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Definitions:

Let $$\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} $$ be the standard normal probability density function (pdf) and $$ \Phi(x) = \int_{-\infty}^x \phi(t) dt = \frac{1}{2}\left[ 1 + \text{erf}\left(\frac{x}{\sqrt{2}} \right) \right] $$ be its cumulative distribution function (cdf).

Problem:

For a side project I am dealing with, as an intermediate step of an algorithm I need to solve definite integrals of the following form:

$$ \int_{-\infty}^{\infty} x^p \phi\left(x\right) \Phi(a + b x) dx $$

for $p = 0,1,2$.

For $p = 0$ and $p = 1$, the solutions are listed on this Wikipedia page. They are:

$$ \int_{-\infty}^{\infty} \phi\left(x\right) \Phi(a + b x) dx = \Phi\left(\frac{a}{t} \right) $$ $$ \int_{-\infty}^{\infty} x \phi\left(x\right) \Phi(a + b x) dx = \frac{b}{t} \phi\left(\frac{a}{t} \right) $$ with $t = \sqrt{1 + b^2}$.

Questions:

  • What is the solution for $p = 2$?
  • More in general, what is the simplest approach to solve these integrals?
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    http://nvlpubs.nist.gov/nistpubs/jres/73B/jresv73Bn1p1_A1b.pdf 4.3.7 – 2017-01-25
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    @tired Holy wow! Why did you not tell me of this soon!? – 2017-01-25
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    @Simply beautiful sorry, just forgot it ;) – 2017-01-25
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    To derive identities from wikipedia for $p=0$ differentiate w.r.t. to $a$, for $p=1$ combine this with an integration by parts. $p=2$ should be solvable by similar methods – 2017-01-25
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    @tired: Thanks for the reference. However, Eq. 4.3.7 seems to apply only for no shift (i.e. $a = 0$ in my notation). – 2017-01-25
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    @lacerbi i know, i intended this as a reference to crosscheck the related generalizations – 2017-01-25
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    @tired Ah dang... I was close...maybe I'll redo my answer later. – 2017-01-25
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    @SimplyBeautifulArt that'd be great! :D – 2017-01-25
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    Let us investigate the integral >$$ I(a,b)=\int_{\mathbb{R}}\phi(x)\Phi(a+bx) $$ now let's denote $'$ the derivative w.r.t $a$. Applying this operation to the above integrals yields, after rescaling ($t=\sqrt{1+b^2}$) $$ I^{'}(a,b)=\frac{1}{\pi t}\int_{\mathbb{R}}\exp\left(-\frac{1}{2}\left(y^2+2\frac{ab}{t}y+a^2)\right)\right) $$ ..... – 2017-01-25
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    ....completing the square gives $$ I^{'}(a,b)=\frac{1}{\pi t}e^{-\frac{1}{2}(a^2-\frac{a^2 b^2}{t^2})}\int_{\mathbb{R}}e^{-\frac{1}{2}y^2}=\frac{\sqrt{2}}{\sqrt{\pi} t}e^{-\frac{1}{2}(a^2-\frac{a^2 b^2}{t^2})} $$ integrating back w.r.t $a$ gives $$ I(a,b)=\frac{1}{2}\text{erf}\left(\frac a{\sqrt2t}\right)+C $$ since $\phi(x)$ is normalized and $I(a,0)=\Phi(a)$ we can fix the constant of integration to be $\frac12$ and we get >$$ I(a,b)=\Phi\left(\frac at\right) $$ The generalisations you are mentioning follow the same line of reasoning, so i leave them to you! – 2017-01-25
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    For $p=1$ essentially the only additional information you need is that the interval of an odd function over an even region equals zero – 2017-01-25
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    @tired: Thanks for the hint. Yes, the basic trick is to derive under the integral, from there it's just a matter of manipulation and matching terms. I was able to re-derive independently the results for $p=0$ and $p=1$, I'll work next on the $p=2$ case. – 2017-01-26

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