I am trying to calculate the expected value of a Normal CDF, but I have gotten stuck. I want to find the expected value of $\Phi( \frac{a-bX}{c} )$ where $X$ is distributed as $\mathcal{N}(0,1)$ and $\Phi$ is the standard normal CDF.

I know I can transform $\frac{a-bX}{c}$ to be a normal random variable $\mathcal{N}\left(\frac{a}{c},\frac{b^2}{c^2}\right)$ where $\frac{b^2}{c^2}$ is the variance of the normal random variable. I'm not sure if this helps though.

I think that the expected value of a CDF is $0.5$ but since $\Phi$ is the CDF of a standard normal CDF and $\frac{a-bX}{c}$ is not standard normal I do not think the expected value should be 0.5. I tried integrating the CDF, but I do not believe I did it correctly.

When $a = -2.3338$, $b = 0.32164$, $c = 0.94686$, I believe the correct answer is approximately $0.009803$. I found this through simulation.

I would appreciate any help or suggestions.

  • 1
    $\begingroup$ See stats.stackexchange.com/questions/61080/… $\endgroup$ – Ray Koopman Jul 23 '13 at 4:30
  • $\begingroup$ Thank you for the response. I now understand how to find the expected value over the entire real line. I am now having trouble finding the expected value given x is in an interval. In particular I am interested in the values when x ranges from t to infinity, where t is a known constant. Thank you for any help or suggestions. $\endgroup$ – Robert Aug 2 '13 at 22:35
  • $\begingroup$ See mathoverflow.net/a/135405/4661. $\endgroup$ – Did Aug 3 '13 at 8:43
  • $\begingroup$ @Did - That's for $t=-\infty$. Do you know of anything for finite $t$? $\endgroup$ – Ray Koopman Aug 3 '13 at 17:10
  • $\begingroup$ @RayKoopman As explained several times in the couple of recent near duplicates on MO and MSE and StatsSE, for some finite $t$ there is no explicit formula. $\endgroup$ – Did Aug 3 '13 at 19:42

Assume that the $\Phi$ function is for some other independent standard normal random variable Y, and simply rewrite the problem like so:

$$E\left(\Phi\left( \frac{a-bX}{c}\right)\right)=P(Y<\frac{a-bX}{c})=P\left( Y+\frac{bX}{c}<\frac{a}{c}\right)$$

Now, since $X$ ~ $N(0,1), \frac{bX}{c}$ ~ $N(0,\frac{b^2}{c^2})$, so $Y+\frac{bX}{c}$ ~ $N(0,1+\frac{b^2}{c^2})$


$$ P\left( Y+\frac{bX}{c}<\frac{a}{c}\right) = P\left( \frac{Y+\frac{bX}{c}}{\sqrt{1+\frac{b^2}{c^2}}}<\frac{a}{c\sqrt{1+\frac{b^2}{c^2}}}\right)=\Phi\left( \frac{a}{c\sqrt{1+\frac{b^2}{c^2}}}\right) $$

Therefor: $E\left(\Phi\left( \frac{a-bX}{c}\right)\right)=\Phi\left( \frac{a}{c\sqrt{1+\frac{b^2}{c^2}}}\right)$


Let $c$ in your question be absorbed into $a$ and $b$, so that you are concerned with $\Phi(a-bX)$. Let $X$ and $Y$ be independent standard normals, and consider their joint pdf $\phi(x,y) = \phi(x)\,\phi(y)$ as a surface covering the $X,\!Y$ plane, where $\phi(\cdot)$ is the standard normal pdf.

For any given $x$, $\Phi(a-bx)\,\phi(x)\,\text{d}x$ is the volume under the joint pdf in the thin slice at $x$ with $Y < a-bx$. Then the conditional expected value of $\Phi(a-bX)$, given that $X$ is in some interval, is just the volume under the pdf in the $X$-interval and below the line $Y = a - bX$, divided by all the volume under the pdf in the $X$-interval.

I know of no closed-form expression for the numerator in the general case where the X-interval is of the form $(t,\infty)$.

  • $\begingroup$ This approach was explained before, on MSE and on other SE sites, and pushed further to give an exact formula. $\endgroup$ – Did Aug 3 '13 at 8:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.