# 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. Jul 27, 2019 at 2:56

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 $$$$\mathbb{E}[\max_i X_i]\leq \frac{\log(n\mathbb{E}[\exp(tX)])}{t}.$$$$ You can then optimize in $$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.

• This is really nice. Do you have a book or article reference for it? Jun 28, 2023 at 15:44

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 $$E[X]$$

Note that it is not crucial to choose the exponential function in J.G.'s answer. The idea indeed generalizes.

Indeed let $$f(x) = |x|^q$$, with $$q \geq 1$$ arbitrary, and let $$M = \max\{X_1, \dotsc, X_N\}$$. By Jensen's inequality, it holds that $$f \left( \mathbb E \bigl[ M \bigr] \right) \leq \mathbb E \Bigl[ f\bigl(M \bigr) \Bigr] \leq \mathbb E \left[ \sum_{n=1}^{N} f\bigl( X_n \bigr) \right] = N \mathbb E \Bigl[ f\bigl( X_1 \bigr) \Bigr],$$ and so we obtain, noting that the inverse function $$f^{-1}$$ is increasing, $$\mathbb E \bigl[ M \bigr] = f^{-1} \Bigl( f \left(\mathbb E \bigl[ M \bigr]\right) \Bigr) \leq f^{-1} \left(N \mathbb E \Bigl[ f\bigl( X_1 \bigr) \Bigr]\right) = N^{\frac{1}{q}} \left(\mathbb E \Bigl[ |X_1|^{q} \Bigr] \right)^{\frac{1}{q}}$$ Thus, even if $$X_1$$ does not have an exponential moment, a good bound can be obtained. Furthermore, notice that the more moments $$X_1$$ has, the better the dependence on $$N$$ in the bound on $$\mathbb E[M]$$ is.

Note that, as in J.G.'s answer, this approach does not require independence, only that the random variables are identically distributed.