Argthtjtr

Assume X = $(X_1, ..., X_n)$ ~ $U(theta, 2theta)$, where $theta in Bbb{R}^+$.

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

$$f_{k;n}(x,y,z) = mathcal{I}(0le xle yle z le 1) (y-x)^{k-2}(z-y)^{n-k-1}.$$

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,

$$mathbb{E}left(X_{(k)}mid X_{(1)}, X_{(n)}right) = X_{(1)} + left(X_{(n)}-X_{(1)}right) frac{k-1}{n-1}.$$

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:

$$mathbb{E}left(sum_{k=1}^n X_{(k)}right) = X_{(1)} + sum_{k=2}^{n-1} left(X_{(1)} + left(X_{(n)}-X_{(1)}right) frac{k-1}{n-1}right) + X_{(n)}.$$

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

Thus

$$eqalign{
mathbb{E}left(sum_{k=1}^n X_{(k)}right) &= (n-1)X_{(1)} + left(X_{(n)}-X_{(1)}right) frac{(n-1)(n-2)}{2(n-1)} + X_{(n)} \
&= frac{n}{2}left(X_{(n)}+X_{(1)}right).
}$$

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

$$eqalign{nmathbb{E}left(X_1mid X_{(1)}, X_{(n)}right) &= mathbb{E}left(X_1right) + mathbb{E}left(X_2right) + cdots + mathbb{E}left(X_nright)\
&= mathbb{E}left(X_{(1)}right) + mathbb{E}left(X_{(2)}right) + cdots + mathbb{E}left(X_{(n)}right) \
&= frac{n}{2}left(X_{(n)}+X_{(1)}right),
}$$

with the unique solution

$$mathbb{E}left(X_1mid X_{(1)}, X_{(n)}right) = left(X_{(n)}+X_{(1)}right)/2.$$

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 $X_{(n)}.$