# Expected value of normal CDF

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 $$\Phi\left( \frac{a-bX}{c} \right)$$ 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.

• Commented Jul 23, 2013 at 4:30
• 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. Commented Aug 2, 2013 at 22:35
• – Did
Commented Aug 3, 2013 at 8:43
• @Did - That's for $t=-\infty$. Do you know of anything for finite $t$? Commented Aug 3, 2013 at 17:10
• @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.
– Did
Commented Aug 3, 2013 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 $$X$$ ~ $$N\left(0,1\right), \frac{bX}{c}$$ ~ $$N\left(0,\frac{b^2}{c^2}\right)$$, so $$Y+\frac{bX}{c}$$ ~ $$N\left(0,1+\frac{b^2}{c^2}\right)$$

So,

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

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

• nice answer .......+1 Commented Feb 20, 2023 at 21:20

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 $(t,\infty)$.

• This approach was explained before, on MSE and on other SE sites, and pushed further to give an exact formula.
– Did
Commented Aug 3, 2013 at 8:25