Showing that $\mathrm{Var}(Y) \leq \mathrm{Var}(X)$

Question

Let $X$ be a random variable with $\mathbb{E}(X^2) <\infty$. Let $Y$ defined as $$Y=\begin{cases}X &\text{ if }& \mid X\mid\leq k \\ 0 &\text{ if }& \mid X\mid > k \end{cases}$$ Show that $\mathrm{Var}(Y) \leq \mathrm{Var}(X)$.

I tried solving it this way:

\begin{align}\mathrm{Var(Y)}&=\int (Y-\mathbb{E}(Y))^2d\mathrm{P}=\int_{\mid x \mid \leq k} (Y-\mathbb{E}(Y))^2d\mathrm{P} +\int_{\mid x \mid > k} (Y-\mathbb{E}(Y))^2d\mathrm{P}\\ &=\mathbb{E}(Y^2)P(({\mid x \mid \leq k})\cup ({\mid x \mid > k}))\\ & \leq \mathbb{E}(Y)^2+\mathbb{E}(X^2)\\ &=\mathbb{E}(X^2)-\mathbb{E}(Y)^2\\ &\leq \mathrm{Var}(X)\end{align}

Is this correct?

Answer 1

Try to do it using definiton : $$\mathbb{Var}(X) = \int_{\mathbb{R}} (X-\operatorname{E}X)^2 \mu(dx) \le \int_{{[-k,k]}}(X - \operatorname{E}X)^2 = \dots$$

Answer 2

I am not sure the statement is true.

Take a discrete variable $X$ such that:

• $X=0.8$ with probability $0.5$ and $X=1.0$ with probability $0.5$.

We have $E[X]=0.9$, $E[X^2]=0.82$ and $Var(X)=0.01$.

Now take $k=0.9$. Than:

• $Y=0.8$ with probability $0.5$ and $Y=0$ with probability $0.5$.

This leads to $E[Y]=0.4$, $E[Y^2]=0.32$ and $Var(Y)=0.16$.

So the variance has in this case incremented: $Var(Y) \ge Var(X)$. Is the counterexample correct ?