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

  • $\begingroup$ What does $X$ stand for? Where does the problem come from? $\endgroup$ – André Nicolas Oct 22 '11 at 5:40
  • $\begingroup$ X is a random variable, I thought it up, trying to calculate the expected value of a cumulative distribution $\endgroup$ – Jess C Oct 22 '11 at 5:57
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Oct 22 '11 at 6:12
  • $\begingroup$ oops hehe, thanks for reminding me =] $\endgroup$ – Jess C Oct 22 '11 at 6:26
  • $\begingroup$ X is normally distributed $\endgroup$ – 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}
  • $\begingroup$ how do you get the first equality? thanks $\endgroup$ – Jess C Oct 22 '11 at 6:32
  • $\begingroup$ Sasha: +1. $ $ $ $ $\endgroup$ – Did Oct 22 '11 at 8:18
  • $\begingroup$ @JessC The first equality is the definition of the cumulative density function, namely $\Phi(y) = \mathbb{P}(Y \le y )$. $\endgroup$ – Sasha Oct 22 '11 at 9:22
  • $\begingroup$ +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? $\endgroup$ – Dilip Sarwate Oct 22 '11 at 21:06
  • $\begingroup$ @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 $\endgroup$ – Sasha Oct 22 '11 at 21:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.