# Why is $\operatorname{Var}(X_{(1)}) = \operatorname{Var}(X_{(n)})$ for i.i.d $X_1, \ldots, X_n \sim U(0,1)$?

This is an order statistics question. I'm using the notation found on https://en.wikipedia.org/wiki/Order_statistic.

We have $$n$$ IID random variables $$X_1, \cdots, X_n$$ that are uniformly distributed on $$[0, 1]$$. $$X_{(1)} = \min(X_1, \ldots, X_n)$$ and $$X_{(n)} = \max(X_1, \ldots, X_n)$$.

I am asked to reason that $$\operatorname{Var}(X_{(1)}) = \operatorname{Var}(X_{(n)})$$ without resorting to calculations. So it seems some kind of intuitive answer is wanted.

I'm given the hint that $$\operatorname{Var}(x) =\operatorname{Var}(1-x)$$ for any random variable $$x$$, but it's not obvious to me how this is helpful.

I know that $$\operatorname{Var}(X_{(1)}) = E[X_{(1)}^2] - E[X_{(1)}]^2 \\ \operatorname{Var}(X_{(n)}) = E[X_{(n)}^2] - E[X_{(n)}]^2$$

We known that $$E[X_1] = \cdots = E[X_n] = 0.5$$.

Intuition tells me that $$E[X_{(1)}]$$ and $$E[X_{(n)}]$$ are symmetrically situated about 0.5, i.e, $$0.5 - E[X_{(1)}] = E[X_{(n)}] - 0.5 \\ 1 = E[X_{(1)}] + E[X_{(n)}]$$ Not sure if this helps. I still can't figure out how $$\operatorname{Var}(x) = \operatorname{Var}(1-x)$$ is helpful. Am I going in the right correction with this expected value reasoning?

Honestly, it's kind of intuitive to me that $$\operatorname{Var}(X_{(1)})=\operatorname{Var}(X_{(n)})$$ by symmetry, but I'm having a hard time putting it into words while incorporating the hint.

• This also follows from the fact that $X_i-\frac12$ has the same distribution as $\frac12-X_i$ (since the distribution is symmetric about $1/2$), so that $X_{(n)}-\frac12$ has the same distribution as $\frac12-X_{(1)}$. – StubbornAtom May 6 '20 at 21:29

Let $$Y_k=1-X_k$$. $$Y_k$$ has the same distribution as $$X_k$$. (define $$Y_{(1)}$$ and $$Y_{(n)}$$ using the same min and max notation).
Therefore $$var(Y_{(1)})=var(X_{(1)})$$. Meanwhile $$Y_{(1)}=X_{(n)}$$ and $$Y_{(n)}=X_{(1)}$$
As a result $$var(X_{(n)})=var(X_{(1)})=var(Y_{(n)})=var(Y_{(1)})$$
• I like this proof. This relies on us knowing that $Y_{(1)} = X_{(n)}$ and $Y_{(n)} = X_{(1)}$, which further relies on us knowing that $E[X_{(1)}] - 0 = 1- E[X_{(n)}]$, the symmetry argument I made earlier. Is this obvious by inspection, or does this need to be proven some how? – David May 6 '20 at 12:29
• $Y_{(1)}$ is the minimum of {$Y_k=1-X_k$} which corresponds to the maximum of {$X_k$} which is $X_{(n)}$. Essentially changing the sign of all $X_k$ switches max and min. – herb steinberg May 6 '20 at 17:31
• Isn't $Y_{(1)}=1-X_{(n)}$ and $Y_{(n)}=1-X_{(1)}$? – StubbornAtom May 6 '20 at 21:28
• @herbsteinberg Isn't $Y_{(1)} = X_{(1)}$? – David May 6 '20 at 23:52