Let $f_{X,Y}(x,y)=\frac{3}{16}xy^2$, $0 \leq x \leq 2$, $0 \leq y \leq 2$. be the joint PDF of X and Y. Find the marginal PDF of $f_X$ and $f_Y$.
Problem 1: Compute $P(X<Y)$
Question 1: I got $\frac{3}{5}$, is it right?
Problem 2: Compute E(Y)
Question 2: I got $E(Y)= \int_0^2 \int_y^2 y \cdot y^2 \,dx \,dy$ , is it the right formula?
Problem 3: Compute Cov(X,Y)
Question 3: \begin{align} &E[(X-E[X])(Y-E[Y]) \\ & = E[(X-\frac{4}{3})(Y-4)] \\ & = E[XY-\frac{4}{3}Y -4X+\frac{16}{3}] \\ & = E[XY]-\frac{4}{3} E[Y]- 4 E[X]+\frac{16}{3} \\&= \int_0^2 \int_0^2 xy \frac{16}{3}xy^2 \,dx\,dy-\frac{4}{3} E[Y]-4E[X]+\frac{16}{3} \end{align} Am I on the right track?
Since the density is just a function of $x$ times a function of $y,$ the two random variables $X$ and $Y$ are independent. That implies the marginals must be $$ f_X(x) = \text{constant}\cdot x, $$ $$ f_Y(y) = \text{constant}\cdot y^2. $$ If you multiply the two constants you should get $3/16$ if the density you gave originally is right, but that doesn't tell you what the two constants separately are. But they can be found by saying the integral of either of these from $0$ to $2$ is $1.$
Since they are independent, the covariance is $0.$
Then you have $$ \operatorname EY = \int_0^2 yf_Y(y)\,dy, $$ but you can also find this by using the joint density, as you did.
\begin{align} \Pr(X<Y) & = \iint\limits_{(x,y)\,:\,0\,\le\, x\,<\,y\,\le\,2} \!\!\! \cdots \,d(x,y) \\[8pt] & = \int_0^2 \left( \int_x^2 \cdots \, dy \right) \, dx \\[8pt] & = \int_0^2 \left( \int_0^y \cdots \, dx \right) \, dy. \end{align} Doing these integrals in either order should be a tractable problem. (There are cases in which one of them is routine and the other is intractable, but in this case either should be routine.)