# Is this stopped martingale uniformly integrable?

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?

• $\Bbb{E}(M_{n+1}|F_n)=M_n\Bbb{E}\left(\frac{1-p}{p}p+\frac{p}{1-p}(1-p)\right)=M_n$ for every $p$ @geetha290krm Jan 25 at 10:04
• Sorry, I mistook $M_n$ for $S_n$. Jan 25 at 10:06
• @geetha290krm no problem. But is the rest correct that $M_n$ is bounded hence uniformly integrable Jan 25 at 10:07
• $(\frac{1-p}{p})^{a+n}$ is not bounded if $\frac {1-p} p >1$ which means $p<\frac 1 2$. Jan 25 at 10:09
• @geetha290krm ah okey but if we would furthermore assume $p<1/2$ then it is true what I did? Jan 25 at 10:11

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 $$Z:=\max\{1,\left(\frac{1-p}{p}\right)^b\}$$.∎
• Thanks a lot for your answer. And if we only want to consider $M_T$ does it also work? Jan 25 at 12:10
• Can I then say that $|S_T|\leq b$ hence $|M_T|\leq \max\left\{\left(\frac{1-p}{p}\right)^b, 1\right\}$ Jan 25 at 12:13
• I would say yes because if $p<1/2$ then $|M_T|$ is bounded by $1$ and otherwise we have the other bound Jan 25 at 12:17
• @user1294729 just note that $S_T\in\{0,b\}$, therefore $M_T\in\{x^0,x^b\}$ for $x=(1-p)/p$, thus $|M_T|\leqslant \max\{x^0,x^b\}$ Jan 25 at 12:24