Fix $1/2<p<1$. Let $X_1,X_2,\dots$ be independent identically distributed random variables with $\mathbb{P}(X_1=1)=p$ and $\mathbb{P}(X_1=-1)=1-p$. Let $S_0=0$ and $S_n=X_1+\dots+X_n$ and $\mathcal{F}_n=\sigma(X_1,\dots,X_n)$ for $n\in\mathbb{N}$. Define $$\varphi(x)=\left(\frac{1-p}{p}\right)^x.$$ I have obtained that $(\varphi(S_n))_{n\geq0}$ is a martingale. Additionally, if for $x\in\mathbb{Z}$ we define $T_x=\inf\{n\geq0:S_n=x\}$, I have obtained that for integers $a<0<b$ we have $$\mathbb{P}(T_a<\infty)=\left(\frac{1-p}{p}\right)^{-a},\hspace{1cm}\mathbb{P}(T_b<\infty)=1.$$ Is $(\varphi(S_n))_{n\geq0}$ uniformly integrable?
My thoughts: I think that naturally I expect for $(\varphi(S_n))$ to not be uniformly integrable, as we can see that $S$ will eventually almost surely hit any arbitrarily large integer. However, I am struggling to show this rigorously as $\varphi(S)$ is $L^1$-bounded and this is the only necessary condition I know of for uniform integrability and I don't see how to use the definition of uniform integrability here. Any advice would be greatly appreciated, thanks!
Best Answer
If the martingale $\left(\varphi\left(S_n\right)\right)_{n\geqslant 0}$ was uniformly integrable, then the optional stopping theorem for uniformly integrable martingale would give that $\mathbb E\left[\varphi\left(S_{T_b}\right)\right]=1$ for each positive $b$. But it can be seen that $\mathbb E\left[\varphi\left(S_{T_b}\right)\right]=\left(\frac{1-p}p\right)^b$.