Compute the Variance of $[min(X^2,\eta_0) - min(X^2,\eta_1)]$

Question

Suppose $$X\sim \mathcal{N}(0,\sigma^2)$$ There is another function $w$ given as $$W=min(X^2,\eta_0)-min(X^2,\eta_1)$$ where $\eta_0$ and $\eta_1$ are constant and $\eta_1> \eta_0$.

Now I want to ask what will be $var(W)$=??

Answer 1

Note that $$W = \begin{cases} \eta_0 - \eta_1 & \text{if } X^2 > \eta_1 \\ \eta_0 - X^2 & \text{if } \eta_1 \ge X^2 > \eta_0 \\ 0 & \text{if } \eta_0 \ge X^2 \end{cases}$$

Calculating the first two moments of $W$ now comes down to just some tedious but simple calculations of integrals containing the Gaussian density (You will presumably need partial integration). The solution will probably not be an elementary function, but contain the CDF $\Phi$ of the (standard) normal distribution.

Note that you can assume $\sigma^2 = 1$, since $$\frac{1}{\sigma^2}W = \min\left(\left(\frac{X}{\sigma}\right)^2, \frac{\eta_0}{\sigma^2}\right) - \min\left(\left(\frac{X}{\sigma}\right)^2, \frac{\eta_1}{\sigma^2}\right)$$

and $\frac{X}{\sigma} \sim \mathcal{N}(0, 1)$.

Answer 2

$E(W)=E(W1_{X^2<\eta_0})+E(W1_{\eta_0\leq X^2<\eta_1})+E(W1_{\eta_1\leq X^2})$

$=E((X^2-X^2)1_{X^2<\eta_0})+E((\eta_0-X^2)1_{\eta_0\leq X^2<\eta_1})+E((\eta_0-\eta_1)1_{X^2\geq\eta_1})$

$=E((\eta_0-X^2)1_{\eta_0\leq X^2<\eta_1})+(\eta_0-\eta_1)P({ X^2\geq\eta_1})$

$=(\eta_0-\eta_1)P({ |X|\geq\sqrt{\eta_1}})+\eta_0P(\eta_0\leq X^2<\eta_1)-E(X^21_{\eta_0\leq X^2<\eta_1})$

$P({ |X|\geq\sqrt{\eta_1}})=P({ X\geq\sqrt{\eta_1}})+P({ X\leq-\sqrt{\eta_1}})$

$E(W^2)=E(W^21_{X^2<\eta_0})+E(W^21_{\eta_0\leq X^2<\eta_1})+E(W^21_{\eta_1\leq X^2})$

$=E((\eta_0-X^2)^21_{\eta_0\leq X^2<\eta_1})+(\eta_0-\eta_1)^2P({ X^2\geq\eta_1})$

$=E((\eta_0^2-2\eta_0X^2+X^4)1_{\eta_0\leq X^2<\eta_1})+(\eta_0-\eta_1)^2P({ |X|\geq\sqrt{\eta_1}})$

All it remains to do is to know the entities of the following form :

$$P({a\leq X<b})$$

$$E(X^21_{a\leq X^2<b})$$

$$E(X^41_{a\leq X^2<b})$$

with $0<a<b$

Moreover, $1_{a\leq X^2<b}=1_{\sqrt{a}\leq X<\sqrt{b}}+1_{-\sqrt{b}\leq X<-\sqrt{a}}$

We know that

$$P({a\leq X<b})=\Phi(\frac{b}{\sigma})-\Phi(\frac{a}{\sigma})$$, with $\Phi$ the cdf of a centered normally distributed variable.

The following can be calculated by integrating by parts. $$E(X^21_{a\leq X<b})=\frac{\sigma^2}{\sqrt{2\pi}}\int_{\frac{a}{\sigma}}^{\frac{b}{\sigma}}{x^2e^{-\frac{1}{2}x^2}dx}$$

and $$E(X^41_{a\leq X<b})=\frac{\sigma^4}{\sqrt{2\pi}}\int_{\frac{a}{\sigma}}^{\frac{b}{\sigma}}{x^4e^{-\frac{1}{2}x^2}dx}$$