Let X and Y have a joint pdf given by
$f_{x,y}(x,y) = \begin{cases} 1 & \text{if } 0<y<1,\text{ } y-1<x<1-y \\ 0 & \text{otherwise} \end{cases}$.
(a) Find Cov(X,Y) and Corr(X,Y). Are X and Y independent?
(b) Find Cov(|X|,|Y|) and Corr(|X|,|Y|).
(c) Find Cov(3|X|, 4|Y|) and Corr(3|X|, 4|Y|).
(d) Find Var(3X - 2Y).
So I know Covariance is E(XY)-E(X)E(Y). And since this is the indicator function, E(XY) should be 1*P(0 < y < 1, y-1 < x < 1-y) + 0*P(otherwise), which is just P(0 < y < 1, y-1 < x < 1-y). From here, I'm not sure if I take the double integral of this function with parameters listed above, or if I am taking the wrong approach. Not sure how to calculate E(X) or E(Y) either. Once I figure this part out, how absolute values will play a role?
A picture is (to me) essential. Draw the line $x+y=1$, the line $y=x+1$. Our pairs $(X,Y)$ live in the region that lies below each of these two lines, and above the $x$-axis. The region is a triangle of area $1$. It has corners $(1,0)$, $(0,1)$, and $(-1,0)$. The hard part is now finished.
(a) We want $E(XY)-E(X)E(Y)$. By symmetry we have $E(XY)=0$ and $E(X)=0$, so the covariance is $0$, and therefore so is the correlation coefficient.
But suppose we don't notice. Then to find $E(XY)$ we need to find the integral of $xy$ over our region.
The region naturally breaks up into the part to the left of $0$ and the part to the right. We get $$E(XY)=\int_{x=-1}^0 \left(\int_{y=0}^{x+1} xy\,dy\right)\,dx+ \int_{x=0}^1 \left(\int_{y=0}^{1-x} xy\,dy\right)\,dx .$$ Calculate. We get $0$.
The random variables $X$ and $Y$ are uncorrelated. They are not independent.
The non-independence is obvious: if we know $X$ is big, then $Y$ must be small. To prove independence fails with minimal computation, note that $\Pr(X\gt 0.9)$ is non-zero, as is $\Pr(Y\gt 0.9)$. But $\Pr((X\gt 0.9)\cap (Y\gt 0,9))=0$.
(b) For the covariance of $|X|$ and $|Y|$, we will need to find $E(|X||Y|)-E(|X|)E(|Y|)$. For the correlation coefficient, we may also need the variance of $|X|$ and of $|Y|$, so we will also need $E(|X|^2)$ and $E(|Y|^2)$. So a whole lot of expectations, a whole lot of integrals.
For $E(|X||Y|)$ we need to integrate $|x||y|$ over our triangle, since the density function is $1$ in the triangle and $0$ elsewhere. When $x$ is between $-1$ and $0$, we have $|x|=-x$, and when $x$ is between $0$ and $1$, we have $|x|=x$. Thus $$E(|X||Y|)=\int_{x=-1}^0 \left(\int_{y=0}^{x+1} -xy\,dy\right)\,dx+ \int_{x=0}^1 \left(\int_{y=0}^{1-x} xy\,dy\right)\,dx .\tag{1}$$ By symmetry the two integrals are equal, so if we wish we can evaluate one of them and double.
We will also need $E(|X|)$ and $E(|Y|)$. For $E(|X|)$, we replace $-xy$ in the left integral of (1) by $-x$, and replace $xy$ in the right integral by $x$. For $E(|Y|)$, both integrands get replaced by $y$. Use of symmetry again cuts the work in half.
The integrals $E(X^2)$ and $E(Y^2)$ use the same kind of double integral.
(c) No more integration! Note that the covariance of $aU$ and $bV$ is $ab$ times the covariance of $U$ and $V$. The correlation coefficient situation is even simpler.
(d) We want $E((3X-2Y)^2)-(E(3X-2Y))^2$. Expand. Or else use a standard formula that expresses the variance of $aX+bY$ in terms of the variance of $X$, the variance of $Y$, and the covariance of $X$ and $Y$. Conveniently the covariance is $0$.