Let $(X_k)_{k\in\mathbb{N}}$ be iid random variables with $\mathbb{P}(X_1=1)=\mathbb{P}(X_1=-1)=\frac{1}{2}$.

Let $Z_n=\prod_{k=1}^n(1+X_k)$, so $Z_n$ a martingale.

Consider $T:=min\{k\ge 1: Z_k=c\}$, $c\in\mathbb{R}$.

Show that $T$ is a stopping time with respect to the natural Filtration $\mathcal{F_n}$ of $X_n$, i.e. $\mathcal{F_n}=\sigma(X_1,...,X_n)$

My idea:

$\{T\le n\}=\bigcup_{k=1}^n \{Z_k=c\}$.

I think I have to prove now that $\{Z_k=c\} \in \mathcal{F_k}$, then one can follow $\bigcup_{k=1}^n\{Z_n=c\}\in \mathcal{F_n}$. Why is it $\mathcal{F_k}$-measurable?

Thanks for help!


Note that

$$Z_k = f(X_1,\ldots,X_k)$$

for $$f(x_1,\ldots,x_k) := \prod_{j=1}^k (1+x_j).$$ Use that $X_1,\ldots,X_k$ are $\mathcal{F}_k$-measurable in order to conclude that $Z_k$ is $\mathcal{F}_k$-measurable.

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  • $\begingroup$ Thanks saz. That means since f is a measurable function (it's continuous), $f(X_1,...,X_k)$ is $\mathcal{F_k}$-measurable? $\endgroup$ – lemontree Sep 15 '14 at 15:39
  • $\begingroup$ @Zitrone Yes, exactly. $\endgroup$ – saz Sep 15 '14 at 16:06

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