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?

If correct, $S_n \to 0$ in $L^1$, but one doesn't even have $S_n \to 0$ in probability. Indeed, $Y_n = \log S_n = \log X_1+\dots+\log X_n$ And $E e^{Y_n} = E S_n = (7/6)^n \to \infty \neq 1$ — Andrew, Nov 22 '23 at 18:54
@Andrew, mmmm, I think I can prove that $S_n\to 0$. Let me add that part too. — Hosein Rahnama, Nov 22 '23 at 18:57
I just proved that $S_n$ doesn't converge to anything. It doesn't even converge weakly to anything as the moment generating function of $\log S_n$ diverges — Andrew, Nov 22 '23 at 18:58
@Andrew: What you proved is indeed the limit of expectation and not the expectation of the limit. :) — Hosein Rahnama, Nov 22 '23 at 19:00
No, I showed the limit doesn't even exist. So there is nothing to even take a limit of — Andrew, Nov 22 '23 at 19:00
Hmm, yes, I think you are right. In that case, there is nothing to show. You have already identified $\lim S_n = 0$, so $E[\lim S_n] = E[0] = 0$ — Andrew, Nov 22 '23 at 19:29
@Andrew: Hmmm, I couldn't convince myself for the last part you said. I mean we don't really have $\lim_{n\to\infty}S_n = 0$. So, this last part seems buggy to me! :) — Hosein Rahnama, Nov 22 '23 at 19:32

Hosein Rahnama · Accepted Answer · 2023-11-23T09:03:41.190

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

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

edited Nov 23 '23 at 09:03

answered Nov 22 '23 at 19:33

Hosein Rahnama

14,882

@Andrew: Can you elaborate on the last part you said. – Hosein Rahnama Nov 22 '23 at 19:44
1

Define $Y(\omega) = \lim S_n(\omega)$ and arbitrarily otherwise. Using your notation, $E[Y]=E[0\mathbf 1_B]+E[Y\mathbf 1_{B^c}]=0$, by the measure theory convention that integrating anything over a measure zero set is zero – Andrew Nov 22 '23 at 20:21
@Andrew: So, we don't care about the values $\lim_{n\to\infty}S_n(\omega)$ for $\omega \notin B$? Is there any guarantee that these values are bounded above? – Hosein Rahnama Nov 22 '23 at 21:05
There are of course, no guarantees that $S_n$ is bounded on $B^c$. Just take $\omega$ with heads, all but finitely often. On the other hand, it is a measure theory convention that $\infty \cdot 0 = 0$. Namely, $E[Y\mathbf 1_{B^c}] \leq \infty \cdot P(B^c) = 0$ – Andrew Nov 22 '23 at 21:18
@Andrew: Thanks. :) I am not much familiar with measure theory. Where can I study about this convention and property you just mentioned? Is there any specific keyword for this? – Hosein Rahnama Nov 22 '23 at 21:36
1

Any book on analysis or probability theory. Personally, I learned from Durrett’s book PTE. A more detailed book is the one by Jacod and Protters. – Andrew Nov 22 '23 at 21:41

The Money Left in an Infinite Gambling Game

1 Answers1

Linked