How to simplify the conditional expectation $E[v_3\mid v_1 < \max\{v_2,v_3\}, v_3=\max\{v_2,v_3\}]$

Question

Suppose $v_1,v_2,v_3$ are three random variables drawn independently from the same distribution $\mathrm{uniform}(0,1)$, is it correct that $$E[v_3\mid v_1 < \max\{v_2,v_3\}, v_3=\max\{v_2,v_3\}] = E[\max\{v_1,v_2,v_3\}]\ ?$$

Answer 1

Yes; though not in general, it is the case for identical and independent distributions (iid). (Sorry, missed that in the first reading.)

$$\mathsf E[V_3\mid V_1<\max\{V_2,V_3\}, V_3=\max\{V_2, V_3\}] \;=\; \mathsf E[V_3\mid V_3=\max\{V_1,V_2, V_3\}]$$

This is not necessarily the same thing as $\mathsf E[\max\{V_1,V_2, V_3\}]$

Indeed:

$$\begin{align} \mathsf E[\max\{V_1,V_2, V_3\}] \; &= \; {\mathsf E[V_1\mid V_1=\max\{V_1,V_2, V_3\}]\;\mathsf P(V_1=\max\{V_1,V_2, V_3\}) + \\ \; \mathsf E(V_2\mid V_2=\max\{V_1,V_2, V_3\}] \;\mathsf P(V_2=\max\{V_1,V_2, V_3\}) + \\ \; \color{blue}{\mathsf E(V_3\mid V_3=\max\{V_1,V_2, V_3\}]} \;\mathsf P(V_3=\max\{V_1,V_2, V_3\})}\end{align}$$

However, in the case of iid random variables, symmetry means that all three of these probability terms evaluate to $1/3$ and all the expection terms are equal, so:

$$\mathsf E[\max\{V_1,V_2, V_3\}] \; = \; \mathsf E(V_3\mid V_3=\max\{V_1,V_2, V_3\}]$$