac{1}{n}\log(F_n) = \frac{1}{n}\log\bigg( \prod_{i=1}^n(1+f(X_i-1)) \bigg) = \frac{1}{n} \sum_{i=1}^n \log(1+f(X_i-1)).$$ Using the law of large numbers, we find that $$G_n \to E[\log(1+f(X-1))] \quad \text{almost surely}.$$ Now, $G := E[\log(1+f(X-1))]$ is what some authors refer to as the “exponential rate of growth” that we should aim at maximizing. Indeed, $G$ seems to be the “exponential rate of growth” at the “infinity,” in some sense.

In our present case, we find that $$G = \log(1+f)p + \log(1-f)(1-p),$$ which—interestingly—is maximized at $f=2p-1$. According to this, we have the largest exponential rate of growth if we choose $f=2p-1$, and this is the value for $f$ at which some authors suggest gamblers place their bets.

Question 1. Why should I care about this? Can you give me some intuitive explanation of why my goal should be to maximize the exponential rate of growth, as opposed to maximizing $E[F_n]$ (i.e. to choose $f=1$), or choosing $f=0.99$ to keep $E[F_n]$ high while avoiding that almost sure bankruptcy? I don’t have a natural feel for what to do and why, in order to get as much growth as possible in the long run.

Question 2. How can it be possible that the exponential rate of growth is maximized at $f=2p-1$, yet $E[F_n]$ is always the highest for $f=1$? That doesn’t make much sense to me. Wouldn’t you expect $E[F_n]$ to eventually become the highest at $f=2p-1$? In other words, how can I we—in the “infinity”—have the greatest growth at $f=2p-1$ but the greatest expectation value at $f=1$?

Question 3. Here is a graph that shows $G$ as a function of $f$ for $p=2/3$. asdf

The maximum is indeed at $2p-1=1/3$. However, something interesting happens around $f\approx 0.618$, when the curve goes below zero. Supposedly, this is the point beyond which we are almost sure to end up with a loss, i.e. end up with less than 1 dollar. Do you understand why this is? Furthermore, it still remains a mystery to me why the curve for $E[F_n]$ wouldn’t show similarly interesting features, instead of being a monotonically increasing function of $f$.

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Nice question -- +1 for the elaborate presentation -- I'll have a go at an answer shortly... –  joriki Dec 12 '11 at 22:07
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On your question 3: No, this isn't the point beyond which we're almost sure to go broke. In fact, as long as $f\ne1$, we're certain never to go broke; it's only when $f=1$ that we almost surely go broke. –  joriki Dec 13 '11 at 2:04
    
@joriki, you’re right. Sorry about that. I have fixed my question accordingly. –  Topology Dec 13 '11 at 13:54
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The quick way to understand why G(f) is eventually decreasing in f is: you are reinvesting the profits of your strategy, which means all of your growth is multiplicative. This has an interesting consequence - a 10% loss followed by a 10% gain actually leaves you out of pocket (because $(1-x)(1+x)=1-x^2 < 1$). The larger $x$ is, the more you are out of pocket. In fact, if $x$ is sufficiently large then one loss is enough to wipe you out - this is why increasing your leverage actually decreases your growth after a certain point. –  Chris Taylor Dec 13 '11 at 19:18
    
@Chris: This is interesting, but only part of the story. I could just as plausibly argue: Because it's multiplicative, you have $(1+x)^2=1+2x+x^2$ if you win twice and $(1−x)^2=1−2x+x^2$ if you lose twice, so even if the $2x$ averages out you still always have the $x^2$, so the larger $x$ is, the better. Neither argument is correct; in fact the $x^2$ terms cancel overall and the growth rate of the expected amount of money is positive iff $p\gt1/2$. It's only by taking logarithms (i.e. calculating a logarithmic utility) before taking expectation values that we get a different answer. –  joriki Dec 14 '11 at 5:39

1 Answer 1

up vote 3 down vote accepted

First, there is some imprecision to clear up. You introduce $X$ as the amount of dollars received. This takes different values with certain probabilities, so it's a random variable. Then you calculate $E[X]$ and the fortune $1+f(X−1)$ after the first play. So far so good.

But then you write $F_n = [1+f(X-1)]^n$. Here you're describing the result of $n$ plays, but using only a single random variable $X$. Since the amount of dollars received may differ from play to play, you need a separate random variable $X_n$ for each play. In each play, your fortune is multiplied by the factor $1+f(X_n-1)$, so your fortune after $n$ plays is

$$F_n=\prod_{k=1}^n\left(1+f(X_n-1)\right)\;.$$

The reason this doesn't invalidate your entire argument is that because each $X_n$ occurs linearly in $F_n$ and they all have the same expectation value $E[X]$, the expectation value does come out to

$$E[F_n]=\left(1+f(E[X]-1)\right)^n\tag 1$$

as you wrote. Incidentally, it wouldn't have come out to this if your $F_n = [1+f(X-1)]^n$ had been correct, since this is non-linear and you can't commute forming the expectation with non-linear operations like you did to arrive at $(1)$.

Now, given that your expectation value was correct, your question about how to choose $f$ remains valid. Your idea for solving this is also interesting and relevant, but it seems to me that by phrasing it in terms of an "exponential growth rate", you're missing what I would regard as the point of it.

This also has to do with the non-commutativity of forming the expectation and non-linear operations. In this case, you're forming the expectation after taking the logarithm. Let's set this up in a slightly different way to throw some light on how this might throw some light on the "paradox".

Money isn't utility. The utility of money diminishes the more money you have. Certainly if you have $\$1,000$ and you get another $\$1,000$, you'll be more happy and have more important needs and wishes to fulfill with this money than if you already had $\$1,000,000$. A reasonable simple model of this might be that the utility of money is roughly logarithmic.

Assuming this model, given that the relationship between money and utility is non-linear, we need to decide whether to form the expected value of the money or of the utility. It doesn't make much sense to first form the expected value of the money and then take its utility, since the expected value isn't an amount of money you would ever have in any of the potential situations, but a linear combination of all these. What we need to do is first for each situation find the utility of the money you'd have in that situation, and then form the expected value of all these utilities.

This is where your calculation of the expected value of $G$ comes in. We have

$$ \begin{eqnarray} U_n &=& \log F_n \\ &=& \log\prod_{k=1}^n\left(1+f(X_n-1)\right) \\ &=& \sum_{k=1}^n\log\left(1+f(X_n-1)\right)\;, \\ \end{eqnarray} $$

and thus

$$ \begin{eqnarray} E[U_n] &=& E\left[\sum_{k=1}^n\log\left(1+f(X_n-1)\right)\right] \\ &=& \sum_{k=1}^nE\left[\log\left(1+f(X_n-1)\right)\right] \\ &=& nE[G]\;. \end{eqnarray} $$

Thus, with each play our expected logarithmic utility increases by $E[G]$, so in this model that's what we should maximize.

You might want to check out the links for the St. Petersburg paradox in the comments under your previous question, Resolving a paradox concerning an expected value (which, by the way, it would have been a good idea to link to in this question so that people can build on what's written there.)

[Edit in response to Question 3:]

You ask why it is that we're almost sure to end up with a loss if $E[G]$ is negative. This has to do with the fact that $E[U_n]$ is the $n$-th multiple of $E[G]$, whereas $E[F_n]$ is the $n$-th power of something. According to the central limit theorem, the sum of independent and identically distributed random variables converges (under suitable circumstances in a suitable sense) to a normal distribution. Since the logarithm of a product is a sum, a product correspondingly converges to a log-normal distribution.

The difference in the behaviour of these two distributions is crucial. The normal distribution is symmetric, its mean is proportional to $n$ and its standard deviation is proportional to $\sqrt n$. As a result, unless the mean of the individual distributions is $0$, we eventually almost surely end up on the same side of $0$ as the expected value of the sum.

The log-normal distribution, on the other hand, is highly asymmetric. You can see in this diagram in the Wikipedia article that the mean can be much higher than the median and the mode. As a result, there's no contradiction in the expected value of the product increasing without bound while the area under the curve above $1$ goes to zero.

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Thanks for the quick reply. Oh, right, you're right about my mistake of using the same random variable $X$ for all plays. I will try to fix my question accordingly soon. I also added a "question 3" to my question. In the meantime, I'll try to grasp your current answer. –  Topology Dec 12 '11 at 23:22
    
I have now substantially improved my question based on your comments. I will think about your “utility” thoughts. Sounds interesting… –  Topology Dec 13 '11 at 0:07
    
I’m marking your response as an answer. Thanks for your insights. However, I should probably mention for the record that I found the theorems in Thorp’s article “Optimal Gambling Systems for Favorable Games” to provide all of my answers in the greatest possible detail. Of course, it would be silly of me to expect someone to regurgitate, or even re-invent, those theorems here. Thanks again! –  Topology Dec 14 '11 at 2:25

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