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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 $X_i=X_i\mathbf{1}_{\{X_i=X_{\max}\}}+X_i\mathbf{1}_{\{X_i<X_{\max}\}}$. But this only left me with $$E[X_i|X_{\max}]=X_{\max}\mathbb{P}(X_i=X_{\max}|X_{\max})+E[X_i\mathbf{1}_{\{X_i<X_{\max}\}}|X_{\max}],$$ 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.

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

2 Answers 2

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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 $$\sum_{i=1}^n E[X_i\mid X_{\max}]=\sum_{i=1}^n E[X_1\mid X_{\max}]=\dfrac{2n+1}{3}X_{\max}$$

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

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

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

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:

$\begin{align}\mathsf E[X_i\mid X_\max] & = {{X_\max\,\mathsf P(X_i=X_\max\mid X_\max)}+{\mathsf E(X_i:{X_i<X_\max}\mid X_\max)\,\mathsf P(X_i<X_\max\mid X_\max)}}\\[1ex]&=\dfrac{X_\max}n+\dfrac{\int_{x<X_\max} xf_{X}(x)\,\mathrm d x}{\int_{x<X_\max}f_X(x)\,\mathrm d x}\cdot\dfrac{n-1}{n}\\[2ex] &~\vdots\\[2ex]&=\dfrac{(1+2n)X_\max}{3n}\end{align}$

You should complete the work yourself to verify the result.

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