# Order statistic expectations and the variance of the original distribution

Is there a proof or theorem out there that says that the difference between the expectations of the order distributions of $n$ draws from a distribution $X$ decreases as the variance of $X$ decreases?

Specifically, I am asking: If X and Y are probability distributions defined over the same range and with the same mean, does Variance(X) < Variance (Y) imply that $\mathbb{E}X_{kn} - \mathbb{E}X_{sn} < \mathbb{E}Y_{kn} - \mathbb{E}Y_{sn}$ where $X_{in}$ is the $i_{th}$ greatest of $n$ draws from $X$, and $k>s$. In fact, I can be even more specific, as I am interested in the difference between the expectation of the mean and the extreme values. So, does $Var(X) < Var(Y) \to \mathbb{E}X - \mathbb{E}X_{1n} < \mathbb{E}Y - \mathbb{E}Y_{1n}$ (and the same for the absolute value of the difference between EX and EXnn)

Also, for the general case: would it be true that for any distributions X and Y defined over the same range, that Variance(X) < Variance (Y) implies $\mathbb{E}X_{kn} - \mathbb{E}X_{sn} < \mathbb{E}Y_{kn} - \mathbb{E}Y_{sn}$

• I fear this question may not be well-posed (yet). For example, some order statistics may have well-defined means, even when the mean (or variance) of the distribution is undefined. Second, the parameterization of a particular family of distributions may not be closely related to the variance. Are you, perhaps, interested in some restricted class of distributions, such as location-scale families or similar? More details would be welcome. – cardinal Oct 11 '11 at 13:37
• @cardinal: I updated to hopefully be more clear, and now I am looking through your link. – OctaviaQ Oct 11 '11 at 15:35
• Consider a $\mathrm{Ber}(p)$ random variable. Then $\mathbb E X - \mathrm X_{1n} = p - p^n$ and this is not monotonic in $p(1-p)$. – cardinal Oct 11 '11 at 18:09

One shouldn't expect general relationships to hold between moments and order distributions.

As an example, consider the family of discrete distributions $f_{(\alpha,\beta,p)}$ for which $\Pr(\alpha)=p$, $\Pr(-\beta)=(1-p)/2$, and $\Pr(0)=(1-p)/2$ ($\alpha$ and $\beta$ are non-negative; $0 \le p \le 1$). I plan to choose $\alpha$ and $\beta$ later and right now just want to investigate whether the variance $V$ and expected range of $n=2$ independent draws, $E$, are increasing or decreasing. ($E$ is, of course, the difference between the first and second order statistics.) It is elementary to compute

$$\frac{dV}{dp} = \frac{(2\alpha+\beta)^2 - \beta^2}{4} - \frac{p}{2}(2\alpha+\beta)^2$$

and

$$\frac{dE}{dp} = 2\alpha - (4\alpha+\beta)p.$$

Consequently we can find simple formulas for where they attain their maxima. Plotting the differences of these locations, as a function of $\alpha/\beta$, suggests looking at the case $\alpha=\beta$: the maximum of $V$ is attained when $p=4/9$ and the maximum of $E$ is attained when $p=2/5$. Consequently, if we fix $\alpha=\beta$ and vary $p$ from $4/9$ to $2/5$, the variance is decreasing while the expected range is increasing. The differences are small but real; for instance, with $\alpha=\beta=1$, the variance and expected range at $p=4/9$ are 0.694444 and 0.895062, respectively, whereas at $p=2/5$, they are 0.69 and 0.9, respectively: one has decreased and the other has increased.

• (+1) For this answer and the other similar one. By happenstance the examples you give in both were quite close to ones I had contrived, but did not have a chance to post. It's always a nice present to see myself thinking along the same lines as you have in the answers you post. :) – cardinal Oct 11 '11 at 18:24