Conditional Expectation of Log-Normal Distribution I want to evaluate a conditional expectation of log-normal distribution. Let $y$ be a log-normal distributed random variable. So $\log(y)\sim N(\mu,\sigma^2)$. I want to calculate $E[y-1|y-1>0].$ If we assume that $X\sim N(\mu,\sigma^2)$, then the problem can be seen as $E[e^x-1|e^x-1>0].$ Does it actually have a closed-form? 
 A: First notice that we can write the last expectation as $\mathbf{E}[e^X|X>0] - 1 = (\int e^x f_{X|X>0}(x)dx)-1$. We will focus on evaluating the integral.
Let $\Phi_{\mu,\sigma}$ be the distribution function of the $\mathcal{N}(\mu,\sigma^2)$. Writting in an informal manner, the density of X|X>0 is given by
$$
\begin{align}
f_{X|X>0}(x) &= \frac{P(X = x,X>0)}{P(X>0)} \\
             &= \mathcal{1}_{(0,\infty)}(x)
\left(\frac{e^{\frac{-(x-\mu)^2}{2\sigma^2}}}{\sqrt{2\pi\sigma^2}}\right)\frac{1}{1-\Phi_{\mu,\sigma}(0)}\quad .
\end{align}
$$
Therefore the integral becomes 
$$\int e^x f_{X|X>0}(x)dx = \frac{1}{1-\Phi_{\mu,\sigma}(0)}\int_0^\infty e^x\frac{e^{\frac{-(x-\mu)^2}{2\sigma^2}}}{\sqrt{2\pi\sigma^2}}dx
$$
Before evaluating it, we do a little algebra with the terms on the exponential function
$$
\begin{align}
 x + \frac{-(x-\mu)^2}{2\sigma^2} &= \frac{-x^2+2x(\mu+\sigma^2)-\mu^2}{2\sigma^2}\\
 & = \frac{-x^2+2x(\mu+\sigma^2)-(\mu+\sigma^2)^2}{2\sigma^2}+\frac{-\mu^2+(\mu+\sigma^2)^2}{2\sigma^2}\\
 & = \frac{-(x-(\mu+\sigma^2))^2}{\sigma^2} + \mu+\frac{\sigma^2}{2}\quad.
\end{align}
$$
Define $\mu^* = \mu+\sigma^2$, we evaluate the integral
$$
\begin{align}
\int e^x f_{X|X>0}(x)dx &= \frac{1}{1-\Phi_{\mu,\sigma}(0)}\int_0^\infty e^x\frac{e^{\frac{-(x-\mu)^2}{2\sigma^2}}}{\sqrt{2\pi\sigma^2}}dx\\
& = e^{\mu+\frac{\sigma^2}{2}}\frac{1}{1-\Phi_{\mu,\sigma}(0)}\int_0^\infty \frac{e^{\frac{-(x-\mu^* )^2}{2\sigma^2}}}{\sqrt{2\pi\sigma^2}}dx\\
&= e^{\mu+\frac{\sigma^2}{2}}\frac{1-\Phi_{\mu^*,\sigma}(0)}{1-\Phi_{\mu,\sigma}(0)}\int_0^\infty \frac{e^{\frac{-(x-\mu^* )^2}{2\sigma^2}}}{\sqrt{2\pi\sigma^2}}\frac{1}{1-\Phi_{\mu^*,\sigma}(0)}dx\\
& = e^{\mu+\frac{\sigma^2}{2}}\frac{1-\Phi_{\mu^*,\sigma}(0)}{1-\Phi_{\mu,\sigma}(0)}
\end{align}
$$
The final equation holds because we are integrating the density of a random variable of the form $X^*|X^*>0$, where $X^* \sim \mathcal{N}(\mu^*, \sigma^2)$. Hence
$$
E[e^X-1|e^X-1>0] = e^{\mu+\frac{\sigma^2}{2}}\frac{1-\Phi_{\mu^*,\sigma}(0)}{1-\Phi_{\mu,\sigma}(0)}-1\quad.
$$
Obs: This is my first answer, I would be grateful if you could told me what I should improve. Thanks! Hope it helps.
