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?

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

1 Answer 1


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 $Z:=\max\{1,\left(\frac{1-p}{p}\right)^b\}$.∎

  • $\begingroup$ Thanks a lot for your answer. And if we only want to consider $M_T$ does it also work? $\endgroup$ Jan 25 at 12:10
  • $\begingroup$ Can I then say that $|S_T|\leq b$ hence $|M_T|\leq \max\left\{\left(\frac{1-p}{p}\right)^b, 1\right\}$ $\endgroup$ Jan 25 at 12:13
  • $\begingroup$ @user1294729 what do you think? We can or we can't? $\endgroup$
    – Masacroso
    Jan 25 at 12:16
  • $\begingroup$ I would say yes because if $p<1/2$ then $|M_T|$ is bounded by $1$ and otherwise we have the other bound $\endgroup$ Jan 25 at 12:17
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    $\begingroup$ @user1294729 just note that $S_T\in\{0,b\}$, therefore $M_T\in\{x^0,x^b\}$ for $x=(1-p)/p$, thus $|M_T|\leqslant \max\{x^0,x^b\}$ $\endgroup$
    – Masacroso
    Jan 25 at 12:24

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