# Conditional expectation of uniform random variable given order statistics

Assume X = $$(X_1, ..., X_n)$$ ~ $$U(\theta, 2\theta)$$, 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$$(\theta, 2\theta)$$ 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 $$1\lt k\lt n.$$ 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

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

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{n\mathbb{E}\left(X_1\mid X_{(1)}, X_{(n)}\right) &= \mathbb{E}\left(X_1\right) + \mathbb{E}\left(X_2\right) + \cdots + \mathbb{E}\left(X_n\right)\\ &= \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

$$\mathbb{E}\left(X_1\mid X_{(1)}, X_{(n)}\right) = \left(X_{(n)}+X_{(1)}\right)/2.$$

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

\begin{align} f_{\theta}(x_1,\ldots,x_n)&=\frac{1}{\theta^n}\mathbf1_{\theta

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, $$E\,[X_1\mid T]$$ is the UMVUE of $$E(X_1)=\frac{3\theta}{2}$$.

Now, $$\frac{1}{\theta}(X_i-\theta)\stackrel{\text{i.i.d}}\sim \mathsf U(0,1)$$, so that $$\frac{1}{\theta}(X_{(n)}-\theta)\sim\mathsf{Beta}(n,1)$$ and $$\frac{1}{\theta}(X_{(1)}-\theta)\sim \mathsf{Beta}(1,n)$$.

Therefore, $$E(X_{(n)})=\frac{n\theta}{n+1}+\theta=\frac{(2n+1)\theta}{n+1}$$ and $$E(X_{(1)})=\frac{\theta}{n+1}+\theta=\frac{(n+2)\theta}{n+1}$$.

Hence,

$$E\left[\frac{1}{2}(X_{(1)}+X_{(n)})\right]=\frac{1}{2(n+1)}\left((n+2)\theta+(2n+1)\theta\right)=\frac{3\theta}{2}$$

This proves that $$\frac{1}{2}(X_{(1)}+X_{(n)})$$ is the UMVUE of $$\frac{3\theta}{2}$$ by Lehmann-Scheffe.

Since UMVUE is unique whenever it exists, it verifies the claim that $$E\,[X_1\mid T]=\frac{1}{2}(X_{(1)}+X_{(n)})$$.

• +1 This answer is nice because it reveals a deeper way to understand the exercise and what it can teach us. – whuber Mar 6 '19 at 15:48