How to show that $\operatorname{var}(X_{(1)}) = \operatorname{var}(X_{(n)})$ for $X_i \sim U(0, 1)$?

Question

Define $Y_i = 1 - X_i$ and $Y_{(k)}$ as the k-th order statistic, then $$ Y_{(1)} = 1 - X_{(n)} $$

Variance is invariant to shifts, so $$ \operatorname{var}(X_{(n)}) = \operatorname{var}(Y_{(1)}) \\ \operatorname{var}(X_{(1)}) = \operatorname{var}(Y_{(n)}) \\ $$

I am stuck at this point, but I think the proof should somehow follow from these two expressions.

We also know that $\operatorname{var}(X_{i}) = \operatorname{var}(Y_{i})$, but I don't think this (directly) implies that $\operatorname{var}(X_{(1)}) = \operatorname{var}(Y_{(1)})$. If it does, then

$$ \operatorname{var}(X_{(n)}) = \operatorname{var}(Y_{(n)}) = \operatorname{var}(X_{(1)}) $$

Answer 1

The idea that you are using is perfect. Just note the fact that $Y_1,Y_2,\dots,Y_n$ also follow iid $U(0,1)$ distribution. Hence, $Y_{(k)}$ and $X_{(k)}$ have the same distribution. But by the definition of $Y$, we have $Y_{(k)} = 1-X_{(n-k+1)}$. Using both these facts we have, $$\mathrm{Var}(Y_{(k)})=\mathrm{Var}(X_{(n-k+1)}){=}\mathrm{Var}(X_{(k)})$$ Putting $k=1$, we get the desired special case.

P.S. Note that this argument is general and holds for every symmetric distribution, symmetric about some $\theta\in\Bbb R$ say. Just define $Y_i=2\theta-X_i$ in that case.

Answer 2

there is a simple and direct way to proceed...

$X_{(1)}$ is the minimum value of $X_1,\dots,X_n$ and $X_{(n)}$ is the maximum

Thus

$$f_{X_{(1)}}(t)=n(1-t)^{n-1}$$

$$f_{X_{(n)}}(t)=n t^{n-1}$$

Now it is easy to calculate all the variances with the definition

After some easy calculations in both cases you get

$$V=\frac{n}{(n+2)(n+1)^2}$$