I'm having trouble with proving the following problem, I think partial expectation should be used to solve this but can't figure it out.
Show that:
$E[X \cdot I(X\leq x)] = e^{\mu + \frac{1}{2}\sigma^{2}} \Phi(\frac{ln(x)-\mu-\sigma^{2}}{\sigma})$
where $I$ is the indicator function and $X$ is log-normal distributed with parameter $\mu$ and $\sigma$, i.e. $X \sim LN(\mu, \sigma^{2})$.
Appreciate if I can get some help with this and hopefully this is sufficient information but comment if you guys need more.
For $Y=log X\sim N(\mu,\sigma^2)$ and $x>0$, $$E[X\cdot I(X\leq x)]=E[e^Y\cdot I(e^Y\leq x)]=E[e^Y\cdot I(Y\leq \log x)]= \int_{-\infty}^{\log x} e^y \dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(y-\mu)^2}{2\sigma^2}}\ dy=\int_{-\infty}^{\log x} \dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(y-\mu)^2-2y\sigma^2}{2\sigma^2}}\ dy$$
Next completing the square in $(y-\mu)^2-2y\sigma^2=(y-(\mu+\sigma^2))^2-2\mu\sigma^2-\sigma^4$ and put it back into the exponent in the above integral. The answer will follow.