Uniform integrability of "exponent martingale"

Question

Suppose that $X_n$ is an iid sequence of random variables with $P[X_i=1]=p$ and $P[X_i=-1]=q:=(1-p)$. Then, $S_n=X_1+\cdots+X_n$ (with $S_0=0$) is a simple random walk.

We can easily check that $T_n=(q/p)^{S_n}$ is a nonnegative martingale. If $p=q=1/2$, this $T_n$ is constant and thus clearly uniformly integrable. However, I'm having trouble verifying if this is uniformly integrable or not in other cases.

We know that $T_n$ converges almost surely to a limit $T$, and that $E[T_n]=1$ for all $n$. Thus we must try to understand the convergence $E[|T_n-T|]$ or $E[T_n]\to E[T]$ to conclude that it is UI or get a contradiction. However, I have not been able to understand the limit $T$ and/or compute its expectation.

Answer 1

This martingale is also known as De Moivre's martingale. As you correctly observed, it is non-negative and hence converges a.s. to some $T$. Depending on the values of $p,q$ there are $3$ cases:

1. $p=q$. You have correctly treated this case, by observing that $T_n$ is constant and equal to $1$ and hence uniformly integrable.
2. $p>q$. Now $S_n\to +\infty$ a.s. and $q/p<1$ hence $$\left(\frac qp\right)^{S_n}\to 0$$ a.s. That is $T=0$ and therefore $$E[T]=0 \neq \lim_n E[T_n]=E[T_0]=(q/p)^0=1$$
3. $p<q$. As case 2. Now $S_n \to -\infty$ a.s. but $q/p>1$.

In sum, the case $p=q$ is the only case where $T_n$ converges in $L^1$ to $T$.