What is the standard deviation of the expected maximum value of a set of n random numbers between 0 and 1?

Question

If I generate $n$ random numbers, $0 < x < 1$, I have found the expected maximum value is $\frac{n}{n+1}$. What is the standard deviation of that expected maximum value?

Answer 1

The $k$th order statistic of a sample of size $n$ from $U[0,1]$ has a Beta distribution with parameters $\alpha=k, \beta=n-k+1$, with mean $\dfrac{k}{n+1}$ and variance $\dfrac{k(n-k+1)}{(n+1)^2(n+2)}$

So the maximum (i.e. the $n$th order statistic) has a Beta distribution with parameters $\alpha=n, \beta=1$, with mean $\dfrac{n}{n+1}$ and variance $\dfrac{n}{(n+1)^2(n+2)}$

The standard deviation of the maximum is then $\sqrt{\dfrac{n}{(n+1)^2(n+2)}}$

Answer 2

You can use $Var(Y)=\mathbb E(Y^2)-[\mathbb E(Y)]^2$

It is $P(Y<x)=x^n$, where $Y$ is the maximum of independent and uniformly distributed random variables. Thus derivative is

$f_Y(x)=n\cdot x^{n-1}$

$$\mathbb E(Y^2)=\int_{-\infty}^{\infty} x^2\cdot f_Y(x) \,dx = \int_{0}^{1}x^2\cdot n\cdot x^{n-1} \, dx=n\int_{0}^{1} x^{n+1} \, dx$$

$$=n\cdot \left[ \frac{1}{n+2}\cdot x^{n+2} \right]_0^1 =\frac{n}{n+2}$$

Finally we have $Var(Y)=\frac{n}{n+2}-\left(\frac{n}{n+1}\right)^2=\ldots$