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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.

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  • $\begingroup$ Yes, thanks, corrected! $\endgroup$
    – gwtw14
    May 2, 2022 at 15:57

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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.

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