The random variables X and Y have a joint density function given by as below. T= min(X,Y) Find the ET=?

Question

The random variables $X$ and $Y$ have a joint density function given by:

$$f(x,y) = \begin{cases} Cxy & \textrm{if }0 \leq x \leq 1, &0\leq y \leq 1\\ 0 & \textrm{elsewhere} \end{cases}$$

$T= \min(X,Y)$. Find the $ET=?$

I can find the constant $C=4$ from the area of the plane, and I know if $T=\min(X,Y)$ then $F(t)= 1-P(x>t)P(y>t)$ so what now? How can i found the expectation of $T$?

Answer 1

First calculate the normalizing constant $C$ obtaining that

$$f(x,y)=4xy=2x\cdot 2y=f_X(x)\cdot f_Y(y)$$

this shows the two marginal densities and consequently the two marginals CDF's are

$$F_X(x)=x^2$$

$$F_Y(y)=y^2$$

then calculate T density

$$f_T(t)=4t(1-t^2)$$

and finally its expectation

$$\mathbb{E}[T]=\int_0^1 4t^2(1-t^2)dt=\frac{8}{15}$$