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I am wondering what is the probability density function for the normal cdf $\Phi (aX+b)$, where $\phi$ is the usual standard normal cumulative distribution function

I want to calculate $\mathbb{E}[\Phi(aX+b)]$ but i am stuck on how to get the distribution. thank you =]

note: X is normally distributed

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    Convnetionally the capital $\Phi$ is the cdf and the lower-case $\varphi$ is the density function. This posting says "where $\phi$ is the[...]cumulative distribution function", after an expression that uses $\Phi$, not $\phi$.2011-10-22

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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} 
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    @DilipSarwate Yes, I agree this would add clarity. I will be able to make the change only in few hours from now. Thanks for the comment and the upvote2011-10-22