# Conditional Expectation of Squared Normal Tail

Let $$X$$ be a normally distributed random variable with mean 0 and variance $$1$$. Let $$\lambda > 0$$. What is the value of the conditional expectation $$E[X^2 | |X|>\lambda]?$$

I got the following answer, where $$\phi$$ is the standard normal pdf, and $$\Phi$$ is the standard normal cdf. $$E[X^2| |X| > \lambda] = 2\lambda \phi(\lambda) + 2(1-\Phi(\lambda)).$$

Here is my attempted Proof: $$E[X^2| |X| > \lambda] = 2\int_\lambda^\infty x^2 \frac{1}{\sqrt{2\pi}} e^{-x^2/2}dx \\ = -4\frac{d}{da} (\sqrt{1/a} \int_\lambda^\infty \frac{dx}{\sqrt{2\pi (1/a)}} e^{-ax^2/2}) |_{a=1} \\ = -4 \frac{d}{da} (\sqrt{1/a} (1-\Phi(\sqrt{a}\lambda))|_{a=1}\\ = -4 (\sqrt{1/a} (-\phi(\sqrt{a}\lambda)\frac{\lambda}{2\sqrt{a}}) + (1-\Phi(\sqrt{a}\lambda) (-\frac{1}{2}a^{-3/2}) |_{a=1})\\ = 2 (\phi(\lambda)\lambda + (1-\Phi(\lambda)).$$

I think this is a standard expression, but I was unable to find an explicit answer for finite $$\lambda>0$$ (I know the limiting behavior as $$\lambda \to \infty$$ is correct) and would like to check that I have the correct form.

• Yes, thanks, corrected! May 2, 2022 at 15:57

The conditional expectation would actually be $$E[X^2 \: | \: |X| > \lambda] = \frac{2}{P(|X|>\lambda)} \int_\lambda^\infty x^2\phi(x) \: dx,$$ so provided that your other calculations are correct, then you should divide your formula by $$2(1-\Phi(\lambda))$$, which gives us that $$E[X^2 \: | \: |X| > \lambda] = \frac{\lambda\phi(\lambda)}{1-\Phi(\lambda)} + 1.$$ This is indeed the correct expression. We can note that $$E[X^2 \: | \: |X| > \lambda] = E[X^2 \: | \: X > \lambda] = \operatorname{Var}[X \: | \: X > \lambda] + E[X \: | \: X > \lambda]^2,$$ where the right hand side involves the variance and expectation for the Truncated Normal Distribution.