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Let $X_1, \ldots, X_n$ a random sample of a Uniform(0,1):

Which is $E(X_1\mid X_{(n)})$ ?

where $X_{(n)}=\max\{X_1,\ldots,X_n\}$

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    $\begingroup$ Hint: compute $E(X_1\mid X_{(n)}, Y)$ where $Y=1\cdots n$ is the index of the variable that attains the maximum value, and apply $E(X_1\mid X_{(n)})=E ( E(X_1\mid X_{(n)}, Y))$ $\endgroup$
    – leonbloy
    Nov 5, 2014 at 18:39
  • $\begingroup$ @leonbloy I don't understand how your hint can help me $\endgroup$
    – rcg90
    Nov 5, 2014 at 18:46
  • $\begingroup$ @leonbloy Excellent hint. rcg90: Sub-hint: The cases $Y=1$ and $Y\ne1$ are to be treated separately. $\endgroup$
    – Did
    Nov 5, 2014 at 18:51
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    $\begingroup$ As so often happens, I am the only person who has up-voted this question after it has receive attention from several people. $\endgroup$ Nov 5, 2014 at 19:07
  • $\begingroup$ Same question:math.stackexchange.com/questions/617415/compute-ex-1y?rq=1. $\endgroup$ Mar 3, 2018 at 13:02

3 Answers 3

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Let $Y$ be the index of the maximum value: $X_Y=X_{(n)}$ (the event of ties is unimportant, it has zero probability). Let $Z=X_{(n)}$

Then

$$E(X_1\mid Z ,Y) = \begin{cases} Z & \text{ if } Y=1 \\ \frac{1}{2}Z & \text{ elsewhere} \\ \end{cases} $$

Then, applying the iterated expectations property, and because $P(Y=k)=1/n$:

$$E(X_1\mid Z)=E( E(X_1\mid Z , Y))=\frac{1}{n}Z+ \frac{n-1}{n}\frac{Z}{2} = \frac{n+1}{n}\frac{Z}{2} $$

Edit: this is not need to answer the question, but (from a comment) the full probability can be obtained (some abuse of notation follows) thus:

$$P(X_1\mid Z ) =\sum_{Y=1}^n P(X_1,Y\mid Z )\\ =\sum_{Y=1}^n P(X_1\mid Y,Z ) P(Y \mid Z)\\ =\delta(X_1 -Z)\frac{1}{n}+\sum_{Y=2}^n P(X_1\mid X_1<Z ) \frac{1}{n}\\ =\frac{1}{n}\delta(X_1 -Z)+ \frac{n-1}{n}U[0,Z] $$

So, yes $P(X_1\mid X_{(n)} )$ can be seen a mixing of a delta with a (truncated) uniform.

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  • $\begingroup$ Is it true that $X_1\mid X_{(n)}$ has a mixed distribution? $\endgroup$ Aug 25, 2018 at 13:59
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    $\begingroup$ @StubbornAtom Yes, see update $\endgroup$
    – leonbloy
    Aug 25, 2018 at 14:24
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You're looking for $E(X_1\mid \max)$. The probability that $X_1=\max$ is $1/n$. So $$ E(X_1\mid \max) = \frac 1 n E(X_1\mid X_1=\max, \max) + \frac{n-1} n E(X_1\mid X_1\ne \max, \max). $$ The first term is easy to find: $E(X_1\mid X_1=\max,\max) = \max$, so we have $$ E(X_1\mid \max) = \frac 1 n \max + \frac{n-1} n E(X_1\mid X_1\ne \max, \max). $$ Then you need to show that the conditional distribution of $X_1$ given $\max$ and given the event that $X_1\ne\max$ is uniform on the interval from $0$ to $\max$.

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I'd like to give an answer based on some concepts and theorems in mathematical statistics, they can be found in Casella and Berger's Statistical Inference. Let $\theta\in(0,+\infty)$ be a parameter and consider $Y_i=X_i\theta\sim U(0,\theta)$. The joint density of $(Y_1,\cdots,Y_n)$ is $f(y_1,\cdots,y_n)=\theta^{-n}1_{y_{(1)}>0}1_{y_{(n)}<\theta}$, here $y_{(1)}=\min y_i$ and $y_{(n)}=\max y_i$. So by factorization theorem, $T(Y)=Y_{(n)}$ is a sufficient statistic for $\theta$. Now note that $Y_{(n)}$ is complete, i.e. for any function $g:\mathbb{R}\to\mathbb{R}$, $Eg(T(Y))=0$ for any $\theta$ implies $P(g(Y)=0)=1$ for any $\theta$,(this can be seen from directly write the expectation as integration), and $2Y_1$ is unbiased for $\theta$, i.e. $E(2Y_1)=\theta$, we know that $E(2Y_1|T(Y))$ is the UMVUE for $\theta$.

Now we note that $E(T(Y))=\frac{n}{n+1}\theta$, we know $\frac{n+1}{n}T(Y)$ is unbiased for $\theta$ and thus $\frac{n+1}{n}T(Y)=E(\frac{n+1}{n}T(Y)|T(Y))$ is the UMVUE for $\theta$. Since the UMVUE is unique (up to almost sure equivalence), we derive that $E(2Y_1|T(Y))=\frac{n+1}{n}T(Y)$ almost surely for any $\theta$. Taking $\theta=1$ gives us the desired result.

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