$Var(\min\{X,Y\}) \leq Var(X)$

extreme-value-analysisprobabilityrandom variablesstatisticsvariance

I have this problem, and the only thing that tells me is that X and Y are random variables that have finite variance, when $\min\{X,Y\} = X$, it is true, since $Var (X) = Var (X)$, I have the problem when I take the case $Var(\min\{X,Y\}) = Y$, clearly $Y <X$, but how do I prove that then $Var(Y)<Var (X)$?

I know that $Var(\min\{X,Y\}) = Var(\max\{X,Y\})$ when they have the same distribution or are symmetric, and I also read that $Var(\min\{X,Y\}) \leq Var(X) + Var(Y)$ but the latter I do not know how to prove it or how are the restrictions of such a statement, I would appreciate your help.

Best Answer

The claim is not true. Put $X=0$ and $Y = U[-1,0]$,.i.e. uniform distribution on $[-1, 0]$.