Let $U$ and $V$ be independent and $\text{unif}[0,1]$ and let $X = \text{min}(U,V)$ and $Y = \text{max}(U,V)$. Find $\text{Cov}[X,Y]$ and comment on its sign.
I know that $\text{Cov}[X,Y]=E[XY]-E[X]E[Y],$ so I need both of the mean-terms. Letting $g(X,Y)=XY$ I get
$$E[XY]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(x,y)f_{X,Y}(x,y) \ dxdy=...?$$
What I don't understand is the min and max functions, I know what they mean but not how to do algebra with them. How do I find the bounds of my integrals?
Am I correct to say that since $U,V\sim\text{Unif}[0,1]$ and independent, then the variables $X$ and $Y$ are also $\sim\text{Unif}[0,1]$ and independent? If this is true, then $f_{X,Y}(x,y)=f_X(x)f_Y(y)=1\cdot 1=1$ and $E(X)=E(Y)=\frac{1}{2}.$ So
$$E(XY)=\int_{0}^{1}\int_{0}^{1}xy\ dxdy=\frac{1}{4}.$$
Finally: $$\text{Cov}[X,Y]=E[XY]-E[X]E[Y]=\frac{1}{4}-\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}-\frac{1}{4}=0.$$
This means that $X$ and $Y$ is independent. But this does feel circular because I assume that $X$ and $Y$ are independent and then I prove what I'm assuming.
Hmm, I got an answer of $Cov(X,Y) > 0$... Am I right or wrong?
First of all, $X,Y$ are not independent, because as @StubbornAtom pointed out, $P(Y>X) = 1$. Alternatively, consider that $P(X>{1\over 2}) = P(U>{1\over 2}, V>{1\over 2}) = {1\over 4}$ but $P(X > {1\over 2} | Y < {1\over 2}) = 0$, directly demonstrating dependence.
So $Cov(X,Y)$ is not trivially zero. To find its value:
$U, V$ are $\text{Unif}[0,1] \Rightarrow E[U] = E[V] = {1 \over 2}$.
$U,V$ are independent $\Rightarrow E[UV] = E[U]E[V] = {1 \over 4}$.
Regardless of $U > V$ or $U < V$, we have $UV = XY$ always. So $E[XY] = E[UV] = {1 \over 4}$.
From integration (not shown here), $E[Y] = {2\over 3}, E[X] = {1\over 3}$.
$Cov(X,Y) = E[XY] - E[X]E[Y] = {1\over 4} - {1\over 3}{2\over 3} = {1\over 36} \neq 0$.