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I want to compute the variance of $\Phi(Z-c)$

where Z is standard normal r.v. , $\Phi$ is the standard normal CDF, and c is a constant.

I tried to start with the expectation but don't know what to do next $$E[\Phi(Z-c)] = \int \Phi(z-c) \phi(z) dz $$

Bowen Deng
  • 11

1 Answers1

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This is only an answer for $E(\Phi(Z-c))$. I want to show

$E(\Phi(Z-c))=\Phi(\frac{-c}{\sqrt{2}})$

   > z<-rnorm(1000000)
   > c<-5
   > mean(pnorm(z-c))
   [1] 0.0002044583
   > pnorm(-c/sqrt(2))
   [1] 0.000203476

Let $Y\sim N(\mu, \sigma^2)$ and $\Phi$ is c.d.f of standard normal and we want to show $E(\Phi(Y))=\Phi(\frac{\mu}{\sqrt{1+\sigma^2}})$.

This is more general that case $E(\Phi(Z-c))$ since $Z-c\sim N(-c,1)$.

$$E(\Phi(Y))=\int_{-\infty}^{+\infty} \Phi(y) \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} dy$$

$$=\int_{-\infty}^{+\infty} \int_{-\infty}^{y} \frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}} dz \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} dy$$

$$=\int_{-\infty}^{+\infty} \int_{-\infty}^{y} \frac{1}{\sqrt{2\pi }}e^{\frac{-z^2}{2}} \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} dz dy$$

$$=\iint_{z<y} \frac{1}{\sqrt{2\pi }}e^{-\frac{z^2}{2}} \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} dz dy$$

$$=P(Z<Y)$$

where $Z\sim N(0,1)$ and $Y\sim N(\mu , \sigma^2)$ are independent. so

$$=P(Z-Y<0)$$ and $Z-Y\sim N(-\mu , 1+\sigma^2)$

$$=P(\frac{Z-Y +\mu}{\sqrt{1+\sigma^2}}<\frac{\mu}{\sqrt{1+\sigma^2}})=\Phi(\frac{\mu}{\sqrt{1+\sigma^2}})$$

Masoud
  • 2,715