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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?

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

1 Answer 1

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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.

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

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