# Calculating a Conditional expectation

My question is the following. Given that we have $$n$$ i.i.d. random variables $$X_1,...,X_n$$ with distribution $$f(x)=\frac{2}{\lambda^2}x\mathbf{1}_{[0,\lambda]}(x)$$, where $$\lambda> 0$$ is some parameter, how do I calculate the the conditional expectation $$E[X_i|X_{\max}],$$ with $$X_{\max}=\max\{X_1,...,X_n\}$$. My idea was writing $$X_i=X_i\mathbf{1}_{\{X_i=X_{\max}\}}+X_i\mathbf{1}_{\{X_i. But this only left me with $$E[X_i|X_{\max}]=X_{\max}\mathbb{P}(X_i=X_{\max}|X_{\max})+E[X_i\mathbf{1}_{\{X_i which doesn't really gelp me I think. I've been given one hint: In the end we should have $$\sum_{i=1}^n E[X_i|X_{\max}]=\frac{2n+1}{3}X_{\max},$$ So my guess is $$E[X_i|X_{\max}]=\frac{2n+1}{3n}X_{\max}.$$ Any suggestions on how to proceed are welcome. Thank you.

• This can be found using the same arguments as used here and its linked posts. Commented May 25 at 15:22

Let $$V=\max\{X_2,\ldots,X_n\}$$. For $$0\leq v\leq \lambda$$ we have

$$\begin{eqnarray} F_V(v) &=& \Pr(V \leqslant v) = \Pr(\max\{X_2,\ldots,X_n\} \leqslant v) \\ &=& \Pr(X_2 \leqslant v,\ldots,X_n \leqslant v) \\ &=& F_X(v)^{n-1} = \left(\dfrac{x}{\lambda}\right)^{2n-2} \end{eqnarray}$$

And so $$f_V(v)=\dfrac{2n-2}{\lambda^{2n-2}}x^{2n-3}$$ for $$0\leq v\leq \lambda$$.

Now, notice that $$X_{\max}=\max\{X_1,\ldots,X_n\}=\max\{X_1,V \}$$ (with $$X_1$$ and $$V$$ independent).

Therefore, $$\begin{eqnarray} \mathsf{E}(X_1\mid X_{\max}=y) &=& \mathsf{E}(X_1\mid \max\{X_1,V\}=y) \\ &=& \mathsf{E}(X_1\mid \max\{X_1,V\}=y,X_1 \leqslant V) \Pr(X_1 \leqslant V) + \\ && \mathsf{E}(X_1\mid \max\{X_1,V\}=y,X_1 > V) \Pr(X_1>V) \\ &=& \mathsf{E}(X_1\mid V=y, X_1\leqslant V) \Pr(X_1 \leqslant V) + \mathsf{E}(X_1\mid X_1=y,X_1>V) \Pr(X_1>V) \end{eqnarray}$$

You can compute these quantities (remember that $$X_1$$ and $$V$$ are independent): $$\mathsf{E}(X_1\mid V=y, X_1\leqslant V)=\mathsf{E}(X_1\mid X_1\leqslant y)=\dfrac{2y}{3}$$ $$\Pr(X_1 \leqslant V)=\dfrac{2n-2}{2n}$$ $$\mathsf{E}(X_1\mid X_1=y,X_1>V)=\mathsf{E}(X_1\mid X_1=y)=y$$ $$\Pr(X_1>V)=\dfrac{1}{n}$$

This means that $$\mathsf{E}(X_1\mid X_{\max}=y)=\dfrac{2n+1}{3n}y=\dfrac{2n+1}{3n}X_{\max}$$

Given that the $$X_i$$ are i.i.d, we get that $$\sum_{i=1}^n E[X_i\mid X_{\max}]=\sum_{i=1}^n E[X_1\mid X_{\max}]=\dfrac{2n+1}{3}X_{\max}$$

But this only left me with $$E[X_i|X_{\max}]=X_{\max}\mathbb{P}(X_i=X_{\max}|X_{\max})+E[X_i\mathbf{1}_{\{X_i which doesn't really gelp me I think.

Mixing the notation slightly, we may also write that as:

\begin{align}\mathsf E(X_i\mid X_\max) & = {{X_\max\,\mathsf P(X_i{\,=\,}X_\max\mid X_\max)}+{\mathsf E(X_i:{X_i

Now, what are those probability masses?   Easy!   Since each of the $$n$$ samples is independent and identically distributed, each has the same probability for being the maximum.

\qquad\begin{align}\mathsf P(X_i{\,=\,}X_\max\mid X_\max)~&=~1/n\\\mathsf P(X_i{\,<\,}X_\max\mid X_\max)~&=~(n-1)/n\end{align}

So therefore:

\begin{align}\mathsf E[X_i\mid X_\max] & = {{X_\max\,\mathsf P(X_i=X_\max\mid X_\max)}+{\mathsf E(X_i:{X_i

You should complete the work yourself to verify the result.