# Expected value of maximum of i.i.d. random variables

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?

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 $$\lim_{p \to 1-}\frac{\mathbb E[M_{n+1}]}{\mathbb E[M_n]} = \lim_{p \to 1-} \frac{1-p^{n+1}}{1-p^n} = \frac{n+1}{n}$$
• How does your formula for $\mathbb E[M_n]$ follow from the definition $\mathbb E[X]=\int_0^\infty xf(x)dx$? – pi66 Jan 25 at 17:02
• I think you get $xF(x)^n\mid_0^\infty - \int_0^\infty F(x)^n dx$. How do you take care of the first term? – pi66 Jan 25 at 18:18
• Use $F(x)^n - 1$ instead of $F(x)^n$. Then $\left.x (F(x)^n-1)\right|_0^\infty = 0$. – Robert Israel Jan 27 at 1:51
• Sorry, I'm still confused. What $f(x)$ do you use then? And in $x(F(x)^n-1)$, if you take $x=\infty$, you get $\infty\cdot 0$ which is undefined, don't you? – pi66 Jan 27 at 10:33