Covariance combined with normal distribution

Question

We have $N_1$ and $N_2$, normal distributed random variables with averages $µ_i=E[N_i]$ and variances $σ_i^2=Var[N_i]$ and $c = Cov(N_1, N_2)$. We want to compute $E[e^{N_1} I(N_2>0)]$, where I is the indicator function. I have looked over lots of definitions of covariance but I don't know which one is applicable in this case.

Answer 1

The declaration that you "have looked over lots of definitions of covariance but don't know which one is applicable in this case" is quite mysterious... but anyway, a canonical representation of the gaussian vector $(N_1,N_2)$ proves useful in this context, that is, the fact that there exists some $a$, $b$ and $d$ such that $$ N_1=aN_2+bN+d, $$ where $N$ is standard normal and independent of $N_2$. Thus, the expectation to compute is $$ E[\mathrm e^{bN}\mathrm e^d\mathrm e^{aN_2}:N_2\gt0]=E[\mathrm e^{bN}]\mathrm e^dE[\mathrm e^{aN_2}:N_2\gt0], $$ where $E[\mathrm e^{bN}]=\mathrm e^{b^2/2}$. Note that $(a,b,d)$ solve the system $$ \mu_1=a\mu_2+d,\qquad\sigma_1^2=a^2\sigma_2^2+b^2,\qquad c=a\sigma_2^2, $$ hence $(a,b,d)$ is known. On the other hand, $N_2$ is distributed as $\mu_2+\sigma_2N$ hence $$ E[\mathrm e^{aN_2}:N_2\gt0]=\mathrm e^{a\mu_2}E[\mathrm e^{a\sigma_2N}:N\gt-\mu_2/\sigma_2]. $$ For every $(u,v)$, $$ E[\mathrm e^{uN}:N\gt v]=\int_v^\infty\mathrm e^{ux}\varphi(x)\mathrm dx=\mathrm e^{u^2/2}\int_v^\infty\varphi(x-u)\mathrm dx=\mathrm e^{u^2/2}\Phi(u-v), $$ hence, finally, $$ E[\mathrm e^{N_1}:N_2\gt0]=\exp(\mu_1+\tfrac12\sigma_1^2)\cdot\Phi(\sigma_2^{-1}(c+\mu_2)). $$