4
$\begingroup$

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 $F$?

$\endgroup$
5
  • 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
6
$\begingroup$

No.

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.