Ciao,
I was making some computation and I've been stucked in this one.
Let $B$ and $b$ be positive contant. We call $\phi(x)$ standard gaussian distribution and $\Phi(x)$ its cumulative function, i.e. $$ \phi(x) = \frac{1}{\sqrt{2 \pi}}e^{-\frac{x^2}{2}} $$ $$ \Phi(x) = \int_{-\infty}^x \phi(s) ds $$
then compute
$$ \int_{-\infty}^{+\infty}x\Phi(x)\phi(Bx - b) dx $$
If it helps I can proove this result: $$ \int_{-\infty}^{+\infty}x\Phi(x)\phi(x) dx = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-x^2}= \frac{1}{\sqrt{2}} $$
Any suggestion or hint will be appreciated, thank you!
Ciao
AM
We have $$ \int_{- \infty}^x ds \ \phi(s) = \int_{- \infty}^0 ds \ \phi(s + x) \ .$$ Using Fubini's theorem, we have $$ \int_{-\infty}^\infty dx \ x \Phi(x) \phi(Bx - b) = \int_{- \infty}^0 ds \int_{-\infty}^\infty dx \ x \phi(s + x) \phi(Bx + b) \ .$$ Computing the inner integral, we have $$ \int_{-\infty}^\infty dx \ x \phi(s + x) \phi(Bx + b) = \frac{(bB - s)e^{- \frac{(Bs + b)^2}{2(1 + B^2)}} }{\sqrt{2 \pi}(1 + B^2)^\frac{3}{2}} \ .$$ We have $$ \int_{- \infty}^0 ds \int_{-\infty}^\infty dx \ x \phi(s + x) \phi(Bx + b) = \int_{0}^\infty ds \ \frac{(bB + s)e^{- \frac{(- Bs + b)^2}{2(1 + B^2)}} }{\sqrt{2 \pi}(1 + B^2)^\frac{3}{2}} \ .$$ From here, I suppose one has a representation in terms of the error function. I do not see any other way to simplify this expression.
If we set $B = 1$ and $b = 0$, then the value of this integral agrees with the value that Anne calculated in the comments.