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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.

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    $\begingroup$ 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. $\endgroup$
    – Xi'an
    Oct 15, 2022 at 19:31
  • $\begingroup$ 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. $\endgroup$
    – rick
    Oct 15, 2022 at 19:35
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    $\begingroup$ 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? $\endgroup$
    – jbowman
    Oct 15, 2022 at 19:57
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    $\begingroup$ 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? $\endgroup$
    – rick
    Oct 15, 2022 at 21:24

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

This can be solved, check https://en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions You will have of course to make transformations, change of variables, etc.

Look up also Owen 1980 cited in the wikipedia article, it contains more indefinite integrals, which is what you need.

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