Conditional Expectation of the Maximum Order Statistic
Let $U_1, U_2, U_3$ be i.i.d. $\text{Uniform}(0,1)$ random variables, and let $U_{(1)} < U_{(2)} < U_{(3)}$ denote their order statistics.
Compute $E[U_{(3)} \mid U_{(1)} = u_1,\, U_{(2)} = u_2]$.
Hints
- Given the two smaller order statistics, what interval must the largest value lie in?
- Conditioned on $U_{(2)} = u_2$, the remaining variable above $u_2$ is $\text{Uniform}(u_2, 1)$. What is its mean?
- Does knowing $U_{(1)} = u_1$ give you any additional information about $U_{(3)}$ beyond what $U_{(2)} = u_2$ already tells you?
Worked Solution
How to Think About It: You know the two smallest values out of three i.i.d. uniforms. The largest must lie somewhere above $u_2$. Since the original variables are i.i.d. $\text{Uniform}(0,1)$, the remaining one -- conditioned on being the largest -- is simply uniform on $(u_2, 1)$. This is a clean application of the fact that, given an order statistic $U_{(k)} = u$, the next order statistic $U_{(k+1)}$ behaves like the maximum of a single $\text{Uniform}(u, 1)$ draw (when there is only one remaining variable above $u$).
Quick Estimate: Take $u_1 = 0.2$, $u_2 = 0.5$. Then $U_{(3)} \mid U_{(2)} = 0.5$ is $\text{Uniform}(0.5, 1)$, so the conditional expectation is $(0.5 + 1)/2 = 0.75$. Notice $u_1$ does not appear in the answer -- once you condition on $U_{(2)}$, the value of $U_{(1)}$ adds no information about $U_{(3)}$.
Approach: Use the conditional distribution of the maximum given the second order statistic.
Formal Solution:
We have three i.i.d. $\text{Uniform}(0,1)$ draws. Given $U_{(2)} = u_2$, exactly one of the three original variables exceeds $u_2$ (namely $U_{(3)}$). Conditional on this variable exceeding $u_2$, and using the fact that it was originally $\text{Uniform}(0,1)$, its conditional distribution is: $U_{(3)} \mid U_{(2)} = u_2 \;\sim\; \text{Uniform}(u_2, 1)$
Note that conditioning additionally on $U_{(1)} = u_1$ provides no further information about $U_{(3)}$, because $U_{(1)}$ and $U_{(3)}$ are conditionally independent given $U_{(2)}$ (the Markov property of order statistics from a continuous distribution).
Therefore: $E[U_{(3)} \mid U_{(1)} = u_1,\, U_{(2)} = u_2] = E[U_{(3)} \mid U_{(2)} = u_2] = \frac{u_2 + 1}{2}$
Answer: $E[U_{(3)} \mid U_{(1)} = u_1,\, U_{(2)} = u_2] = \dfrac{u_2 + 1}{2}$.
Intuition
Order statistics from i.i.d. continuous distributions have a beautiful Markov property: given $U_{(k)}$, the statistics $U_{(k+1)}, U_{(k+2)}, \ldots$ are conditionally independent of $U_{(1)}, \ldots, U_{(k-1)}$. This means the "past" order statistics below $U_{(k)}$ are irrelevant for predicting what happens above $U_{(k)}$. Intuitively, once you know the $k$-th smallest value, the remaining values above it are like fresh uniform draws on a truncated interval.
This property generalizes nicely: for $n$ i.i.d. $\text{Uniform}(0,1)$ variables, $E[U_{(k+1)} \mid U_{(k)} = u] = \frac{(n-k)u + 1}{n - k + 1}$, which is a weighted average of $u$ and