# 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?

• I'd be inclined to evaluate $E[Y|Y>1]-1$ (which is equivalent, via properties of expectation). Apr 5, 2019 at 9:18
• It does have a closed form in terms of the exponential function and the error function (which is linearly equivalent to the Normal CDF).
– whuber
Apr 5, 2019 at 14:30

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.

• +1. Your answer is very clear and well explained. Welcome to our site!
– whuber
Apr 5, 2019 at 16:24