# Expectation of maximum of n i.i.d random variables

I have $$n$$ i.i.d. random variables, $$X_1,..., X_n$$ which follow some arbitrary distribution. Based on experiments in Python with various distributions, it seems that $$\mathbb{E}(\max(X_1,...,X_n))$$ is a linear (or seemingly close to linear) function of $$\mathbb{E}(X_i)$$. It is indeed linear for some examples where it is possible to get a closed form solution for $$\mathbb{E}(\max(X_1,...,X_n))$$ or a good approximation.

Expected value of $\max\{X_1,\ldots,X_n\}$ where $X_i$ are iid uniform.

Expectation of the maximum of i.i.d. geometric random variables

I wonder if this is the case more generally? Is there some way to prove it?

• For continuous non-negative IIDRVs $X_n$ with arbitrary common CDF $F(x)$ we cannot say much better than $\mathbb{E}(M)=\int_0^\infty (1-F(x)^n)dx$, where $M=\max\{X_1,\dotsc, X_n\}$, as written in the second link. The answer by zjm covers the general case. – Nap D. Lover Jul 27 at 2:56

This is question is related to something called order statistics in probability theory. You can read more about them here. For $$n$$ iid variables $$X_1, …, X_n$$ with cumulative density function $$F$$ and density function $$f$$, the density function of the maximum is:

$$f_{max}(x) = nf(x)F(x)^{n-1}$$

Then this implies the expected value would be:

$$E[X_{max}] = \int_{-\infty}^{\infty} nxf(x)F(x)^{n-1} dx$$

I don't see any linear relationship here in general between $$E[X_{max}]$$ and $$E[X]$$

A general technique to get a bound that is often pretty decent is to use the MGF if you have it: for all $$t\geq 0$$: \begin{align} \exp(t\mathbb{E}[\max_i X_i])&\leq \mathbb{E}[\exp(t\max_i X_i)]\\ &\leq\mathbb{E}[\sum_{i=1}^n \exp(t X_i)] \\ &= n\mathbb{E}[\exp(t X)], \end{align} so $$\begin{equation} \mathbb{E}[\max_i X_i]\leq \frac{\log(n\mathbb{E}[\exp(tX)])}{t}. \end{equation}$$ You can then optimize in $$t\geq 0$$ to get a decent upper bound. For instance, doing this with Gaussians with variance $$\sigma^2$$ would show $$\mathbb{E}[\max_i X_i]\leq \sigma\sqrt{2\log n}$$, which turns out to be right up to a constant.

It seems the idea generalizes. Say $$E(max(X_1,X_2,...,X_n))=X_j =E(X_j \geq X_1 ,X_j >X_2,...,X_j>X_n)$$ Then(By assumed independence of the $$X_i$$) Let $$f_i$$ be the pdf. of $$X_i$$ :$$P(X_j \geq X_1 ,X_j >X_2,...,X_j>X_n)= P(X_j >X_1)P(X_j > X_2).....P(X_j >X_n) = (\int_{- \infty}^{x_j} f_idx_i)^n$$ and then $$f_x= \frac {d}{dt}(F_x)=$$ ( By chain rule) $$n( \int_{-\infty}^{\infty}f_idx_i)^{n-1}$$ , so that the expected value is $$n\int_{-\infty}^{\infty} x \int_{x_j}^{\infty} f_idx_i$$

I don't see how this is linear on $$E(X_i):=\int _{-\infty}^{\infty}x_if_idx_i$$