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Let X be normally distributed random variable with some given mean and variance. Is there an expression for the expected value of X, conditional on X^2 being greater than some constant c?

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    Well, $X^2$ has a Chi-square distribution with 1 degree of freedom. maybe you can start from that distribution to solve $P(X^2>c)$ to obtain the distribution of your variable and then compute the expected value. – 2017-01-05
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    Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, [this link](http://meta.math.stackexchange.com/a/9960) might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – 2017-01-05

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$\Pr(X^2 \gt c) = \Pr(X \gt \sqrt{c})+ \Pr(X \gt \sqrt{-c})$ so this is related to a truncated normal distribution though you are looking to truncate inside rather than outside.

So I would have thought that if $\alpha=\dfrac{-\sqrt{c}-\mu}{\sigma}$ and $\beta=\dfrac{\sqrt{c}-\mu}{\sigma}$ than the mean you are looking for will be $$E[X \mid X^2 \gt c]=\mu + \dfrac{\phi(\beta)-\phi(\alpha)}{1-\Phi(\beta)+\Phi(\alpha)}\sigma$$ where $\phi(x)$ is the density of a standard normal and $\Phi(x)$ is the cumulative distribution function of a standard normal