# Conditional expectation of normal cdf

Suppose $$X \sim N(0, \sigma^2)$$, is it possible to evaluate $$E[\Phi(\frac{aX + b}{c}) | X > k]$$ in closed form, where $$\Phi$$ is the standard normal cdf?

The motivation comes from that it is possible to evaluate something like $$E[\Phi(\frac{aX + b}{c})]$$ with a closed-form expression. But not sure if something similar holds for the conditional expectation case.

• Since$$\Phi(\frac{aX+b}{c})=\mathbb E[\mathbb I_{Y<\frac{aX+b}{c}}]|X|$$with $Y$ standard Gaussian independent from $X$, the computation of $$\mathbb E[\mathbb I_{X>k}\mathbb I_{Y<\frac{aX+b}{c}}]|$$may prove feasible. Oct 15, 2022 at 19:31
• I agree. thanks, but is it feasible to express everything analytically, say in terms of $\Phi(\cdot)$? The expectation with the two indicators does not seem straightforward to me that it can be written in a closed form immediately.
– rick
Oct 15, 2022 at 19:35
• Are you sure this is what you want to do? $(aX+b)/c \sim \text{N}(b/c, a/c)$, so basically you're running a $\text{N}(\mu,\sigma)$ variate through a $\text{N}(0,1)$ CDF, not through what I would expect - a $\text{N}(\mu, \sigma)$ CDF. What's the situation that's causing you to want to do this? Oct 15, 2022 at 19:57
• I did not specify what $a$, $b$, and $c$ are. So, it can be anything that is appropriate. I thought this formulation is just a general formulation that I am considering some linear transformation of $X$ insider $\Phi(\cdot)$, although the $c$ is probably not needed. Does that clarify?
– rick
Oct 15, 2022 at 21:24

$$E\left[\Phi\left(\frac{aX + b}{c}\right) | X > k\right] = \left[1-\Phi(k/\sigma)\right]^{-1}\int_{k}^{\infty}\frac 1 {\sigma}\phi\left(x/\sigma\right)\Phi\left(\frac{ax + b}{c}\right)dx$$