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I am wondering how you compute the conditional expectation of a normal distributed variable X with mean and variance known. So more specifically:

E(X|x<0)

Is there a specific formula for this calculation?

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  • $\begingroup$ The answer is the usual definite integral that defines the unconditional expectation $\int_{-\infty}^{\infty}x\phi(x;\mu,\sigma^2)dx$, but this time the upper limit of the integral is $0$ instead of $\infty$. Here $\phi(x;\mu,\sigma^2)$ is the normal density with your mean and variance. $\endgroup$ Commented Jan 25, 2022 at 15:10
  • $\begingroup$ @RichardHardy Do you maybe have a link to a paper or website where there is elaborated on this method? Somehow I cannot find any information about it. $\endgroup$
    – Marcelle
    Commented Jan 25, 2022 at 15:20
  • $\begingroup$ en.wikipedia.org/wiki/Truncated_normal_distribution $\endgroup$ Commented Jan 25, 2022 at 15:21
  • $\begingroup$ We have many threads on truncated Normal distributions, from which you can obtain the information you need: see this site search. $\endgroup$
    – whuber
    Commented Jan 25, 2022 at 15:42

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