Finding the conditional expectation of a max/min random variable 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).
 A: 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:...
$$\begin{align}\mathsf E(X_2{-}X_1\mid X_1{=}x)&= {\mathsf E(X_2{-}X_1\mid X_1{=}x, Z_2{>}Z_1)\,\mathsf P(Z_2{>}Z_1)\\+\mathsf E(X_1{-}X_2\mid X_2{=}x, Z_1{>}Z_2)\,\mathsf P(Z_1{>}Z_2)}\\&=1\end{align}$$
