let $(X_n)$ be a sequence of i.i.d. random variables with $\Bbb{P}(X_1=1)=p$ and $\Bbb{P}(X_1=-1)=1-p$ (s.t. $p\in (0,1)\setminus\{1/2\}$) then define for $a>0$ the sequence $S_0=a$ and $S_n=S_{n-1}+X_n$ for every $n\geq 1$. Consider the stopping time $T=\inf\{n\geq 0: S_n\in \{0,b\}\}$ where $b>a$ is fixed. I define the process $$M_n=\left(\frac{1-p}{p}\right)^{S_n}$$The question is, is $M_{n\wedge T}$ a uniformly integrable martingale with respect to $F_n=\sigma(X_1,…,X_n)$.
I know that if $M_n$ is a uniformly integrable martingale then for all stopping times $T$ also $M_{n\wedge T}$ is a uniformly integrable martingale with terminal value $M_T$. So I want to show that $M_n$ is a uniformly integrable martingale.
I could show that $M_n$ is a martingale but now it remains to prove that it is uniformly integrable. I know that if the martingale is bounded it is uniformly integrable but $$\left|M_n\right|=\left|\left(\frac{1-p}{p}\right)^{S_n}\right|\leq \left(\frac{1-p}{p}\right)^{a+n}$$so it is clearly bounded and hence uniformly integrable.
Does this proof works? Or am I wrong?
Best Answer
Note that $|S_{n\,\land\, T}|\leqslant b$, therefore $|x^{S_{n\,\land\, T}}|\leqslant \max\{1,x^b\}$ for every $x\geqslant 0$, therefore $\{M_{n\,\land\, T}\}_{n\in \mathbb{N}}$ is a collection uniformly integrable as all functions are dominated by the constant function $Z:=\max\{1,\left(\frac{1-p}{p}\right)^b\}$.∎