Is there a proof/theorem that states that $$\mathrm{Var}(X_{kn}) \leq \mathrm{Var}(X)$$ where $X$ is a continuous random variable with distribution $F$, and $X_{kn}$ is the $k^{th}$ of $n$ order statistics from a sample of size $n$ from $F$?

  • 2
    $\begingroup$ A word on notation: I have made some edits to this post. While everyone is entitled to their own approach to mathematical notation, it's often useful to have conventions. It is quite uncommon to denote a distribution by the letter $X$ or $Y$, and it is quite common to denote a random variable by those letters. $\endgroup$
    – cardinal
    Oct 11 '11 at 18:18
  • $\begingroup$ @cardinal -- thanks for the clarification! Not knowing conventions one of the many trials of being self-taught. I will try to use the conventional notation moving forward. $\endgroup$
    – OctaviaQ
    Oct 11 '11 at 18:21
  • $\begingroup$ @cardinal -- actually, the mistake I made was in thinking I could take the variance of a random variable, as opposed to its distribution. Should I have said $\mathbb{V}F_{kn} \leq \mathbb{V}F, X \sim F, X_{kn}\sim F_{kn} =$ kth of $n$ draws from $F$ ? $\endgroup$
    – OctaviaQ
    Oct 11 '11 at 19:38
  • $\begingroup$ You can speak of taking the variance of a random variable and this is quite common. In fact, one of the nice things about the mathematical theory of expectation is that you can often find the variance of a random variable without having any explicit information about its distribution. :) $\endgroup$
    – cardinal
    Oct 11 '11 at 20:42
  • $\begingroup$ @cardinal. Oh.. ok.. and now I see you were referring to the last F, which you changed (I must have had an X). Thanks! $\endgroup$
    – OctaviaQ
    Oct 11 '11 at 20:48


Let $X$ have a Bernoulli($p$) distribution and take $n=2$. The probability that the maximum $X_{12}$ equals $1$ is $1 - (1-p)^2$. Because the maximum also is a Bernoulli variable, its variance equals $p(2-p)(1-p)^2$. Comparing this to the variance of the original distribution, $p(1-p)$, shows that the variance of $X_{12}$ exceeds the variance of $X$ for $0 \lt p \lt (3-\sqrt{5})/2$ (about 0.382).

Although $X$ in this example is not continuous, simply adding a small continuous well-behaved value to it (such as a Normal$(0, \varepsilon^2)$ variate with sufficiently small $\varepsilon$) will make it continuous and not appreciably change the relationships among these variances.

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    $\begingroup$ That point, $p = (3-\sqrt{5})/2$ is also the fixed point for iteratated minimax of $n=2$, where you are taking maximum of 2 minimums of 2 draws from $U(0,1)$ (e.g. the solution to $x==(1-(1-x)^n)^n$. Iterations of minimax increase $F(x)$ (the cdf) for values below $p$, and decrease it for values above $p$. Any thoughts on what might be the relationship between the two? $\endgroup$
    – OctaviaQ
    Oct 15 '11 at 14:55

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