Expected Value of Max of IID Variables

Question

What is the expected value of the maximum of 500 IID random variables with uniform distribution between 0 and 1?

I'm not quite sure of the technique to go about solving something like this. Could anyone point me the right direction?

Thanks

Answer 1

Suppose the maximum is $X_{500}$, then $$P(X_{500}\le x)=P(X_i \le x ,i=1,2,...,500)$$ Note that this is so because if the maximum is less than $x$ , then every other order statistic is less than $x$. Now since the $X_i's$ are IID, it follows that; $$P(X_{500}\le x)=\prod_{i=1}^{500} P(X_i\le x)=x^{500}$$ which is the CDF and so the PDF is $500x^{499}$ (which is obtained by differentiation). Now the expected value of the maximum is found as follows; $$E[X]=\int _0^1 x (500x^{499})dx=\int _0^1 500x^{500}dx=\frac {500}{501}$$

Answer 2

There is one trick useful for uniform distribution.

The trick: If you have 500 independent uniform random variables on $[0;1]$ then you may think that you have 501 independent uniform random variables on a circumference with unit length. You just consider the first random variable as a cut-point, which transforms the circumference into a unit segment $[0;1]$.

The average distance between points on the circumference is $1/501$ and ...

the average length of 500 intervals is equal to $500/501$.

Answer 3

George Pólya says "Is there a simpler problem of the same type that you do know how to solve?"

The way to go about solving something like this is to look at simpler cases first. What's the expected value of the maximum of one IID uniform random variable? What is the expected value of the maximum of two?

Mouse over if you want the actual answer.

I believe the answer is $\frac{500}{501}$.

Answer 4

Let $Y$ be the maximum. Then $P(Y\le y)=y^{500}$ (if $0\le y\le 1$). So now we know the cumulative distribution function of $Y$, and therefore the density, and therefore the mean.

Answer 5

This can be solved by taking repeated expectations. Fix $n$ (500 in your example), and let $Y_k = \mathrm{max}\,\left\{X_1, \ldots, X_k\right\}$ for $1 \leq k \leq n$. Set $R_k = \mathbb E\left(Y_k^{n + 1 - k}\right)$.

Then $R_1 = \int_0^1 x^n\,dx = \frac{1}{n + 1}$, and for $k > 1$, $$R_k = \mathbb E\left(\mathbb E Y_k^{n+1-k}\right|Y_{k-1}).$$

Now, $Y_k$ is $X_k$ if $X_k \geq Y_{k-1}$, and $Y_{k-1}$ otherwise. Thus

$$ R_k = \mathbb E \left(\int_0^{Y_{k1}}Y_{k-1}^{n+1-k}\,dx + \int_{Y_{k-1}}^1x^{n+1-k}\,dx\right) = \mathbb E\left(Y_{k-1}^{n+2-k} + \frac{1 - Y_{k-1}^{n+2-k}}{n+2-k}\right),$$ which in turn gives $$R_k = \frac{1}{n+2-k}+\frac{n+1-k}{n+2-k}R_{k-1}.$$

Using this expression for $R_k$ in terms of $R_{k-1}$, it is a matter of basic algebra to prove via induction that

$$R_k = \frac{1}{n+2 - k} + (n+1 -k)\sum_{i=n+2-k}^n \frac{1}{i(i+1)}.$$

Then $$\mathbb E Y_n = R_n = \frac{1}{2} + \sum_{i=2}^n \frac{1}{i(i+1)} = \sum_{i=1}^n \frac{1}{i(i+1)},$$ and it is again easy to prove by induction that the sum on the right equals $\frac{n}{n+1}$.