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I saw the identity below but I'm not sure how to derive it.

$$E[X\mid X>K] = \mu + \sigma \frac{\phi(z)}{\Phi(-z)} \text{ where } z = \frac{K-\mu} \sigma$$

I'm stuck at the following step:

$$E[X\mid X>K] = \frac{E[X 1_{X>K}]}{E[1_{X>K}]} = \frac{\int_K^\infty x f_X(x)\,dx}{\int_K^\infty f_X(x)\,dx} = \frac{\int_K^\infty x f_X(x)\,dx}{\Phi\left(-\frac{K-\mu} \sigma\right)} $$

Could anyone help with this? many thanks

Michael Hardy
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  • Where did this step come from? $$E[X|X>K] = \frac{E[X 1_{X>K}]}{E[1_{X>K}]}$$ – mhdadk Nov 14 '21 at 12:53
  • You may be interested in this post (or maybe more relevant: the posts that are linked to closing it as duplicate): https://stats.stackexchange.com/questions/499657/bivariate-normal-and-truncated-expectation. Also maybe search on roy models for example to get this post https://stats.stackexchange.com/questions/60013/properties-of-bivariate-standard-normal-and-implied-conditional-probability-in-t. – Jesper for President Nov 14 '21 at 14:16
  • Because the question remains the same when $X$ and $K$ are negated ($X$ still has a Normal distribution), thereby reversing the inequality, this is the same as the duplicate. – whuber Nov 14 '21 at 16:01

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