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Assume X = $(X_1, ..., X_n)$ ~ $U(theta, 2theta)$, where $theta in BbbR^+$.

How does one calculate the conditional expectation of $E[X_1|X_(1),X_(n)]$, where $X_(1)$ and $X_(n)$ are the smallest and largest order statistics respectively?

My first thought would be that since the order statistics limit the range, it is simply $(X_(1)+X_(n))/2$, but im not sure if this is correct!

Consider the case of an iid sample $X_1, X_2, ldots, X_n$ from a Uniform$(0,1)$ distribution. Scaling these variables by $theta$ and translating them by $theta$ endows them with a Uniform$(theta, 2theta)$ distribution. Everything relevant to this problem changes in the same way: the order statistics and the conditional expectations. Thus, the answer obtained in this special case will hold generally.

Let $1lt klt n.$ By emulating the reasoning at https://stats.stackexchange.com/a/225990/919 (or elsewhere), find that the joint distribution of $(X_(1), X_(k), X_(n))$ has density function

$$f_k;n(x,y,z) = mathcalI(0le xle yle z le 1) (y-x)^k-2(z-y)^n-k-1.$$

Fixing $(x,z)$ and viewing this as a function of $y,$ this is recognizable as a Beta$(k-1, n-k)$ distribution that has been scaled and translated into the interval $[x,z].$ Thus, the scale factor must be $z-x$ and the translation takes $0$ to $x.$

Since the expectation of a Beta$(k-1,n-k)$ distribution is $(k-1)/(n-1),$ we find that the conditional expectation of $X_(k)$ must be the scaled, translated expectation; namely,

$$mathbbEleft(X_(k)mid X_(1), X_(n)right) = X_(1) + left(X_(n)-X_(1)right) frack-1n-1.$$

The cases $k=1$ and $k=n$ are trivial: their conditional expectations are, respectively, $X_(1)$ and $X_(k).$

Let's find the expectation of the sum of all order statistics:

$$mathbbEleft(sum_k=1^n X_(k)right) = X_(1) + sum_k=2^n-1 left(X_(1) + left(X_(n)-X_(1)right) frack-1n-1right) + X_(n).$$

The algebra comes down to obtaining the sum $$sum_k=2^n-1(k-1) = (n-1)(n-2)/2.$$

Thus

$$eqalign
mathbbEleft(sum_k=1^n X_(k)right) &= (n-1)X_(1) + left(X_(n)-X_(1)right) frac(n-1)(n-2)2(n-1) + X_(n) \
&= fracn2left(X_(n)+X_(1)right).
$$

Finally, because the $X_i$ are identically distributed, they all have the same expectation, whence

$$eqalignnmathbbEleft(X_1mid X_(1), X_(n)right) &= mathbbEleft(X_1right) + mathbbEleft(X_2right) + cdots + mathbbEleft(X_nright)\
&= mathbbEleft(X_(1)right) + mathbbEleft(X_(2)right) + cdots + mathbbEleft(X_(n)right) \
&= fracn2left(X_(n)+X_(1)right),
$$

with the unique solution

$$mathbbEleft(X_1mid X_(1), X_(n)right) = left(X_(n)+X_(1)right)/2.$$

It seems worth remarking that this result is not a sole consequence of the symmetry of the uniform distribution: it is particular to the uniform family of distributions. For some intuition, consider data drawn from a Beta$(a,a)$ distribution with $a lt 1.$ This distribution's probabilities are concentrated near $0$ and $1$ (its density has a U or "bathtub" shape). When $X_(n)lt 1/2,$ we can be sure most of the data are piled up close to $X_(1)$ and therefore will tend to have expectations less than the midpoint $(X_(1)+X_(n))/2;$ and when $X_(1)gt 1/2,$ the opposite happens and most of the data are likely piled up close to $X_(n).$

The following is not a proof but a verification of the desired result once you know that $(X_(1),X_(n))$ is a complete statistic for $theta$ :

Joint pdf of $X_1,X_2,ldots,X_n$ is

beginalign
f_theta(x_1,ldots,x_n)&=frac1theta^nmathbf1_theta<x_(1),x_(n)<2theta
\&=frac1theta^nmathbf1_frac12x_(n)<theta<x_(1)quad,,thetainmathbb R^+
endalign

So $T=(X_(1),X_(n))$ is a sufficient statistic for $theta$. It can be shown that $T$ is also a complete statistic by proceeding along these lines.

Then by Lehmann-Scheffe theorem, $E,[X_1mid T]$ is the UMVUE of $E(X_1)=frac3theta2$.

Now, $frac1theta(X_i-theta)stackreltexti.i.dsim mathsf U(0,1)$, so that $frac1theta(X_(n)-theta)simmathsfBeta(n,1)$ and $frac1theta(X_(1)-theta)sim mathsfBeta(1,n)$.

Therefore, $E(X_(n))=fracnthetan+1+theta=frac(2n+1)thetan+1$ and $E(X_(1))=fracthetan+1+theta=frac(n+2)thetan+1$.

Hence,

$$Eleft[frac12(X_(1)+X_(n))right]=frac12(n+1)left((n+2)theta+(2n+1)thetaright)=frac3theta2$$

This proves that $frac12(X_(1)+X_(n))$ is the UMVUE of $frac3theta2$ by Lehmann-Scheffe.

Since UMVUE is unique whenever it exists, it verifies the claim that $E,[X_1mid T]=frac12(X_(1)+X_(n))$.