# Conditional expectation of a normal variable given a lower bound [duplicate]

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

• Where did this step come from? $$E[X|X>K] = \frac{E[X 1_{X>K}]}{E[1_{X>K}]}$$ Nov 14, 2021 at 12:53
• 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/…. Nov 14, 2021 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, 2021 at 16:01