# Inequality for Uniform Distribution

Let $$X_1,..,X_n$$ be a random independent sample $$X_1,…,X_n$$ from a Uniform$$[0,\theta]$$ distribution, $$\theta \in [0, \infty)$$, with probability density function $$f(x;\theta) = \begin{cases} 1/\theta, & 0 \le x \le \theta \\ 0, & \text{otherwise} \end{cases}$$ and
$$X_{(n)}=max(X_1,X_2,…,X_n)$$ and $$X_{(1)}=min(X_1,...,X_n)$$.

I know that $$X_{(1)}$$ has the same distribution as $$\theta - X_{(n)}$$

and that $$\hat{\theta}=X_{(1)}+X_{(n)}$$ is an unbiased estimator of θ.

I want to show that $$Var(X_{(1)}+X_{(n)}) \le 4Var(X_{(n)})$$.

My work: I know that $$Var(X_{(n)})= \frac{\theta^2n}{(n+1)^2(n+2)}$$.

Next: $$Var(X_{(1)}+X_{(n)}) = E[(X_{(1)}+X_{(n)})^2]-E[X_{(1)}+X_{(n)}]^2= E[X_{(1)}^2]+E[2X_{(1)}X_{(n)}]+E[X_{(n)}^2]-\theta^2= E[X_{(1)}^2]+E[2(\theta-X_{(n)})X_{(n)}]+E[X_{(n)}^2]-\theta^2 =E[X_{(1)}^2]+2\theta E[X_{(n)}]-E[X_{(n)}^2]-\theta^2.$$

I also calculated that: $$E[X_{(1)}^2]= \frac{2\theta^2}{(n+1)(n+2)}$$ and $$E[X_{(n)}^2]=\frac{2\theta^2}{(n+2)}$$.

• I wonder whether you actually have to calculate anything. Since $(\frac{n+1}{n})X_{(n)}$ has the minimum variance among all unbiased estimators of $\theta$, I can say that $\operatorname{Var}(X_{(1)}+X_{(n)})\ge (\frac{n+1}{n})^2\operatorname{Var}(X_{(n)})$. Maybe something similar can be said for your result. – StubbornAtom May 7 at 21:32

Third equality in your calculation of the variance of the sum of order statistics is wrong. Yes, $$X_{(1)}$$ has the same distribution as $$\theta - X_{(n)}$$, and $$X_{(1)}X_{(n)}$$ does not have the same distribution as $$(\theta-X_{(n)})X_{(n)}$$.
You can calculate either $$\mathbb E[(X_{(1)}+X_{(n)})^2]$$ or $$\mathbb E[X_{(1)}X_{(n)}]$$ using pdf of joint distribution of $$X_{(1)}$$ and $$X_{(n)}$$: $$f_{X_{(1)},X_{(n)}}(x,y) = \frac{n(n-1)(y-x)^{n-2}}{\theta^n}\cdot \mathbb 1_{\{0\leq x\leq y\leq \theta\}}$$ The resulting variance should be $$Var(X_{(1)}+X_{(n)})=\frac{2\theta^2}{(n+1)(n+2)}$$.