# Conditional expectation to de maximum $E(X_1\mid X_{(n)})$

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

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

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

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

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.

• Is it true that $X_1\mid X_{(n)}$ has a mixed distribution? Aug 25, 2018 at 13:59
• @StubbornAtom Yes, see update Aug 25, 2018 at 14:24

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 $\max$.

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.