Expectation of the maximum as the number of random variables goes to infinity?

Question

Suppose $v_1,v_2,...,v_n$ are $n$ i.i.d. continuous random variables with the range $[\underline v,\bar v]$, does the expectation of the maximum of the random variables $E[\max v_i, i=1,2,\dots,n]$ go to $\bar v$ as $n$ goes to infinity?

Thanks guys!

Answer 1

The CDF of $Y_n=\max_{1\le i\le n}\{v_i\}$ is

$$F_{Y_n}(t)=[F_v(t)]^n\rightarrow 1\{t=\bar v\}\text{ as }n\rightarrow \infty$$

which means that $Y_n\xrightarrow{d} \bar v$.Using the a.s. representation there is a sequence $\{Y_n'\}$ where each $Y_n'$ has the same distribution as $Y_n$ and $Y_n'\xrightarrow{a.s}\bar v$. Since $|Y_n'|\le |\underline v|\vee|\bar v|$, the bounded convergence theorem implies that

$$\lim_{n\rightarrow\infty}\mathbb{E}Y_n'=\mathbb{E}\lim_{n\rightarrow\infty}Y_n'=\bar v$$

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Alternatively, since $Y_n\le\bar v$ a.s. and for any $\epsilon>0$

$$\sum_{n=1}^\infty P\{Y_n\le \bar v-\epsilon\}=\sum_{n=1}^\infty [F_v(\bar v-\epsilon)]^n<\infty$$ $Y_n\xrightarrow{a.s.}\bar v$ and the same argument (about expectations) applies to $Y_n$ directly.

Answer 2

In general we have for a nonnegative random variable $X$: $$\mathbb{E}X=\int_{0}^{\infty}1-F_{X}\left(x\right)dx$$

Let $X_{1},X_{2},\cdots$ be iid and nonnegative with a CDF $F$ that satisfies:

• $F\left(a\right)=1$ for some $a>0$
• $F\left(x\right)<1$ if $x<a$

Then $M_{n}:=\max\left\{ X_{1},\cdots,X_{n}\right\} $ is a nonnegative random variable with CDF $F^{n}$ so:

$$\mathbb{E}M_{n}=\int_{0}^{\infty}1-F\left(x\right)^{n}dx$$ Here $\lim_{n\rightarrow\infty}1-F\left(x\right)^{n}=1$ if $x<a$ and $\lim_{n\rightarrow\infty}1-F\left(x\right)^{n}=0$ otherwise so that: $$\lim_{n\rightarrow\infty}\mathbb{E}M_{n}=\int_{0}^{a}1dx=a$$

Here we apply the dominated convergence theorem. The nonnegative integrands are dominated by the integrable function $1_{[0,a]}$.

This can also be used if the $X_{n}$ are not necessarily nonnegative but satisfy a condition like $X_{n}\in\left[\underline{v},\overline{v}\right]$ a.s..

In that case just look at the nonnegative $X_{n}-\underline{v}$.