A gambler's fortune converges 0 or 1? (Martingale convergence theorem)

Question

Let $ X_0 = 1 , X_n = \prod_{i=1}^n Y_i, \space (Y_n : nonnegative \space independent, \space E[Y_i]=1) \space \Rightarrow \space then, \space \{X_n\} \space is \space martingale. $

Knowing this, Now let's consider a gambler who bets a fraction q of his fortune $(0\lt q\lt1)$ with each toss of a fair coin$(p=q=1/2)$ starting with $1.

Let his fortune $ X_n \space after \space n \space tosses \prod_{i=1}^n (1 + \delta_i\mathbf q), \space where \space \delta_i = \pm1 \space $

And, My textbook says that

according to the following Martingale convergence theorem

Let $\{X_n\} \space be \space submartingle \space satisfying \space \space \sup E[|X_n|]\lt\infty\Rightarrow \space then, \space There \space exists \space a \space random \space variable \space X_\infty\space to \space which \space \{X_n\} \space converges \space with \space probability \space one, \space P\{\lim_{n\to\infty}X_n=X_\infty\}=1$

a gambler's fortune $X_n$, $\space \sup E[|X_n|] = E[X_n]=1$, so that $\space X_n$ tends to a finite limit as $n\rightarrow\infty$, which must be ZERO, since every other state is transient.

I really don't get it at all. I think it's 1(one dolloar) because it's martingale and $E[X_n]=1$ . Why ZERO ?

Answer 1

Just because $EX_n=1$ and $X_n \to X$ with probability $1$ you cannot say that $X=1$.

I am not sure about the argument in the book but a simple argument shows that $X=0$: If $\prod_{i=1}^n (1 + \delta_i\mathbf q) \to X>0$ then $1 + \delta_n\mathbf q =\frac {\prod_{i=1}^n(1 + \delta_i\mathbf q)} {\prod_{i=1}^{n-1} (1 + \delta_i\mathbf q)} \to \frac X X =1$. But then $\delta_i\mathbf q \to 0$ which is impossible since $\delta_n =\pm 1$.

Answer 2

In the most extreme case, the gambler bets all of his/her fortune in each step. then $X_n$ is a Martingale that equals $2^n$ with probability $2^{-n}$ and otherwise is zero. So clearly $X_n$ tends to zero a.s.

To understand the rate of convergence to 0 in the general case, consider $\log X_n$. It is a sum of $n$ i.i.d. variables, each taking two values, with mean
$$M=\frac{\log(1+q)+\log(1-q)}{2}= \frac{ \log(1-q^2)}{2}<0 \,,$$ so the law of large numbers implies that almost surely $$\frac 1n \log(X_n) \to M<0 $$ so $X_n =\exp(n(M+o(1)))$ tends to zero exponentially fast.