Maximum of Three

First find the CDF of the maximum.

$P(\max \le x)=x^3$, so the density is $3x^2$.

1. 1. Find the distribution

   All three observations must be at most $x$.

   $$F_M(x)=P(M\le x)=x^3$$

2. 2. Differentiate and integrate

   The density is $3x^2$ on $[0,1]$.

   $$\mathbb{E}[M]={\int }_0^{1}x\cdot 3x^{2}dx=\frac{3}{4}$$

Answer: 3/4