$x_1$ and $x_2$ are uniformly distributed in $[0,1]$ with correlation $\rho$.
What is $E[min(x_1,x_2)]$?
I know in a simple case when $x_1$ and $x_2$ are independent, then I can get the cdf of $min(x_1,x_2)$ first and then break the joint distribution down. Next get the pdf and then the expectation of $min(x_1,x_2)$.
But I have no idea how to break the cdf if they are no longer independent.
Best Answer
HINT
The formula for the CDF of the minimum of two iid variables $$ P(X_{min} \le x) = 1 - P(X_{min}>x) = 1-P(X>x)^2$$ generalizes to $$ P(X_{min} \le x) = 1 - P(X_{min}>x) = 1 - P(X_1>x, X_2>x) $$ in terms of the joint distribution of $(X_1,X_2).$
EDIT
Actually, come to think of it, is the problem even well-defined? "Two uniform variables with correlation $\rho$" does not define a unique joint distribution. So unless the distribution of the min is somehow independent of everything except the correlation, I don't know that this is doable. Someone correct me if I'm wrong. Does it only depend on $\rho$ (or is there some obvious implied choice of the joint distribution like bivariate normal in the case with two normals?)
EDIT 2
For example, if $U_1\sim U(0,1)$ and conditional on $U_1,$ $U_2 = U_1$ with probability $1/2$ and $1-U_1$ with probability $1/2,$ then $U_1$ and $U_2$ are uncorrelated $U(0,1)$'s. We can compute $E(\min(U_1,U_2)) = 3/8.$ However, if $U_1$ and $U_2$ are independent $U(0,1)$'s (also uncorrelated of course), then $E(\min(U_1,U_2)) = 1/3.$ So the correlation is not enough to determine this quantity.