Distribution of the normal cdf

Solution 1:

Let $X$ and $Y$ be the standard normal random variables. Then $$ \mathbb{E}(\Phi(a X + b)) = \mathbb{E}( \mathbb{P}( Y \le a x + b \vert X = x ) ) = \mathbb{P}(Y- a X \le b ) $$ But the combination $Z = Y-a X$ also follows normal distribution (being a linear combination of normals), with zero mean and variance $\mathbb{E}((Y-a X)^2) = 1 + a^2$. Hence $$ \mathbb{E}(\Phi(a X + b)) = \Phi\left(\frac{b}{\sqrt{1+a^2}}\right) $$

Here is numerical checks:

In[14]:= With[{a = 3., 
  b = 1/2}, {NExpectation[CDF[NormalDistribution[], a x + b], 
   x \[Distributed] NormalDistribution[]],
  CDF[NormalDistribution[], b/Sqrt[1 + a^2]]}]

Out[14]= {0.562816, 0.562816}