The question is:
Suppose that X and Y are independent Normally distributed random variables, $$X\sim N(a,\sigma_1^2)$$ $$Y\sim N(b,\sigma_2^2)$$ and $Z = \rho X + \sqrt{(1-\rho^2)}Y$, find $E[max(0,e^Z-e^Y)]$.
My solution is as follows: $(Z,Y)$ has joint normal distribution: $$\mu_Z = \rho a + \sqrt{1-\rho^2}b$$ $$\sigma_Z^2 = \rho^2\sigma_1^2 + (1-\rho^2)\sigma_2^2$$ $$\mu_Y = b$$ $$\sigma_Y^2 = \sigma_2^2$$ $$cov(Z,Y) = \sqrt{1-\rho^2}\sigma_2^2$$
the desired expectation can be written as: $$E[max(0,e^Z-e^Y)] = E[e^Z 1_{Z\geq Y}] - E[e^Y 1_{Z\geq Y}]$$ to compute $E[e^Z 1_{Z\geq Y}]$, I use the density of truncated normal distribution: $$P(Z\leq z|Z\geq y) = E[1_{Z\leq z}|1_{Z\geq y}] = \frac{E[1_{Z\leq z}1_{Z\geq y}]}{P(Z\geq y)}$$ $$P(Z\leq z|Z\geq y) = \frac{\Phi(\frac{z-\mu_Z}{\sigma_Z})-\Phi(\frac{y-\mu_Z}{\sigma_Z})}{1-\Phi(\frac{y-\mu_Z}{\sigma_Z})}$$ take derivative with respect of z, we have distribution density function on $z\in [y,+\infty)$: $$f_{Z|Z\geq y}(z) = \frac{\frac{1}{\sigma_Z}\phi(\frac{z-\mu_Z}{\sigma_Z})}{1-\Phi(\frac{y-\mu_Z}{\sigma_Z})}$$ Then $$E[e^Z 1_{Z\geq Y}] = E[E[e^Z|1_{Z\geq Y},Y]] = \int_{\infty}^{-\infty}\int_{y}^{\infty}e^z \frac{\frac{1}{\sigma_Z}\phi(\frac{z-\mu_Z}{\sigma_Z})}{1-\Phi(\frac{y-\mu_Z}{\sigma_Z})} dz \phi(y) dy$$ $$= \int_{-\infty}^{\infty}\frac{1-\Phi(\frac{y-\mu_Z}{\sigma_Z}-\sigma_Z)}{1-\Phi(\frac{y-\mu_Z}{\sigma_Z})}\phi(y)dy$$
This is where I got stuck.
I have not try the approach to use the joint distribution of $(Y,Z)$ to directly compute $E[e^Z 1_{Z\geq Y}]$.
Thank you for your answers.
