# distribution of the normal cdf

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 =]

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What does $X$ stand for? Where does the problem come from? – André Nicolas Oct 22 '11 at 5:40
X is a random variable, I thought it up, trying to calculate the expected value of a cumulative distribution – Jess C Oct 22 '11 at 5:57
The expected value of a function $f(X)$ of a random variable $X$ depends in general on the distribution of $X$, and not only the mean of $X$. There was no specification made in the post about the distribution of $X$, only about what $f$ was. – André Nicolas Oct 22 '11 at 6:12
oops hehe, thanks for reminding me =] – Jess C Oct 22 '11 at 6:26
X is normally distributed – Jess C Oct 22 '11 at 6:26

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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how do you get the first equality? thanks – Jess C Oct 22 '11 at 6:32
Sasha: +1.   – Did Oct 22 '11 at 8:18
@JessC The first equality is the definition of the cumulative density function, namely $\Phi(y) = \mathbb{P}(Y \le y )$. – Sasha Oct 22 '11 at 9:22
+1 indeed! But may I request that the first inequality be expanded out a little e.g. as in $$E[\Phi(aX+b)] = \int_{-\infty}^{\infty}\Phi(ax+b)\phi(x)dx = \int_{-\infty}^{\infty}P(Y \leq aX + b\mid X = x)\phi(x)dx = P(Y \leq aX + b)\ldots...$$ so as to make the connection very clear? – Dilip Sarwate Oct 22 '11 at 21:06
@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 upvote – Sasha Oct 22 '11 at 21:23