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I'm working through the proof of a lognormal random variable and am having some difficulty in moving through it. I understand the following:

Our CDF is $\Phi(\frac{logx - \mu}{\sigma})$, and thus our PDF is $\phi(\frac{logx-\mu}{\sigma})$/$\sigma y$. From that, we can find the momment generating function as follows:

$E(Y^n)$ = $\int_0^{\infty}\frac{x^n\phi(\frac{logx-\mu}{\sigma})}{\sigma x}dx$

However, after that, I'm a bit lost towards exactly what to do. I understand what the final answer is and how I should get there, but I don't know exactly how to complete the square/substitute in this integral. Can anyone help me out here?

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The lognormal doesn't have an MGF; the integral needs to converge for $t$ in a neighborhood of 0, but the integral for $E(e^{tX})$ is not defined on the positive side.

(edit: see the correction i kjetil's answer; of course it does have an MGF, just for $t<0$ -- indeed I originally mentioned that above, but my claim that you need it for some neighborhood above 0 doesn't apply to everything you might want an mgf for)

For some details, see the Wikipedia article on the lognormal distribution

Interestingly, the lognormal is an example of a distribution with a finite moment sequence that is not characterized by that set of moments (i.e. there are other distributions with the same sequence of moments). If the MGF existed in a neighborhood of 0 this could not occur.

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  • $\begingroup$ understood, but then how would I find E(Y^n)? Am I not doing it in the right way? $\endgroup$ – Anna Jan 30 at 2:14
  • $\begingroup$ That would be a new question ... "what's a good way to find $E(Y^n)$ for a lognormal?" (rather than "how do I do this integral?"). Fortunately that question is already answered in the section that the link in my answer takes you to. ... though there are other ways to do it (e.g. you can do it from the MGF of the normal pretty easily; that's how I'd do it) $\endgroup$ – Glen_b Jan 30 at 12:01
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Other answers to this question claims that the moment generating function (mgf) of the lognormal distribution do not exist. That is a strange claim. The mgf is $$\DeclareMathOperator{\E}{\mathbb{E}} M_X(t) = \E e^{tX}. $$ And for the lognormal this only exists for $t\le 0$. The claim is then that the "mgf only exists when that expectation exists for $t$ in some open interval around zero. Well, some important theorems about mgf's depend on such an assumption, so the mgf of the lognormal distribution might lack some properties guaranteed by such theorems, but still be useful.

The existence of papers about the lognormal mgf do suggest that some useful properties there are! This papers often talk about the Laplace transform not mgf, but that is only a parameter change from $t$ to $-t$.

I will come back here with a more complete answer, but for the moment I will just point to some papers. On the Laplace transform of the Lognormal distribution by Søren Asmussen, Jens Ledet Jensen and Leonardo Rojas-Nandayapa. There is no exact formula for the mgf, but that paper gives good approximations. Laplace Transforms of Probability Distributions and Their Inversions Are Easy on Logarithmic Scales by A. G. Rossberg.

Finally, Accurate Computation of the MGF of the Lognormal Distribution and its Application to Sum of Lognormals by C. Tellambura and D. Senaratne, and the paper Uniform Saddlepoint Approximations and Log-Concave Densities by Jens Ledet Jensen uses saddlepoint approximations for the lognormal as an example.

So much for the mgf not existing! And now I discovered that all this (and more) was stated earlier by Cardinal Existence of the moment generating function and variance. For instance, from what Cardinal proves there, one can conclude that the lognormal do not have exponentially decaying tails. (Which is one of the properties that follows from existence of mgf in an open interval containing zero).

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Suppose $X\sim\mathcal{LN}(\mu,\sigma^2)$ is your lognormal variable. Then $E(X^n)$ is not the moment generating function; it is the $n$th order raw moment of $X$ about zero. Remember that the moment generating function of a lognormal distribution does not exist but all moments do exist.

By definition, $Y=\ln X\sim \mathcal N(\mu,\sigma^2)$.

Hence, using the mgf of a univariate normal distribution, $$E(X^n)=E(e^{nY})=e^{n\mu+n^2\sigma^2/2}$$

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