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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.

Mary
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    You should not look at the variance of X or Y bur rather the variance of them conditional on being lower. In any case, I guess it is simpler to just evaluate the integral and compute the variance analitically. – Pedro Ignacio Martinez Bruera Nov 04 '21 at 05:12
  • It's not correct to write "If $\min{X,Y} = X$ then is true, since $Var (X) = Var (X)$", because is this case you imply conditional variance but write just variance. Example: by this logic: "Bern(p) takes values 0 and 1, if it's equal to 0, then Var0=0, in other case Var1=0, hence Var(Bern(p))=0", that is false, of course. – Botnakov N. Nov 04 '21 at 07:03
  • https://math.stackexchange.com/q/4129387/321264 – StubbornAtom Nov 04 '21 at 07:34

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

Botnakov N.
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