Show that $X_{\tau}=2b(\tau p-1)$

Question

Problem Let $0<b<1$ and let $Z_1, Z_2,...$ be independent random variables with distribution $$P(Z_n=-b)=p, \: \: P(Z_n=b)=1-p, \:\: p\in(0,1)$$ for $n=1,2...$ Let $\mathscr{F}_n=\sigma(Z_1,...,Z_n), n\geq 1 $ be a filtration and define

$$X_n=\displaystyle\sum_{i=1}^{n}Z_i-nb(1-2p) \: for \: n>1,$$ $$\tau=min\{n\geq 1|Z_n=-b\}$$

Show that $X_\tau =2b(\tau p-1)$

Attempt:

I know that $X_\tau :=\displaystyle \sum_{n=0}^{\infty}X_n \mathbb{1}_\{\tau=n\},$ therefore if $$\tau=n \Rightarrow X_\tau =X_n$$

Furthermore $$\{\tau = n\} \Leftrightarrow \{Z_1=...=Z_{n-1}=b, Z_n=-b\}$$

Maybe i can use this conclusion for something (?)

I feel i need go down this road: $$X_\tau = \displaystyle \sum_{i=1}^{\tau}Z_i - \tau b(1-2p)$$ But i'm having trouble seeing how to both justify it and how it will help me onwards.

Any hints?

Answer 1

You have almost finished it. When $\tau=n$ we have $$X_{\tau}=X_n =[(n-1)b-b]-nb(1-2p)$$ $$=-2b+2nbp=-2b+2\tau bp =2b(\tau p-1).$$