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.

  • 5
    $\begingroup$ Look at this answer for a technique that will likely help you find $E\left[\Phi\left(\frac{X+c}{d}\right)\right]$. Perhaps the same idea can be adapted to find $E\left[\Phi^2\left(\frac{X+c}{d}\right)\right]$, from which two results you can get $\text{Var}\left(\Phi\left(\frac{X+c}{d}\right)\right)$. $\endgroup$ May 13, 2016 at 19:22

1 Answer 1


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.

  • 1
    $\begingroup$ Looks healthy ! $\endgroup$
    – wolfies
    Sep 28, 2016 at 16:57

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