$x_1$ 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.



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 $(X_1,X_2).$


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?)


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.

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  • $\begingroup$ Yes, this is exactly what I'm thinking of! $\endgroup$ – James LT Dec 12 '17 at 1:53
  • $\begingroup$ Actually in this special case, correlation=0 means independence, which is not true generally.. $\endgroup$ – James LT Dec 12 '17 at 1:53
  • $\begingroup$ In what special case? Uniforms with correlation zero implies independence? I don't think that's true for any marginal distribution. $\endgroup$ – spaceisdarkgreen Dec 12 '17 at 1:58
  • $\begingroup$ Like $U_1$ is $U(0,1),$ and conditional on $U_1,$ $U_2 = U_1$ with probability $1/2$ and $1-U_1$ with probability $1/2$? That's a counterexample, right? $\endgroup$ – spaceisdarkgreen Dec 12 '17 at 2:11
  • $\begingroup$ And also in this counterexample, we have $E(\min(U_1,U_2)) = 3/8 $ whereas it's $1/3$ in the independent case, which shows the problem is not well-posed (in the case of zero correlation, anyway). $\endgroup$ – spaceisdarkgreen Dec 12 '17 at 3:06

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