The Money Left in an Infinite Gambling Game

Question

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. Suppose that $\Omega$ is defined as below \begin{equation*} \Omega = \Big\{ \omega = (\omega_1, \omega_2, \dots) \big|\, \omega_i \in \{H, T\}, i \in \mathbb{N} \Big\} = \{H, T\}^{\infty}, \end{equation*}

Let $\mathcal{F}$ be the set of all subsets of $\Omega$ which forms a $\sigma$-algebra and $\mathbb{P}$ to be a probability distribution defined over $\mathcal{F}$. Suppose $X$ is a random variable such that $X(H)=2$ and $X(T)=\frac{1}{3}$ with uniform distribution over $\mathcal{X}=\{2, \frac{1}{3}\}$. Consider a sequence of random variables $X_i$ which are independent and identically distributed with $X$. Define $S_n(\omega) = X_1(\omega_1)\dots X_n(\omega_n)$. This can be interpreted as an infinite gambling game where at each step you toss a fair coin with heads doubling your money and tails costing you $\frac{2}{3}$ of your money. So, $S_n$ is a fraction of your intial money at step $n$.

It seems that finally you loose all your money! I want to prove that

$$\mathbb{E}(\lim_{n\to\infty}S_n) =0.$$

I am new to limit theorems in probability so I am not sure how should go about this. Intuitively, this seems right as for a sequence of length $2n$ for large $n$ we have $S_{2n}(\omega) \approx \Big(\frac{2}{3}\Big)^n$.

Here is my effort. By the strong law of large numbers we have

$$\mathbb{P}\Big(\lim_{n\to\infty} \log_2 S_n^{\frac{1}{n}} = \mathbb{E}(\log_2 X)\Big) = 1.$$

Using the continuity of logarithm, we can conclude

$$\mathbb{P}\Big(\lim_{n\to\infty} S_n^{\frac{1}{n}} = 2^{\mathbb{E}(\log_2 X)}\Big) = 1.$$

Furthermore, we have

\begin{equation*} \mathbb{E}(\log_2 X) = \sum_{x \in \mathcal{X}} p(x) \log_2 x = \frac{1}{2} \log_2\frac{1}{3} + \frac{1}{2}\log_2 2 = \frac{1}{2} \log_2\frac{2}{3} = \log_2\sqrt{\frac{2}{3}} \approx -0{.}2925 < 0. \end{equation*}

So the above equation reads as

$$\mathbb{P}\Big(\lim_{n\to\infty} S_n^{\frac{1}{n}} = \sqrt{\frac{2}{3}} \Big) = 1.$$

How should I go further?

Answer 1

Define

$$ A = \Big\{ \omega = (\omega_1, \omega_2, \dots) \in \Omega \big|\, \lim_{n\to\infty} S_n(\omega)^{\frac{1}{n}} = \sqrt{\frac{2}{3}} \Big\}. $$

Let $\omega \in A$. Choose any $\epsilon$ such that $\epsilon < 1 - \sqrt{\frac{2}{3}}$. Using the definition of limit and noting $S_n(\omega) \ge 0$ gives

$$0 \leq S_n(\omega) < \Big( \sqrt{\frac{2}{3}} + \epsilon \Big)^n,$$

for sufficiently large $n$. Letting $n\to\infty$, we conclude that $S_n(\omega)\to 0$. So, if we define that

\begin{equation*} B = \Big\{ \omega = (\omega_1, \omega_2, \dots) \in \Omega \big|\, \lim_{n\to\infty} S_n(\omega) = 0 \Big\}, \end{equation*}

then $\omega \in B$. This in turn implies that $A \subseteq B$ which leaves us with $1 = \mathbb{P}(A) \leq \mathbb{P}(B)$. Consequently, we have $\mathbb{P}(B) = 1$ which reads as

\begin{equation*} \mathbb{P}\Big( \lim_{n\to\infty} S_n = 0 \Big) = 1. \end{equation*}

It then follows by some "measure theoretic arguments" that

$$ \mathbb{E}\Big( \lim_{n\to\infty} S_n \Big) = 0. $$