# Expected value of $\max\{X_1,\ldots,X_n\}$ where $X_i$ are iid uniform. [duplicate]

Let $X_i\sim\mathrm{Uniform}(0,\theta)$ be iid, what is $E[\max\{X_1,\ldots,X_n\}]$? Apparently the answer is $$\frac{n}{n+1}\theta,$$ but I do not see why?

It seems intuitive in that you would "expect" them to be spaced out evenly, hence the maximum would be $\frac{n}{n+1}$-of-the-way along the interval, but how can we prove this mathematically? I feel like it should be simple, but evidently $$E[\max\{X_1,\ldots,X_n\}]\neq\max\{E[X_1],\ldots,E[X_n]\}.$$ Thanks.

## 2 Answers

Your intuition is correct.

To see this mathematically, suppose $X_1, \ldots, X_n$ are independent and uniformly distributed and $M_n = \max(X_1,X_2,\ldots,X_n).$

The distribution function of the maximum is the joint probability that $X_k \leq x$ for all $k.$ This is a product of marginal probabilities since the variables are independent.

$$F_M(x)=P(M_n \leq x) =P(X_1 \leq x,\ldots,X_n \leq x)=(x/\theta)^n$$

for $0 \leq x \leq \theta$.

Also $F_M(x) = 0$ for $x < 0$ and $F_M(x) = 1$ for $x > \theta$.

Hence, the probability density function on $[0,\theta]$ is

$$f_M(x)=F'_M(x)=nx^{n-1}\theta^{-n}$$

and the expected value is

$$E(M_n) = \theta^{-n}\int_0^{\theta}xnx^{n-1}\, dx=\frac{n}{n+1}\theta.$$

• What I don't get is why the PDF $p(m) \neq p(X_{1} \leq m, ..., X_{n} \leq m)$ and instead it is the CDF? [Note: $m$ is the maximum's variate -- same as $x$]. Feb 29 at 12:38

Here is a brute force answer:

With appropriate scaling, we can take $\theta =1$ without loss of generality.

Let $\sigma$ be a permutation of $1,...,n$ and $A_\sigma = \{ x \in [0,1]^n | x_{\sigma_1} > \cdots > x_{\sigma_n} \}$.

Note that $p ([0,1]^n \setminus (\cup_\sigma A_\sigma ) ) = 0$, and $E (\max(X_1,...,X_n) 1_{A_\sigma} (X_1,...,X_n) ) = E (\max(X_1,...,X_n) 1_{A_{\sigma'}}(X_1,...,X_n) )$ for any two permutations $\sigma, \sigma'$.

Hence $E \max(X_1,...,X_n) = n! E (\max(X_1,...,X_n) 1_{A_\sigma} (X_1,...,X_n))$ for any $\sigma$.

Let $\sigma$ be the identity permutation, then if $(X_1,...,X_n) \in A_\sigma$ we have $\max(X_1,...,X_n) = X_1$ and so \begin{eqnarray} E (\max(X_1,...,X_n) 1_{A_\sigma} (X_1,...,X_n)) &=& \int_{x_1=0}^1 \int_{x_2=0}^{x_1} \cdots \int_{x_n=0}^{x_{n-1}} x_1 dx_n \cdots dx_1 \\ &=& {1 \over (n+1)(n-1)!} \end{eqnarray} Hence $E \max(X_1,...,X_n) = { n! \over (n+1)(n-1)!} = {n \over n+1}$.