How do I compute this integral of standard normally distributed variables?

Question

I tried to apply this approach: Question on probability of two random variables for example $p(X < a, X < Y-b)$

to one of my problems where in my case we have Gaussian random variables and $a=0$. I ended up with the following term, but I don't know how to continue.

Denote by $Y \sim N(0,1)$ a standard Gaussian random variable with distribution function $G(y)$, and by $c$ and $b$ some constants. I want to compute the integral

$\int_0^\infty \frac{1}{2}erf(\frac{cy}{\sqrt2 b})dG(y) $

How do I do that? I know that $G(y)=\frac{1}{2}(1+erf(\frac{y}{\sqrt2}))$, but I am stuck.

Answer 1

Firstly, note that the error function is given by: $$\text{erf} (z) = \frac{2}{\sqrt{\pi}} \int_{0}^{z} e^{-t^{2}} dt$$

Knowing this, it should be trivial to expand $\text{erf} \left( \frac{cy}{b \sqrt{2}}\right)$ and $\text{erf} \left( \frac{y}{\sqrt{2}}\right)$.

Now, simply find the derivative of $G(y)$ to get $dG = G'(y) \ dy$, substitute $dG$ and $\text{erf} \left( \frac{cy}{b \sqrt{2}}\right)$ in your integral and integrate wrt $y$.

You might find it useful to note that $\text{erf}(\infty) = 1$.

$\frac{dG}{dy} = \frac{1}{\sqrt{2 \pi}} e^{-y^{2}/2}$

$\int \text{erf} (z) = z \cdot \text{erf}(z) + \frac{e^{-z^{2}}}{\sqrt{\pi}} + C$