Randomness maximised

From the pdf of $X$ , we know that $X_i \sim \mathcal{U}(0,1)$ $\forall i$ . Given this, the cdf of $Y_n$ , where $Y_n = \max\limits_{1 \leq i \leq n} {X_i}$ is:

$F(y) = \left(\int_0^y dt\right)^n = \left( t |_{0}^y\right)^n = y^n, 0 \leq y \leq 1$

The pdf of $Y_n$ is given by the derivative of the its cdf, which is $f(y) = ny^{n-1}, 0 \leq y \leq 1$ .

Thus, $E \left[ Y_n \right] = \int_0^1 ny^n dy = \frac{ny^{n+1}}{n+1} \bigg|_0^1 = \frac{n}{n+1}$ . And given this, $E_n = \left( \frac{n}{n+1} \right)^n$ .

$\mathcal{E} = \lim\limits_{n\to\infty} \left( \frac{n}{n+1} \right)^n$

$= \lim\limits_{n\to\infty} \left( 1 - \frac{1}{n+1} \right)^n$

$= \lim\limits_{m\to\infty} \left( 1 - \frac{1}{m} \right)^{m-1}$

$= \lim\limits_{m\to\infty} \frac{\left( 1 - \frac{1}{m} \right)^m}{\left( 1 - \frac{1}{m} \right)}$

$= \frac{e^{-1}}{1} = \frac{1}{e}$

Thus, $\lfloor 1000 \mathcal{E} \rfloor = \lfloor \frac{1000}{e} \rfloor = 367$ .

Without loss of generality let us consider the case when :- $\displaystyle X_{n}>X_{n-1}>X_{n-2}>.....>X_{2}>X_{1}$

In this case $\displaystyle \text{max} \{X_{1},X_{2}...X_{n}\}=X_{n}$

Now calculating the expected value we have :- $\displaystyle \int_{0}^{1}\int_{X_{1}}^{1}\int_{X_{2}}^{1}......\int_{X_{n-1}}^{1} X_{n}\,dX_{n}\,dX_{n-1}\,....\,dX_{3}\,dX_{2}\,dX_{1}=\frac{n}{(n+1)!}$

(The integration is not very difficult. You can maybe take the case for n= 3 or 4 and then prove by induction).

Now there are $n!$ such ordered n-tuples of $X_{1},X_{2}...X_{n}$ . So there are $n!$ ways to arrange them in descending order. Each of them will have equal probabilities and hence equal expected values for the maximum of the n random variables. Hence by total probability we have the expected value is $\displaystyle n!\cdot \frac{n}{(n+1)!} = \frac{n}{n+1}$ .

So raising it to the nth power and taking the limit as $n\to\infty$ we have the answer as $\displaystyle \frac{1}{e}$ .

(Note we can also do it by considering the minimum of the n variables . In that case the integral would be a little easier. Then we can just substract it from 1 to get the expected value of maximum of n variables.)