Let $Z$ and $W$ be two independent exponential random variables with mean $λ$. What is $E(Z|Z > W)$?

Question

Can someone help me find out the answer to this question?
What I already know is that:

$E(Z|W) = \int zf_{Z|W = w}(Z|W) dx$

your help is appreciated. thanks.

Answer 1

We have:$$\mathbb{E}\left(Z\mid Z>W\right)P\left(Z>W\right)=\mathbb{E}Z\mathbf{1}_{Z>W}$$

Here $P\left(Z>W\right)=\frac{1}{2}$ because $Z,W$ are continuous random variables having the same distribution.

So we find:$$\mathbb{E}\left(Z\mid Z>W\right)=2\mathbb{E}Z\mathbf{1}_{Z>W}$$

We can find $\mathbb{E}Z\mathbf{1}_{Z>W}$ by finding:

$$\mathbb{E}Z\mathbf{1}_{Z>W}=\int_{0}^{\infty}\int_{0}^{z}zf_{Z}\left(z\right)f_{W}\left(w\right)dwdz=\int_{0}^{\infty}zf_{Z}\left(z\right)F_{W}\left(z\right)dz=\lambda\int_{0}^{\infty}ze^{-\lambda z}\left(1-e^{-\lambda z}\right)dz$$

Try to find the integral yourself.