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Let $X_n$ be a (random) fortune of a gambler after the $n$th round of gambling, which evolves as follows. After the $n$th round, his fortune $X_n$ depends on the fortune $X_{n-1}$ from the prior round as follows:

$X_n$= $Z_n$ $X_{n-1}$ for $n=1,2,\ldots$

where $X_0 = 1$ is the gambler's initial investment. The variables $Z_n$ ($n\geq1$), are independent identically distributed discrete random variables with pmf given by

$$p(z) = \begin{cases} 1/3 & \text{ if } z=3 \\ 2/3 & \text{ if } z=1/3 \\ 0 & \text{ o.w.} \end{cases}$$

Q)What is the expected fortune $\mathbb{E}[X_n]$ for arbitrary $n$?

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    A interesting supplementary question is finding $\Pr(X_n \ge X_0)$ or at least an approximation to it. The two answers together are counter-intuitive, especially for large $n$ and they teach a valuable lesson for gamblers and financial investors. – 2017-02-09
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    So for example you have $E[X_{1000}] \gt 10^{87}$ but $\Pr(X_{1000} \ge 1) \lt 10^{-26}$ – 2017-02-09
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    @henry good point. This is an example of a so called "gambler's ruin" problem. Sometimes, that counter intuitive result is impresissely stated as "Persistently taking beneficial chances is never beneficial at the end." Meaning, in a case such as this, if you played forever, you'd eventually go broke. – 2017-02-10
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    @David: one answer is not to bet/invest all your money all the time. For example, bet/invest a twelfth of your net wealth each time, and $E[X_{1000}] \gt 90 000 000$ while $\Pr(X_{1000} \ge 1) \gt 0.99999$, which is much safer if not so potentially profitable – 2017-02-10
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    @Henry yes, good point, but this is assuming you can find somebody with infinite wealth to keep making bad bets for eternity! I've yet to find that mark, but I'll keep looking :) – 2017-02-10

2 Answers 2

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Have you tried looking computing the values of $\mathbb{E}(X_n)$ for small values of $n$? For example, $$\mathbb{E}(X_1) = \mathbb{E}(Z_1X_0),$$ by independence of $Z_1$, we have $$\mathbb{E}(Z_1X_0) = \mathbb{E}(Z_1)\mathbb{E}(X_0) = \left(3\cdot\frac{1}{3} + \frac{1}{3}\cdot \frac{2}{3}\right) \cdot 0 = \frac{11}{9}.$$

Similarly, $$\mathbb{E}(X_2) = \mathbb{E}(Z_2X_1) = \mathbb{E}(Z_2)\mathbb{E}(X_1) = \frac{11}{9} \cdot \frac{11}{9} = \left(\frac{11}{9}\right)^2.$$

In general, $$\mathbb{E}(X_n) = \left(\frac{11}{9}\right)^n.$$ As a very simple exercise you could try proving this by induction. Since I have already demonstrated the base case, you would simply need to show that $$\mathbb{E}(X_n) = \left(\frac{11}{9}\right)^n \Rightarrow \mathbb{E}(X_{n+1}) = \left(\frac{11}{9}\right)^{n+1}$$ to complete the proof. Alternatively, you could show (again by very simply induction) that $$\displaystyle X_n = \prod_{i=1}^nZ_i.$$ Then it would follow by independence that $$\mathbb{E}(X_n) = \mathbb{E}\left(\prod_{i=1}^nZ_i\right) = \prod_{i=1}^n\mathbb{E}(Z_i).$$

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Note that $$ X_n=Z_1\cdot Z_2\cdot\ldots\cdot Z_n. $$ Since $Z_i$ are independent $\mathbf{E}X_n=(\mathbf{E}Z_i)^n=\left(1\frac{2}{9}\right)^n.$