Is it true that $\text{Var}(X)=\mathbb E[(X-Y)_{+}^{2}]$ for $Y$ iid to $X$

Question

Is it true that $\text{Var}(X)=\mathbb E[(X-Y)_{+}^{2}]$ if $X$ and $Y$ are independent identically distributed?

Note that $(X-Y)_{+}^{2}$=$(X-Y)^{2}\chi_{\{X\geq Y\}}+0\chi_{\{X< Y\}}$

Why is this true? Any proof?

Answer 1

Hint: $E(X-Y)^{2}_{+}=E(X-Y)^{2}I_{X \geq Y}=E(Y-X)^{2}I_{Y \geq X}$ because $(X,Y)$ is i.i.d. Hence, $E(X-Y)^{2}_{+}=E(X-Y)^{2}I_{Y \geq X}$ which shows that $E(Y-X)^{2}I_{Y \geq X}=\frac 1 2 E(X-Y)^{2}$. Now you can complete the proof by expanding $(X-Y)^{2}$.

[It should be noted that $E(X-Y)^{2}I_{X=Y}=0$]