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.

  • $\begingroup$ 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 $\frac12-X_{(1)}$. $\endgroup$ – 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 $var(X_{(n)})=var(X_{(1)})=var(Y_{(n)})=var(Y_{(1)})$

  • $\begingroup$ 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? $\endgroup$ – David May 6 '20 at 12:29
  • $\begingroup$ $Y_{(1)}$ 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. $\endgroup$ – herb steinberg May 6 '20 at 17:31
  • $\begingroup$ Isn't $Y_{(1)}=1-X_{(n)}$ and $Y_{(n)}=1-X_{(1)}$? $\endgroup$ – StubbornAtom May 6 '20 at 21:28
  • 1
    $\begingroup$ @StubbornAtom You are correct. The variance conclusion still holds, since var(x)=var(1-x) for any random variable x. $\endgroup$ – herb steinberg May 6 '20 at 21:45
  • $\begingroup$ @herbsteinberg Isn't $Y_{(1)} = X_{(1)}$? $\endgroup$ – David May 6 '20 at 23:52

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