R.v. $X$, $E[X^2]<\infty$. How to prove that $\mathrm{Var}[X\land a] \le \mathrm{Var}[X]$ for any real number $a$?

Question

As the title says, It looks simple, however not very intuitive. I finished it with the step function converging to it. However, it seems too complicated and I wonder would there be some elementary proof.

Answer 1

If $a\geq \mu=\mathrm E(X)$ we have $$ \mathrm {Var} (X\wedge a) \leq \mathrm E[(X\wedge a-\mu)^2] \\ = \mathrm E[(X-\mu)^2\mathbf 1_{X<a}] +\mathrm E[(a-\mu)^2\mathbf 1_{X\geq a}] \\ \leq \mathrm E[(X-\mu)^2\mathbf 1_{X<a}]+ \mathrm E[(X-\mu)^2\mathbf 1_{X\geq a}] =\mathrm{Var}(X). $$

If $a\leq \mu$ we have $$ \mathrm {Var} (X\wedge a) \leq \mathrm E[(X\wedge a-a)^2] \\ = \mathrm E[(a-X)^2\mathbf 1_{X<a}] +\mathrm E[(a-a)^2\mathbf 1_{X\geq a}] \\ \leq \mathrm E[(\mu-X)^2\mathbf 1_{X<a}] \leq \mathrm {Var}(X). $$

Answer 2

Since $X=X\wedge a+(X-a)^+$, then \begin{align*} \mathsf{var}[X]&=\mathsf{var}[X\wedge a]+\mathsf{var}[(X-a)^+]+2\cdot\mathsf{cov}(X\wedge a,(X-a)^+)\\ &\ge \mathsf{var}[X\wedge a]+2\cdot\mathsf{cov}(X\wedge a,(X-a)^+). \end{align*} Meanwhile \begin{align*} &\mathsf{cov}(X\wedge a,(X-a)^+)=\mathsf{cov}(X\wedge a-a,(X-a)^+)\\ &\qquad =-\mathsf{cov}((X-a)^-,(X-a)^+)\\ &\qquad=-\mathsf{E}[(X-a)^-(X-a)^+]+\mathsf{E}[(X-a)^-]\mathsf{E}[(X-a)^+]\\ &\qquad=\mathsf{E}[(X-a)^-]\mathsf{E}[(X-a)^+]\ge 0, \end{align*} therefore $$ \mathsf{var}[X]\ge \mathsf{var}[X\wedge a]. $$