I got stuck when trying to solve the integration of the expectation value of the conditional probability distribution which is Normal. I tried to use integration by parts but I was not able to move on after I set dv=
$$1/\sqrt(2\pi)e^{(-1/2)(y-x)^2}dy$$
I wonder may I understand how to move on?
The formula is:
$$E[Y|X]=\int_{-\infty}^{\infty} y \times 1/\sqrt(2\pi) e^{(-1/2)(y-x)^2}dy$$
I know the latter part is normal distribution
Put $z=y-x$. To evaluate $\int (z+x) e^{-z^{2}/2} \, dz$ split it into two terms. In the first term use the fact that $z e^{-z^{2}/2}$ is the derivative of $-e^{-z^{2}/2}$. The value of the second term should be clear.