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Let $n$ be a fixed positive integer. For a distribution $F$ on positive real numbers, let $a_n(F)$ be the expected value of the maximum of $n$ i.i.d. random variables drawn from $F$, and let $a_{n+1}(F)$ be the expected value of the maximum of $n+1$ i.i.d. random variables drawn from $F$. What is $$\max_{F}\frac{a_{n+1}(F)}{a_n(F)}?$$

For example, if $F$ is the uniform distribution on $(0,1)$, then $a_n(F)=\frac{n}{n+1}$ and $a_{n+1}(F)=\frac{n+1}{n+2}$. My guess is that the maximum is $\frac{n+1}{n}$. How can we show it, or is there a theorem stating this?

pi66
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1 Answers1

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If $X_1, \ldots, X_n$ are iid on the positive reals with cdf $F$, the cdf of $M_n = \max(X_1, \ldots, X_n)$ is $$ \mathbb F_n(x) = \mathbb P(M_n \le x) = \mathbb P(\text{all } X_n \le x) = F(x)^n $$ and so $$ \mathbb E[M_n] = \int_0^\infty (1 - F(x)^n)\; dx $$

Now for any $0 \le t \le 1$ and positive integer $n$, $$1 - t^{n+1} \le \frac{n+1}{n} (1 - t^n)$$ since $g(t) = (n+1) (1 - t^n) - n (1-t^{n+1})$ is nonincreasing on $[0,1]$ (its derivative is $n(n+1)(t^{n-1}-t^n) \le 0$). Thus it is indeed true that $$\mathbb E[M_{n+1}] \le \frac{n+1}{n} \mathbb E[M_n]$$

To see that the bound is sharp, consider a Bernoulli distribution with parameter $p \to 1-$. We have $$\lim_{p \to 1-}\frac{\mathbb E[M_{n+1}]}{\mathbb E[M_n]} = \lim_{p \to 1-} \frac{1-p^{n+1}}{1-p^n} = \frac{n+1}{n} $$

Robert Israel
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