# 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 =]

-

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
show 1 more comment

## 1 Answer

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}
````
-

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
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8596932888031006, "perplexity_flag": "middle"}