martingale stopping time is same as total number of events?

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$X:=\{X_n\}_{n\ge 1}$ is a sequence of i.i.d random variables with \begin{align*} \mathbb P\left(X_1=1\right)=p\in (0,1)\setminus \{\frac{1}{2}\} \text{ and } \mathbb P\left(X_1=-1\right)=q=(1-p) \end{align*} Define $S_n=\sum\limits_{k=1}^{n} X_k$. Show that

\item [a)] $Y_n=\left(\frac{q}{p}\right)^{S_n}$ is a martingale with respect to natural filtration of $X$.

\item[b)] $a, b\in \mathbb Z, a<0<b$ and define $\tau:=\inf\{ n\ge 0: S_n\in \{a,b\}\}$. Then show that $\left(Y_{\tau\wedge n}\right)_{n\ge 0}$ is uniformly integrable.

\item[c)] Assuming $\mathbb P\left(\tau<\infty\right)=1$, show that \begin{align*} \mathbb P\left(S_{\tau}=a\right)=\frac{1-\left(\frac{q}{p}\right)^b}{\left(\frac{q}{p}\right)^a-\left(\frac{q}{p}\right)^b} \end{align*}

I did (a) perfectly, I am not able to understand (b) at all, and what I need to show and how?

for (c) I am able to prove \begin{align*} \mathbb P\left(\tau=\tau_a\right)=\frac{1-\left(\frac{q}{p}\right)^b}{\left(\frac{q}{p}\right)^a-\left(\frac{q}{p}\right)^b} \end{align*} but I don't know how to show what is asked. Thanks for help!

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(b) $\{Y_{\tau\wedge n}\}$ is bounded (and, thus, u.i.): $$ \sup_{n\ge 1}|Y_{\tau\wedge n}|\le \max((q/p)^b,(q/p)^a). $$ (c) $S_{\tau}=a$ on $\{\tau=\tau_a\}$ ($\because\tau=\tau_a\wedge \tau_b$).