I struggle with showing the following.
Let $\tau$ be an exponentially distributed random variable, that means $\mathbb P(\tau > x)=\mathrm{exp}(-x)$ and $\xi$ a bounded random variable.
Show that $$\mathbb E\left[\xi\frac{\mathbb 1_{(x<\tau\leq y)}}{\tau}\right]=0\quad \text{for all $x,y$ with $0<x<y$}$$
implies
$$\mathbb E\left[\xi\mid\tau\right]=0.$$
Maybe you can help me with that.
Thank you very much!
Best Answer
The result is true for any positive random variable $\tau$. Exponential distribution is note required. $E(\xi |\tau)=0$ iff $E(\frac {\xi} {\tau} |\tau)=0$ iff $E(\frac {\xi} {\tau} I_A)=0$ for any set $A$ of the type $\{\tau \in E\}$ where $E$ is a Borel set in $\mathbb R$. It is enough to prove this when $E$ is an open interval and in this case what we need is exactly the hypothesis.
PS: there is no guarantee that $\frac {\xi} {\tau}$ is integrable. To avoid this problem we can replace $\xi$ by $\xi I_{\{\tau >\epsilon\}}$ in the proof and let $\epsilon\to 0$ at the end.