# Lower bound on the variance of the maximum of random variables.

I'm trying to figure out a lower bound on the variance of the follow: $$\max_i(x_1,x_2,...x_m)$$. Where $$x_i, i \in m$$ are independent random variables that shares the same variance $$\sigma_x$$.

I know we can upper bound the above quantity to be: $$\text{Var}[max_i(x_1,x_2,...x_m)] \leq \sum^m_i \text{Var}[x_i]$$, but what about a lower bound? Also, is there a guarantee that this is the tightest lower bound?

Also, can we find a bound for $$\text{Var}[max_i(a_1,a_2,...a_m),max_i(b_1,b_2,...b_m),...,max_i(z_1,z_2,...z_m)]$$

• Could it be the variance of one variable? Commented May 20, 2020 at 23:44
• @herbsteinberg I was thinking it could the variance of the minimum RV and because they are equal, then it's $\sigma_x$ but not sure about this. I'm also looking in mit.edu/~dbertsim/papers/MomentProblems/orderstatistics.pdf Commented May 20, 2020 at 23:54
• For $\{X_i\}$ i.i.d. with $P[X_i=1]=P[X_i=-1]=1/2$ we have $Var(X_i)=1$ for all $i$, but $Var(\max[X_1, ..., X_n]) = 1 - (1 - (1/2)^{n-1})^2 \rightarrow 0$ as $n\rightarrow\infty$. Commented May 21, 2020 at 2:09
• @xsari3x it is just the calculation that $\max[X_1, ..., X_n] = -1 w.p. (1/2)^n$. and 1 otherwise; to add to this discussion, max of $n$ iid standard gaussians has a variance that decays of order 1/(1+log(n)); this answer explains this phenomenon much better than I can, and also has a lower bound for this specific instance stats.stackexchange.com/questions/229073/…
– E-A
Commented May 21, 2020 at 5:52
• Just a suggestion : with markov : $Var(max) \geq a^2 \Big(1 - P(x_1 \leq a + E(max))^n\Big)$ for all $a > 0$.
– jvc
Commented May 23, 2020 at 18:47

Unfortunately, for $$m\ge 2$$, there is no lower bound in terms of the variances of $$X_i$$ (except the trivial one).
@Michael explained that the variance may vanish as $$m\to\infty$$, but this can happen even for fixed $$m\ge 2$$. Indeed, let $$\mathrm{P}(X_i=1/\sqrt{p(1-p)}) = 1- \mathrm{P}(X_i=0) = p$$ so that $$\mathrm{Var}(X_i) = 1$$. Then, $$\mathrm{P}\left(\max_{1\le i\le m} X_i=\frac{1}{\sqrt{p(1-p)}}\right) = 1- \mathrm{P}\left(\max_{1\le i\le m} X_i=0\right)= 1 - p^m,$$ so $$\mathrm{Var}\left(\max_{1\le i\le m} X_i\right) = \frac{p^m(1-p^m)}{p(1-p)}\to 0, p\to 0+.$$
Under the assumption that the variables are bounded by some $$C$$, it may be possible to get a lower bound depending on $$C$$ (and $$m$$).