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I enter a casino with $X_0 = 1$ dollar. I then keep playing a game: With probability $\frac{1}{3}$, my money triples. With probability $\frac{2}{3}$, my money decreases by a factor of 3. Let $X_n$ be how many dollars I have after playing this game $n$ times. Hence, $X_{n+1} = 3X_n$ with probability $\frac{1}{3}$, and $X_{n+1} = \frac{X_n}{3}$ with probability $\frac{2}{3}$. Why does $X_n$ converge to 0 as $n$ approaches $\infty$?

I got that $X_n$ converges to $\infty$ (or does not converge):

$ E(X_n) = \frac{1}{3}(3)E(X_{n-1}) + \frac{2}{3}\frac{1}{3}E(X_{n-1}) \\ E(X_n) = \frac{11}{9}E(X_{n-1}) $

And since $\frac{11}{9} > 1$, shouldn't $E(X_n)$ diverge as $n$ approaches $\infty$?

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While $E(X_n)=(\frac{11}9)^n\to \infty$, we have $P(X_n<\epsilon)\to 1$ for any given $\epsilon>0$. Thus the limit$X:=\lim_{n\to\infty} X_n$ is a random variable with $P(X=0)=1$ and hence $E(X)=0$.

This is an example where $E(\lim X_n)\ne \lim E(X_n)$.

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    Note that with random variables, we usualy talk about weak convergence. For convergence against $X$ with $P(X=0)=1$ it is sufficient that \lim_{n\to\infty}P(X_n<\epsilon)=1 and \lim_{n\to\infty}P(X_n<-\epsilon)=0. These facts (especially the former) should become obvious if you note that $\log_3 X_n$ follows a $B(n,\frac13)$ binomial distribution with $-\frac n3$.2012-12-10
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Introduce the gain random variable $X_n$ which satisfies $X_0=1$ and $X_{n+1} = Y_n X_n$ where $Y_n$ is new random process of the independent draws which is $3$ with probability $1/3$ or $1/3$ with probability $2/3$. Then arguing by recursion you can see that $ \log X_{n+1} = \sum_{i=1}^n \log Y_n. $ Calculate the expectation of the $ \log Y_n$: it is $-\frac{1}{3} \log 3$. Then use the law of large numbers to see that $1/n \log(X_{n+1})$ converges to the expectation of $\log Y_n$ (in probability). This shows that $\log X_{n+1}$ converges in probability to $-\infty$. Taking exponentials give the result.

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    Something seems to have gone wrong with your tex...2012-12-09