Finding the conditional expectation of a max/min random variable

Question

Problem

Suppose that $ Z_1 $ and $ Z_2 $ are independent with common density $$ f_Z(z) = \begin{cases} e^{-z} & \text{ if } z > 0 \\ 0 & \text{ otherwise }. \end{cases} $$ Let $ X_1 = \min\left\{ Z_1, Z_2 \right\} $ and $ X_2 = \max\left\{ Z_1, Z_2 \right\} $. Compute $ \mathrm{E}\left[ X_2 - X_1 | X_1 = x_1 \right] $.

Attempted Solution

Proceed with using the law of iterated expecation with the following finite partition of the sample space \begin{gather*} D_1: \left\{ Z_1 > Z_2 \right\} \\ D_2: \left\{ Z_2 > Z_1 \right\} \\ D_3: \left\{ Z_1 = Z_2 \right\} \end{gather*} We should have \begin{align*} \mathrm{E}\left[ X_2 - X_1 | X_1 = x_1 \right] &= \mathrm{E}\left[ X_2 - X_1| D_1, X_1 = x_1 \right] \Pr(D_1) \\ &\quad+ \mathrm{E}\left[ X_2 - X_1| D_2, X_1 = x_1 \right] \Pr(D_2) \\ &\quad+ \mathrm{E}\left[ X_2 - X_1| D_3, X_1 = x_1 \right] \Pr(D_3) \\ &= \mathrm{E}\left[ X_2 - X_1| Z_1 > Z_2, X_1 = x_1 \right] \Pr(Z_1 > Z_2) \\ &\quad+ \mathrm{E}\left[ X_2 - X_1| Z_2 > Z_1, X_1 = x_1 \right] \Pr(Z_2 > Z_1) \\ &= \mathrm{E}\left[ Z_1 - Z_2 | Z_2 = x_1 \right]\Pr(Z_1 > Z_2) \\ &\quad+ \mathrm{E}\left[ Z_2 - Z_1 | Z_1 = x_1 \right](1 - \Pr(Z_1 > Z_2)) \\ &= \left( \mathrm{E}\left[ Z_1 | Z_2 = x_1 \right] - x_1 \right)\Pr(Z_1 > Z_2) \\ &\quad+ \left( \mathrm{E}\left[ Z_2 | Z_1 = x_1 \right] - x_1 \right)(1 - \Pr(Z_1 > Z_2)). \end{align*} Where we have employed that $P(D_3|X_1 = x_1) = 0$ since $ Z_1 = Z_2 $ occurs with measure $0$ probability and that $\Pr(Z_2 > Z_1) = 1 - \Pr(Z_1 \ge Z_2) = 1 - \Pr(Z_1 > Z_2)$. Notice that $ Z_1, Z_2 $ are identical and independent with mean $ 1 $ (since they are exponential distribution with $ \lambda = 1 $, hence $ \mathrm{E}[Z_i] = \frac{1}{\lambda} = 1 $). Therefore we have \begin{gather*} \mathrm{E}\left[ Z_1 | Z_2 = x_1 \right] = \mathrm{E}[Z_1] = 1 \\ \mathrm{E}\left[ Z_2 | Z_1 = x_1 \right] = \mathrm{E}[Z_2] = 1 \end{gather*}

We can conclude that \begin{align*} \mathrm{E}\left[ X_2 - X_1 | X_1 = x_1 \right] &= \left( \mathrm{E}\left[ Z_1 | Z_2 = x_1 \right] - x_1 \right)\Pr(Z_1 > Z_2) + \left( \mathrm{E}\left[ Z_2 | Z_1 = x_1 \right] - x_1 \right)(1 - \Pr(Z_1 > Z_2)) \\ &= \left( 1 - x_1 \right)\Pr(Z_1 > Z_2) + \left( 1 - x_1 \right)(1 - \Pr(Z_1 > Z_2)) \\ &= 1 - x_1 \end{align*}

Question

Now here is the issue, the answer is supposed to be $ \mathrm{E}\left[ X_2 - X_1 | X_1 = x_1 \right] = 1 $. Both a friend and I have gone through the attempted solution above but could not find the error with it, yet it gives the wrong answer. Can anyone help out with identifying where the error is and how you might modify the approach to get the right answer?

I'd also be open to hear if Math StackExchange could find a different solution method (preferably one that is as succinct).

Answer 1

Your evaluation of $\mathsf E(X_2{-}X_1\mid X_1{=}x, Z_2{>}Z_1)$ is awry.

The event of $\{X_1{=}x, Z_2{>}Z_1\}$ is the event of $\{Z_1{=}x,Z_2{>}x\}$. Also under that condition, then $X_2-X_1$ will equal $Z_2-Z_1$

$$\begin{align} \mathsf E(X_2{-}X_1\mid X_1{=}x, Z_2{>}Z_1) &= \mathsf E(Z_2{-}Z_1\mid Z_1{=}x,Z_2{>}x) \\ &=\mathsf E(Z_2\mid Z_2{>}x,Z_1{=}x)-\mathsf E(Z_1\mid Z_2{>}x,Z_1{=}x) \end{align}$$

Next use that the $Z_\star$ are independent, and recall that exponential distributed random variables have the memoryless property.

$$\begin{align} \mathsf E(X_2{-}X_1\mid X_1{=}x, Z_2{>}Z_1) &=\mathsf E(Z_2\mid Z_2{>}x)-\mathsf E(Z_1\mid Z_1{=}x) \\ &= \mathsf E(Z_2\mid Z_2{>}x)-x\\ &= \mathsf E(Z_2)\\&=1 \end{align}$$

Likewise $\hspace{12ex}\mathsf E(X_2{-}X_1\mid X_1{=}x, Z_1{>}Z_2)=1$

Therefore:...

$$\begin{align}\mathsf E(X_2{-}X_1\mid X_1{=}x)&= {\mathsf E(X_2{-}X_1\mid X_1{=}x, Z_2{>}Z_1)\,\mathsf P(Z_2{>}Z_1)\\+\mathsf E(X_1{-}X_2\mid X_2{=}x, Z_1{>}Z_2)\,\mathsf P(Z_1{>}Z_2)}\\&=1\end{align}$$