Expectation with indicator function $E[I(X > 0)e^{X}]$

Question

How do we compute the expectation of: $$E[I(X > 0)e^{X}]$$ where X is a standard normal? the expectation of $e^X$ is $e^{\frac{1}{2}}$ and that of $I(X > 0)$ is $\frac{1}{2}$ but how do I bring is together?

Thank you!

Answer 1

We can use the definition of expectation, complete the square, do a change a variable and express it in terms of CDF of standard normal distribution.

\begin{align} E[I(X>0)e^X] &= \int_0^\infty e^x \frac1{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\, dx \\ &=\int_0^\infty \frac1{\sqrt{2\pi}} \exp\left( -\frac{x^2-2x}2\right) \, dx \\ &= \int_0^\infty \frac1{\sqrt{2\pi}} \exp\left( -\frac{(x-1)^2-1}2\right) \, dx \\ &= e^\frac12\int_0^\infty \frac1{\sqrt{2\pi}} \exp\left( -\frac{(x-1)^2}2\right) \, dx \\ &= e^{\frac12} \int_{-1}^\infty \frac1{\sqrt{2\pi}} \exp\left( -\frac{x^2}2\right) \, dx \\ &= e^\frac12 Pr(Z > -1)\\ \end{align}