We want, using a simplified version for now:
$$ I_1 = \int_{-\infty}^\infty \Phi(x)\phi(x)\,dx $$And we have:
$$ \frac{d}{dx} \Phi(x) = \phi(x) $$Using integration by parts, with:
So:
$$ I_1 = \int_{-\infty}^\infty \Phi(x)\phi(x)\,dx \\ = [\Phi(x)^2]_{-\infty}^\infty - \int_{-\infty}^\infty \phi(x) \Phi(x) \, dx $$Therefore:
$$ 2I_1 = 1 $$And so:
$$ I_1 = \frac{1}{2} $$Try now for:
$$ I_2 = \int_{-\infty}^\infty \Phi(\lambda x)\phi(x)\, dx $$We can try:
We need to find the derivative of $u(x)$, ie $\frac{d}{dx}\Phi(\lambda x)$
Looking at wikipedia, https://en.wikipedia.org/wiki/Normal_distribution , we have:
$$ \Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x \exp\left(-\frac{t^2}{2}\right)\,dt $$We also have the error function $\def\erf{\text{erf}}\erf(x)$, defined as the probability of a random variable with normal distribution of mean 0 and variance $1/2$ falling in the range $[-x, x]$; that is:
$\Phi(x)$ and $\erf(x)$ are closely related, namely:
$$ \Phi(x) = \frac{1}{2} \left( 1 + \erf \left( \frac{x}{\sqrt{2}} \right) \right) $$"For a generic normal distribution with density $f$, mean $\mu$ and deviation $\sigma$, the cumulative distribution function is:
$$ F(x) = \Phi \left( \frac{x - \mu}{\sigma} \right) = \frac{1}{2} \left( 1 + \erf \left( \frac{x - \mu}{\sigma \sqrt{2}} \right) \right) $$Thinking about this, from the definition of what is the CDF for the Gaussian, we can form the following identity:
$$ \Phi\left( \frac{x - \mu}{\sigma} \right) = \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^x \exp\left( -\frac{(t - \mu)^2}{2\sigma^2} \right) \, dt $$substitue in $r = \frac{ t - \mu}{\sigma}$. Therefore:
$$ t = \sigma r + \mu $$Therefore:
$$ dt/dr = \sigma $$For the limits, we have $t_1 = -\infty$, and $t_2 = x$. Applying the subsitution $r = \frac{ t - \mu}{\sigma}$, we have:
$$ r_1 = \frac{t_1 - \mu}{\sigma} = \frac{-\infty - \mu }{\sigma} = -\infty $$And:
$$ r_2 = \frac{t_2 - \mu}{\sigma} = \frac{x - \mu}{\sigma} $$So:
$$ \Phi\left(\frac{x-\mu}{\sigma}\right) = \frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\frac{x-\mu}{\sigma}} \exp \left( - \frac{r^2}{2} \right) \sigma \, dr $$... which is at least consistent
We also have, from the wikipedia page:
$$ \mathcal{N}(x \mid \mu, \sigma^2) = \frac{1}{\sigma} \phi \left( \frac{x - \mu}{\sigma} \right) $$Therefore:
$$ \frac{d}{dx} \Phi\left( \frac{x - \mu}{\sigma} \right) = \frac{1}{\sigma} \phi\left(\frac{x - \mu}{\sigma}\right) $$Therefore, $u'(x) = \frac{d}{dx}\Phi(\lambda x) = \lambda \phi(\lambda x)$
So we have:
$$ I_2 = \left[ \Phi(\lambda x) \Phi(x) \right]_{-\infty}^\infty - \lambda \int_{-\infty}^\infty \phi(\lambda x) \Phi(x)\,dx $$substitute $u=\lambda x$, therefore $x = u / \lambda$, and $dx = 1/\lambda\, du$, So:
That didnt work
Let's try integration by parts again, with:
Therefore:
$$ I_2 = 1 - \lambda [\Phi(x)\frac{1}{\lambda}\Phi(\lambda x)]^\infty_{-\infty} + \lambda \int_{-\infty}^\infty \phi(x) \frac{1}{\lambda}\Phi(\lambda x)\,dx $$= 1 - 1 + I_2
...so that doesnt work either
Question: "How to calculate $\int \phi(x) \Phi(\frac{x - b}{a})\,dx$?"
Compared to what we want, we dont need the $-b$ bit, but we will need to generalize from $\phi(x)$ to $\mathcal{N}(x \mid \mu, S)$. However, let's work through the derivation for this particular form first.
"We have $\phi(x) = \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{t^2}{2}\right)$ and $\Phi(x) = \int_{-\infty}^x \phi(t)\,dt$. We try to compute:
$$ I(\mu,\sigma) = \int \phi(x) \Phi\left(\frac{x - \mu}{\sigma} \right)\,dx $$"Using the dominated convergence theorem, we are allowed to take the derivative with respect to $\mu$ inside the integral. We have:
$$ \partial_\mu I(\mu, \sigma) = \int \phi(x) \left( - \frac{1}{a}\right) \phi \left( \frac{x - \mu}{\sigma} \right) \, dx $$At this point I get stuck, since it's an indefinite integral. I assume that the answer assumes that it's a definite integral, from $-\infty$ to $\infty$, but it's an assumption. Looking around at other answers:
Looking at https://mathoverflow.net/questions/127086/integral-of-the-product-of-normal-density-and-cdf , which is similar:
Start with:
$$ I = \def\A{\mathbb{A}} \def\B{\mathbb{B}} \int_{-\infty}^\infty \Phi\left( \frac{f - \A} {\B} \right)\phi(f)\, df $$Taking derivative wrt $\A$:
Looking at $E_1 = \phi\left(\frac{f - \A}{\B}\right) \phi(f)$:
$$ E_1 = \frac{1}{2\pi} \exp \left( - \frac{1}{2} \left( \frac{f^2 - 2f\A + \A^2 + \B^2f^2} {B^2} \right) \right) $$Therefore $E_1$ is:
$$ \frac{1}{2\pi} \exp \left( -\frac{1}{2\B^2} \frac{\A^2\B^2}{1 + \B^2} \right) \exp \left( - \frac{1}{2\B^2} \left( f\sqrt{1 + \B^2} - \frac{\A}{\sqrt{1 + \B^2}} \right)^2 \right) $$Make change of variable:
$$ y = f\frac{\sqrt{1 + \B^2}}{\B} - \frac{A}{\B\sqrt{1 + \B^2}} $$Therefore:
$$ dy = \frac{\sqrt{1 + \B^2}}{\B}\,df $$For the limits, we have $f_1 = -\infty$, and $f_2 = \infty$
$\sqrt{1 + \B^2}$ is always positive. $\B$ is always negative. Therefore:
$$ y_1 = +\infty, y_2 = -\infty $$Therefore:
$$ \partial_\A I = \frac{-1}{\B}\int_{+\infty}^{-\infty} \frac{1}{2\pi} \exp \left( - \frac{\A^2} {2(1+\B^2)} \right) \exp \left( - \frac{1}{2} y^2 \right) \frac{\B}{\sqrt{1 + \B^2}} \, dy $$Now we need to re-integrate back up again, since we currently have $\partial_\A I$, and we need $I$.
Since we dont have limits, we'll need to find at least one known point.
We have the following integral:
$$ I = \frac{1}{\sqrt{1 + \B^2}} \int \phi\left( \frac{\A}{\sqrt{1 + \B^2}} \right) \,d\A $$... where $C$ is a constant of integration
Looking at the original integral, we had/have:
$$ I = \int_{-\infty}^\infty \Phi\left( \frac{f - \A}{\B} \right) \phi(f) \, df $$We can see that, given that $\B$ is negative, as $\A \rightarrow \infty$, $\Phi\left(\frac{f-\A}{\B}\right) \rightarrow \Phi(\infty) = 1$.
Therefore, as $\A \rightarrow \infty$, $\int_{-\infty}^\infty \Phi(\cdot)\phi(f)\,df \rightarrow 1$
Meanwhile, looking at the later expression for $I$, ie:
$$ I= \Phi\left( \frac{\A}{\sqrt{1 + \B^2}} \right) + C $$... as $\A \rightarrow +\infty$, $\Phi\left( \frac{\A}{\sqrt{1 + \B^2}} \right) \rightarrow 1$
But we know that as $\A \rightarrow +\infty$, $I \rightarrow 1$.
Therefore, $C = 0$
Therefore:
$$ I = \Phi \left( \frac{\A} {\sqrt{1 + \B^2}} \right) $$