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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

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  • $\begingroup$ Where did this step come from? $$E[X|X>K] = \frac{E[X 1_{X>K}]}{E[1_{X>K}]}$$ $\endgroup$
    – mhdadk
    Nov 14, 2021 at 12:53
  • $\begingroup$ You may be interested in this post (or maybe more relevant: the posts that are linked to closing it as duplicate): stats.stackexchange.com/questions/499657/…. Also maybe search on roy models for example to get this post stats.stackexchange.com/questions/60013/…. $\endgroup$ Nov 14, 2021 at 14:16
  • $\begingroup$ 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. $\endgroup$
    – whuber
    Nov 14, 2021 at 16:01

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