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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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. – Richard Hardy Jan 25 '22 at 15:10
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@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. – Marcelle Jan 25 '22 at 15:20
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https://en.wikipedia.org/wiki/Truncated_normal_distribution – Richard Hardy Jan 25 '22 at 15:21
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We have many threads on truncated Normal distributions, from which you can obtain the information you need: see [this site search](https://stats.stackexchange.com/search?q=trunc*+normal+dist*+var*++score%3A2). – whuber Jan 25 '22 at 15:42