I'm trying to find $\int_{\frac{a-b}{B}}^\infty\Phi\left(tA+ABx\right)\phi(x)\,dx$ where
$A = \frac{\sqrt{\gamma_{3}+\sigma_3^2}}{\gamma_{3}},\ B = \frac{\gamma_{2}}{\sqrt{\gamma_{2}+\sigma_{2}^2}},\ t=b-c$.
This integral arrives from trying the find $E[\max\{X,Y,a\}]$, where $X\sim N(b,B^2) $, $Y\sim N(c,\frac{1}{A^2})$, independent. I've tried differentiating w.r.t. $A$ but after solving the integral, I can't integrate back w.r.t. $A$. This is trying to follow the strategy from How can I calculate $\int^{\infty}_{-\infty}\Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw$.
Is there a closed form solution for this?
I realise that it is similar to rainbow options where they calculate $\max\{X,Y,k\}$ but $X$ and $Y$ follow lognormal distribution. The closed formed solution for that involves the bivariate normal distribution.
Best Answer
You have $tA + ABx$, while the answer you linked is differentiating with $\frac{x-\mu}{\sigma}$.
The answer you linked provides the solution (when integratingn over $-\infty, \infty$)
$$\int_{-\infty}^\infty \Phi \left( \frac{x-\mu}{\sigma} \right) \phi (x) dx = \Phi \left( \frac{-\mu}{\sqrt{1+\sigma^2}} \right)$$
It seems to me you can just transform your integral into that form and apply the answer they provided.
Write
$$tA + ABx = ABx - (-tA) = \frac{x - (-tA)(AB)^{-1}}{(AB)^{-1}} = \frac{x-\mu}{\sigma}$$
Where
$$\mu = (-tA)(AB)^{-1}$$
$$\sigma = (AB)^{-1} = \frac{1}{AB}$$
Then you now have an integral of the form used in that question. So following the logic of that answer, the integral should be
$$\Phi \left( \frac{tA(AB)^{-1}}{\sqrt{1+(AB)^{-2}}} \right) = \Phi \left( \frac{tA}{\sqrt{(AB)^2+1}} \right)$$
Now, in your example you have a lower bound that is not $-\infty$. So, using the transformation above, you should be able to replicate the steps taken in that answer (or similar, as their are multiple on this site) with your lower bound, provided a closed form solution exists.
Perhaps you can run some basic simulations in R or Python to verify this is correct.