Expectation of the product of a Log-normal and an Indicator Function I'm trying to show that $$E\left[ e^{\mu+\sigma Z} \mathbf{1}_{\{Z>-d\}} \right] = e^{\mu+\frac{\sigma^2}{2}} \Phi(d+\sigma),$$ where $Z$ is the standard normal.
As standard results I have $$E\left[e^{\mu+\sigma Z}\right] = e^{\mu + \sigma^2/2}$$ and $$E\left[ \mathbf{1}_{\{Z>-d\}} \right] = \Phi(d).$$
Then I take the product (since they're independent) to get something that's almost the required result, but I'm not sure why it should be $\Phi(\sigma+d)$ rather than just $\Phi(d)$.
 A: You can rewrite $E\left[ e^{\mu+\sigma Z} \mathbf{1}_{\{Z>-d\}} \right] = e^{\mu+\frac{\sigma^2}{2}} \Phi(d+\sigma)$ as 
$$E\left[ e^{\mu+\sigma Z} \mathbf{1}_{\{Z>-d\}} \right] = Pr(Z>-d)E\left[ e^{\mu+\sigma Z} | Z >-d \right] + Pr(Z<-d)E\left[0\right] = Pr(Z>-d)E\left[ e^{\mu+\sigma Z}  | Z >-d\right] $$ 
Let $X = e^{\mu+\sigma Z}$ be the log-normal variable. Then, the previous expression can be rewritten as
$$E\left[ e^{\mu+\sigma Z} \mathbf{1}_{\{Z>-d\}} \right]  = Pr(X>e^{\mu - \sigma d})E\left[ X  | X>e^{\mu - \sigma d} \right],$$ 
which is the partial expectation of a lognormal variable with parameters $\mu$ and $\sigma^2$. Log-normal partial expectations have a closed-form solution as
$$g(k) = \int_k^\infty \!x{\ln\mathcal{N}}(x)\, dx
            = e^{\mu+\tfrac{1}{2}\sigma^2}\, \Phi\!\left(\frac{\mu+\sigma^2-\ln k}{\sigma}\right)$$
In our problem, $k= e^{\mu - \sigma d}$, so 
$$E\left[ e^{\mu+\sigma Z} \mathbf{1}_{\{Z>-d\}} \right] = e^{\mu+\tfrac{1}{2}\sigma^2}\, \Phi\!\left(\frac{\mu+\sigma^2-\ln k}{\sigma}\right)=e^{\mu+\tfrac{1}{2}\sigma^2}\, \Phi\!\left(d + \sigma\right)$$ 
