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I am trying to calculate the expectation $$E[e^{cX}]$$ for arbitrary $c<0$ (for $c>0$ the expectation is infinite) if $X$ is lognormally distributed, i.e. $\log(X) \sim N(\mu, \sigma)$.

My idea was to write the expectation as an integral, but I did not see how to proceed: $$E[e^{cX}] = \frac{1}{\sqrt{2\sigma\pi}}\int_0^\infty \frac{1}{x}\exp\left(cx - \frac{(\log x - \mu)^2}{2\sigma^2}\right)dx$$

I also tried the Itô formula (the actual task is to find $E[e^{cX_T} \mid X_t = x]$ where $X$ is a geometric Brownian motion, but it reduces to the problem above because we are looking at a Markov process), but that did not look very promising either. Can anybody help me?

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    $\begingroup$ You should consider editing your question (by clicking the "edit" link at lower left) to add the self-study tag to this question. $\endgroup$
    – Alexis
    Commented Jul 28, 2014 at 17:38
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    $\begingroup$ This exists only as a formal power series which has no closed-form expression. $\endgroup$
    – whuber
    Commented Jul 29, 2014 at 1:35
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    $\begingroup$ Thank you so much! Although this is not what I hoped for, it also proves my professor wrong. And that is an accomplishment of its own ;-) $\endgroup$ Commented Jul 29, 2014 at 19:19

1 Answer 1

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What you want is the moment generating function of a lognormal variable, which is known to be a hard problem. Alternatively, this is the Laplace transform, which is your expression with $c$ replaced by $-c$. You should have a look at https://en.wikipedia.org/wiki/Log-normal_distribution which do have some useful information.

The paper "On the Laplace transform of the lognormal distribution" by Søren Asmussen, Jens Ledet Jensen and Leonardo Rojas-Nandayapa do give the following approximation, which they investigate in detail. Let $X$ be lognormal with parameters $(\mu, \sigma^2)$, which means that $X=e^Y$ with $Y \sim N(\mu, \sigma^2)$. The Laplace transform is $$ E(\exp(-\theta e^y) = e^{-\theta \mu} E(\exp(-\theta e^{Y_0}) $$ where $Y_0 \sim N(0,\sigma^2)$. So we consider the Laplace transform $L(\theta) = E(\exp(-\theta e^{Y_0})$. Then they give the approximation to $L(\theta)$: $$ \frac1{\sqrt{1+W(\theta \sigma^2)}}\exp\left\{ -\frac1{2\sigma^2} W(\theta \sigma^2)^2 - \frac1{\sigma^2} W(\theta \sigma^2) \right\} $$ where $\theta$ is nonnegative. Here $W$ is the Lambert W function, see https://en.wikipedia.org/wiki/Lambert_W_function . (Then the paper looks into the quality of this approximation, and compares it to older approximations).

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