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I start with \$1. After one iteration of a game, one of the following $m$ outcomes occurs:

  • With probability $p_1$, my wealth multiplies by $r_1$;
  • With probability $p_2$, my wealth multiplies by $r_2$;
  • ...
  • With probability $p_m$, my wealth multiplies by $r_m$.

Assume $r_i > 0$ for all $i$ and $\sum_{i = 1}^{m}{p_i} = 1$.

Let $W_N$ be a random variable that denotes my wealth after $N$ iterations of the game.

If $N$ is large, what will be the median of $W_N$?

I am reading a book (Safe Haven by Mark Spitznagel) that claims that for large $N$ the median will converge to $$M(W_N) = (\prod_{i=1}^m {r_i}^{p_i} )^N$$

Unfortunately he skips all the math and just mentions this is true "under fairly robust statistical convergence". I'm a bit lost and would like to find a more formal proof.

Can you help?

Thanks!

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    $\begingroup$ Have you tried applying the Central limit Theorem to $\log {W_n}$? $\endgroup$ Commented Oct 26 at 22:30
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    $\begingroup$ @Eric The CLT does not actually demonstrate the result. It only demonstrates that the result is plausible. $\endgroup$ Commented Oct 26 at 23:22
  • $\begingroup$ For a binomial lattice it would work as the median path, but since the median is discrete is wouldn't match the median path on many trials, see this $\endgroup$ Commented Oct 27 at 0:56
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    $\begingroup$ I have a problem with the "converge" term. Stating that something converges for big $N$, lets me think that that the value it converges to can not depend on $N$: I expect to converge to a number (or to a function of some parameters, but definitely not on $N$). I might be wrong, but I see "can be approximated" as a better description. Furthermore, that median can easily grow without bound as $N$ increases, so I don't like the "converge" term at all. $\endgroup$ Commented Oct 27 at 7:01
  • $\begingroup$ @nicola I would usually assume in this situation that the precise statement intended is "is asymptotic to" (but per the answer, that's not true). $\endgroup$ Commented Oct 27 at 19:07

1 Answer 1

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The precise statement as made is false. But if you insert appropriate hand waving, it becomes morally true.

The key idea, as a comment gave, is to look at the $\log$ of your net worth. This turns repeated multiplication into repeated addition. And we have very strong theorems about repeated addition.

In particular, let $X$ be the random factor that your wealth is multiplied by each time, and $X_n$ be its actual value in the $n$th step. Then, $$ \mathbb E(\log(X_n)) = \mathbb E(\log(X)) = \sum\limits_{i=1}^{m} p_i \log(r_i) $$ And now $$ \begin{align} \mathbb E(\log(W_N)) &= \sum\limits_{n=1}^{N} \mathbb E(\log(X_n)) \\ &= \sum\limits_{n=1}^{N} \sum\limits_{i=1}^{m} p_i \log(r_i) \\ &= N \sum\limits_{i=1}^{m} p_i \log(r_i) \\ &= \sum\limits_{i=1}^{m} \log(r_i^{Np_i}) \\ &= \log(\prod\limits_{i=1}^m r_i^{Np_i}) \\ &= \log((\prod\limits_{i=1}^m r_i^{p_i})^N) \end{align} $$ The strong law of large numbers implies that as $N$ goes to $\infty$, $\frac{\log(W_N)}{N}$ goes to $\mathbb E(\log(X_n))$.

The central limit theorem talks about how a random sample spreads out. In particular it says that the distribution of $\frac{\log(W_N) - N \mathbb E(X)}{\sqrt{N}}$ is approximately normally distributed with a mean of $0$ and variance $\text{Var}(X)$. This distribution has, of course, a median of $0$.

But the question is how approximately this approximates. So let's pull out a counterexample.

Let's let $p_1 = 1 - \sqrt{0.5}$, $r_1 = 1$, $p_2 = \sqrt{0.5}$, and $r_2 = 4$. Then $\mathbb E(\log(X)) = \sqrt{2}\log(2)$. Also $W_N$ must also be a power of 4.

Obviously, $\sqrt{2}$ is an irrational. For any $N$, so must $N\sqrt{2}$ be. It's a standard fact that for any irrational $\alpha$, the fractional parts of $N\alpha$ are dense in $(0, 1)$. Pick $N$ with that fractional part close to $0.5$. Then $2N\sqrt{2}$ is close to an odd integer. And now whatever the median is, it must be around $\log{2}$ away from the expected value.

This provides a concrete distribution where for large $N$, $(\prod\limits_{i=1}^m r_i^{p_i})^N$ may be off from the median by a factor close to $2$. Therefore the statement as written is false.

But various references, for example Series Approximation Methods in Statistics, prove with Edgeworth series that the error in the central limit theorem is at most $O(\frac{1}{\sqrt{n}})$. Which corresponds to the median value of $\frac{\log(W_N) - N \mathbb E(X)}{\sqrt{N}}$ being off from the expected median by $O(\frac{1}{\sqrt{n}})$. This corresponds to the median value of $\log(W_N)$ being off by at most $O(1)$. Which means that the median value of $W_N$ is off by at most a constant factor. (In the example above, that factor was 2.)

So the stated result is almost true - for a given distribution and large $N$, it is off by at most a constant factor.

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    $\begingroup$ A simpler counterexample is just flipping a coin and either halving or doubling. Then, when $n$ is odd, $W_n$ is either $\leq 1/2$ or $\geq 2$, so the median will forever be far from $1$ for odd $n$. $\endgroup$ Commented Oct 27 at 1:24
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    $\begingroup$ @Eric That's the one I thought of first, but many people define the median of an even number of things as the average of the middle 2. Which allows the median to land at 1 again. So I had to look for an example where the median couldn't be split that way. $\endgroup$ Commented Oct 27 at 1:33
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    $\begingroup$ Ehh, but then it’s still $(0.5+2)/2=1.25\neq 1$. If you really want the geometric mean to also be $\neq 1$, you could do $1/3$ of the time $1/4$ and $2/3$ times $2$ which when $n$ is $1 \mod 3$ give either $\leq 1/4$ or $\geq 2$ $\endgroup$ Commented Oct 27 at 1:59
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    $\begingroup$ @Eric I hadn't thought of that, because my brain was thinking in logs.. I went to irrational probabilities first. Regardless of how you get there, discretization errors can be $O(1)$, $\endgroup$ Commented Oct 27 at 2:18
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    $\begingroup$ Amazing! Thank you very much for this detailed answer. This was extremely helpful and fun! $\endgroup$ Commented Oct 27 at 20:43

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