Suppose, $X\sim N(\mu,\sigma^2)$.
Can anyone help in finding the following : $\text{Var } \bigg(\Phi\big(\frac{X + c}{d}\big) \bigg)$ ? Here, c and d are positive.
Here, $\Phi(x)$ is the "Cumulative Distribution Function" of the above-mentioned normal distribution.
Thanking you in advance.
Expanding on the comment by Dilip Sarwate, using this list of integrals of Gaussian functions gives
\begin{align*} E \left[ \Phi\left(\frac{X+c}{d}\right)\right] & = \int \frac{1}{\sigma} \phi\left(\frac{x-\mu}{\sigma}\right) \, \Phi\left(\frac{x+c}{d}\right) \, dx \\[8pt] & = \int \phi\left(x\right) \, \Phi\left(\frac{\sigma x+ \mu + c}{d}\right) \, dx\\[8pt] & = \Phi \left( \frac{\mu+c}{\sqrt{\sigma^2 + d^2}}\right) \end{align*} and
\begin{align*} E \left[ \Phi\left(\frac{X+c}{d}\right)^2\right] & = \int \frac{1}{\sigma} \phi\left(\frac{x-\mu}{\sigma}\right) \, \Phi\left(\frac{x+c}{d}\right)^2 \, dx \\[8pt] & = \int \phi\left(x\right) \, \Phi\left(\frac{\sigma x+ \mu + c}{d}\right)^2 \, dx\\[8pt] & = \Phi \left( \frac{\mu+c}{\sqrt{\sigma^2 + d^2}}\right) - 2T \left( \frac{\mu+c}{\sqrt{\sigma^2 + d^2}}, \frac{d}{\sqrt{2\sigma^2+d^2}}\right) \end{align*} where $T$ is Owen's T function.
From these expressions, $\text{Var}\left( \Phi\left(\frac{X+c}{d}\right) \right)$ follows.
