Calculating Expectation Value of a dependent random variable

Question

Given the density function $f_{XY}(x,y) = 6x$y, when $0 \leq x \leq 1$ and $0 \leq y \leq \sqrt{x}$,

how do I calculate $\mathbb{E}(Y)$?

I've tried using the law of Total Expectation, but I'm not sure I got a logical result.

Answer 1

The marginal density of $Y$ is given by $F_Y(y)=\int f_{XY} (x,y) dx=\int_{y^{2}}^{1} 6xy dx$ since $y \leq \sqrt x$ translates to $x \geq y^{2}$. Hence $f_Y(y)= 3y(1-y^{4})$ for $ 0 \leq y \leq 1$. Now $EY=\int_0^{1} yf_Y (y) dy$. I will let you carry out the integration.

You can also find $EY$ directly from the formula $EY=\iint yf(x,y)dy$ where the integral is a double integral over the given region.