Suppose $U_1, \ldots, U_n$ are i.i.d. random variables with $U_1$ distributed uniformly on the interval $(-1, 1)$. Compute $\mathbb{E}(U_1 + \ldots + U_n |\max(U_1, \ldots, U_n) = t)$ for $t \in (-1, 1)$.
So my attempt was to try to compute $\mathbb{E}(U_1 |\max(U_1, \ldots, U_n) = t)$, writing $U_1 = U_1 1_{\{U_1 < t\}} + U_1 1_{\{U_1 \geq t\}}$ but I do not think this is a right approach. Generally speaking, in my opinion I do not feel intuitively the conditional expectation in a satisfactory way and hence my problems with this question, I guess.
Best Answer
For any $i=1,\ldots,n$, we have, using the Total Law of Expectation, conditioning on the value of $U_i$,
\begin{eqnarray*} && \\ E(U_i\vert \max\{U_1,..,U_n\}=t) &=& E(U_i\vert U_i=t\cap \max\{U_1,..,U_n\}=t)P(U_i=t\vert \max\{U_1,..,U_n\}=t) \\ && + E(U_i\vert U_i\neq t\cap \max\{U_1,..,U_n\}=t)P(U_i\neq t\vert \max\{U_1,..,U_n\}=t) \\ && \\ &=& t\cdot \dfrac{1}{n}\; + \;\dfrac{t+1}{2}\cdot \dfrac{n-1}{n}. \end{eqnarray*}
By the linearity of expectation, \begin{eqnarray*} E(U_1+\cdots+U_n\mid \max\{U_1,..,U_n\}=t) &=& \sum_{i=1}^n{E(U_i\mid \max\{U_1,..,U_n\}=t)} \\ && \\ &=& \sum_{i=1}^n{\left(t\cdot \dfrac{1}{n}\; + \;\dfrac{t+1}{2}\cdot \dfrac{n-1}{n}\right)} \\ && \\ &=& t + (n-1)\dfrac{t+1}{2}. \end{eqnarray*}
Note:
This result is intuitive: one value of $t$ and the remaining $n-1$ values taking the average value in interval $(-1,t)$.