Evaluation of an expectation involving a product of normally distributed rvs

Question

This should be not too hard but I am stuck on this for 2 hours now... I have a normally distributed rv with mean 0 and variance 1 and I want to calculate $$E[X\cdot(1_{X\in A}X - 1_{X\in\complement A}X)],$$ where $1_B$ denotes the indicator function of the event $B$. I started by using the identitiy $1_B = 1 - 1_{\complement B}$, i.e. $$ 1_{X\in\complement A} = 1 - 1_{X\in A}$$ which yields $$ E[X^21_{X\in A} - X^2 + X^21_{X\in A}] = 2E[X^21_{X\in A}] - 1$$ since $E[X^2] = 1$ as the mean of $X$ is zero and variance $1$. Now, $$E[X^21_{X\in A}] = \int X^21_{X\in A}\ dP = \int_AX^2\ dP = \frac{1}{2\pi}\int_A x^2\exp(-0.5x^2)\ dx$$ since $X$ is normally distributed. Are these steps so far correct? Because if I proceed with a specific $A$, e.g. $A = \{|X|\leq c\}$, I get $$\begin{align*} E[X^21_{X\in A}] &= \frac{1}{\sqrt{2\pi}}\int_{-c}^cx^2\exp(-0.5x^2)\ dx \\ &= 0.5\mathrm{erf}(\sqrt{0.5}c) - 0.5\mathrm{erf}(-\sqrt{0.5}c) \\&=\mathrm{erf}(\sqrt{0.5}c). \end{align*}$$ According to Wikipedia (https://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent#An_asymmetric_example), for $c\approx 1.54$, this expression should be approximately $0.5$ (so that $2E[X^21_{X\in A}] - 1 = 0$). But $\mathrm{erf}(\sqrt{0.5}\cdot 1.54)\approx 0.876$. Is there an error in my reasoning or just an error In the calculations?

Answer 1

As Niki just mentioned, your expression for $E[X^2 \cdot 1_A]$ in terms of an integral is incorrect. To complement his answer, I derive a correct expression by integrating by parts: \begin{align*} \frac{1}{\sqrt{2\pi}} \int_{-c}^{c} x^2 e^{-x^2/2} \ dx &= \frac{1}{\sqrt{2\pi}} \int_{-c}^{c} x \ d\left( -e^{-x^2/2} \right) \\ &= -xe^{-x^2/2}\Big\vert_{-c}^c + \frac{1}{\sqrt{2\pi}} \int_{-c}^{c} e^{-x^2/2} \ dx \\ &= -ce^{-c^2/2} - \left(-(-c)e^{-(-c)^2/2} \right) + \frac{1}{\sqrt{2\pi}} \int_{-c}^{c} e^{-x^2/2} \ dx \\ &= - \frac{2c}{\sqrt{2\pi}} e^{-c^2/2} + \frac{1}{\sqrt{2\pi}} \int_{-c}^{c} e^{-x^2/2} \ dx \\ &= - \frac{2c}{\sqrt{2\pi}} e^{-c^2/2} + \mathrm{erf}(\tfrac{c}{\sqrt{2}}). \end{align*}

Answer 2

You're using the normal distribution as your variable but getting the integral wrong. A normally distributed (standard) random variable has probability density function: $$f(x) = \frac 1{\sqrt{2\pi}} e^{-x^2/2}$$ So that: $$E[X^2 \cdot 1_A] = \frac 1{\sqrt{2\pi}}\int_A x^2 e^{-x^2/2} dx$$ Which does not equal your expression with the error function for $A=[-c, c]$.