In Casella and Berger (2002) I found a proof for the moment-generating function (mfg) of a lognormal distribution not being existent (see exercise 2.36 on page 81 and the answer provided here on page 2-12).

Starting point is the following lognormal pdf (with $\mu = 0$ and $\sigma^2 = 1$):

$f(x) = \int_0^\infty \frac 1{2\pi x}e^{-(ln(x))^2/2}dx$

The mgf of the lognormal distribution is therefore

$M_x(t) = \int_0^\infty \frac {e^{tx}}{2\pi x}e^{-(ln(x))^2/2}dx$

In reference to l'Hopital rule, they then point out that

$\lim \limits_{x \to \infty} e^{tx-(ln(x))^2} = \infty.$

They conclude the proof by stating that for any $k > 0$ there is a constant $c$ such that

$\int_k^\infty \frac {e^{tx}}{x}e^{-(ln(x))^2/2}dx\ge c\int_k^\infty \frac 1xdx = c\ln|_k^\infty = \infty.$

Hence, the $M_x(t)$ does not exist.

I think I get the proof. Key is when $t>0$ and $x \to\infty$ then $e^{tx}$ tends to blow up (also see here).

What I do not get is what constant $c$ is supposed to represent. It must be related to $e^{tx}e^{-(ln(x))^2/2}$. However, why can one integrate out any term based on $e^{tx}e^{-(ln(x))^2/2}$ if it is dependent on $x$? Can we integrate out such a term if if we evaluate it accordingly? For example, something along the lines of $()|_k^{\infty}$?

  • $\begingroup$ Consider $t<0$ then you have a momentum genrating function which exists and from which you can generate the moments by differentiating in the vicinity of $t = 0_ {-}$. $\endgroup$ – Dr. Wolfgang Hintze Feb 27 '19 at 17:36
  • $\begingroup$ I quote "Key is when t>0". Again, why do you assume the (unfortunate) case that $t\gt0$? In the definition of the mgf (en.wikipedia.org/wiki/Moment-generating_function) it is just said that $t$ must be real. Hence, contrary to your statement, the mgf of the lognormal distribution exists, viz. for $t<=0$. $\endgroup$ – Dr. Wolfgang Hintze Feb 28 '19 at 2:18
  • $\begingroup$ @Dr.WolfgangHintze: thanks for pointing this out. $t>0$ is really not necessary I assume. In fact, in mgfs, $t$ is generally considered a place holder. If we want to determine a moment using a mgf then we even would evaluate the integral at $t=0$. Thus, so I assume, the key is $ x \to \infty$. If this is what you had in mind then I will edit my question accordingly. $\endgroup$ – DomB Feb 28 '19 at 7:03
  • $\begingroup$ @ DomB (1) No, the "key" is not $x\to\infty$ but $ t\le0$. And (2) to get the moments the integral is to be evaluated (its derivatives) in the vicinity of $t=0$ but from below (as I have written in my 1st comment). I think I edit my answer to include more precisely what I mean. $\endgroup$ – Dr. Wolfgang Hintze Feb 28 '19 at 9:59

The $c$ represents a positive real number such that $$\color{blue}{e^{tx}e^{-(\ln x)^2/2} \ge c\quad \forall x\ge k}$$ (the $c$ can depend on $t$, but is independent of $x$). Make sure you can explain/show why such a $c$ exists! (It's related to the fact that for any $t>0$, we have $e^{tx}e^{-(\ln x)^2/2} \to \infty$ as $x\to \infty$.) Because of this inequality, since $k > 0$, we get the integral inequality obtained in your post.

  • $\begingroup$ Thanks for this. You mentioned that $c$ is independent of $x$. But why is that? Is it because we can plug $k$ into $e^{tx}e^{-(ln(x))^2/2}$ and that way it is not dependent on $x$ anymore? $\endgroup$ – DomB Feb 28 '19 at 5:53
  • $\begingroup$ Basically you should be able to find (or show there exists) a positive lower bound $c$ for $f(x):= e^{tx}e^{-(\ln x)^2/2}$, that is, some $c > 0$ such that $f(x) \ge c$ for all $x\ge k$. (For example, you could take $c$ to be the minimum value of $f(x)$ for $x\ge k$, if you could show that this minimum value exists and is positive.) This $c$ is then independent of $x$. The reason we want it to be independent of $x$ is essentially that the thing we are integrating simplifies to $c/x$, which is easy to integrate and make statements about. $\endgroup$ – Minus One-Twelfth Feb 28 '19 at 5:57
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    $\begingroup$ I got it. You plug in $k$ and that way it becomes the lower bound. Now it makes perfect sense. Cheers $\endgroup$ – DomB Feb 28 '19 at 6:42


I'd like to point out more clearly what I addressed in my comments.

Namely, that the "proof" of non existence of a moment generating function of the lognormal distribution is wrong. Simply because a mgf exists and will be provided below.

The formula for a moment generating function in question is

$$g(t) = \int_0^\infty f(x) e^{t x}\,dx\tag{e1}$$

where the pdf of the lognormal distribution is given by $(1)$ below.

Now the general definition of a moment generating function (mgf) can be found in (https://en.wikipedia.org/wiki/Moment-generating_function) and reads

"The moment-generating function of a random variable $X$ is the expectation of the random variable $e^{t X}$, where $t\in \mathbb {R}$, and wherever this expectation exists "

Notice that it is just stated that $t$ is a real number but nothing is said about its sign.

Hence we look at $(e1)$ and ask when "this expectation exists".

The answer is that the integral must be convergent, and convergence of the integral requires $t\le0$. The case $t\gt0$ was already studied in the answers of others and was shown to violate the condiditon that the expectation exists (because the integral diverges). Hence the formula for $t\gt0$ does NOT define a mgf. And consequently it can't be used to "prove" that the lognormal distribution has no mgf.

Here is a true mgf for the lognomal distribution.

Letting $t=- s$ we assume that $s\ge0$ an write

$$g(s) = \int_0^\infty f(x) e^{-s x}\,dx\tag{e2}$$

Now since the $f(x)$ is positive and $e^{-s x}\le1$ we have $g(s)\le m(0) = 1$. Hence the integral exists and therefore it is a valid definition of the mgf.

Here's a graph of the mgf obtained by numerically integrating over $x$

enter image description here

The moments (calculated in my original post) are retrieved by the expansion

$$g(s) = 1+\frac{(-s)}{1!}m(1) + \frac{s^2}{2!} m(2)+...+ \frac{(-s)^k}{k!} m(k)+ ...\tag{e3}$$

Notice that this series is divergent and should be treated as a formal series which has to be cut off at some convenient point $k$.

It would be desirable to have a closed expression for mgf but up to now I have only derived some approximate formulae which are not very enlightning.

Original post

You can also start from the moments and ask which type of generating function could be possible and find that the standard mgf does not fit into this scheme.

The pdf of the lognormal distribution is

$$f(x) = \frac{1}{x \sqrt(2 \pi) } \exp \left(-\frac{1}{2} \log ^2(x)\right)\tag{1}$$

The k-th moment defined as

$$m(k) = \int_0^\infty x^k f(x)\tag{2}$$

is given by

$$m(k) = e^{\frac{k^2}{2}}\tag{3}$$

A moment generating function $g(t)$ would be a sum like

$$g_d(t) = \sum_{k=0}^\infty m(k) \frac{t^k}{d(k)} \tag{4}$$

where the denomintor $d(k)$ must be chosen so that the sum converges.

These examples

$$g_{d=1}(t) = \sum_{k=0}^\infty m(k) t^k\tag{5a} $$

$$g_{d=k!}(t) = \sum_{k=0}^\infty \frac{m(k)}{k!} t^k\tag{5b} $$

do not work in our case, as $\lim_{k\to\infty}{ m(k) /d(k)} \to \infty$, and hence the sums diverge for any $t>0$.

The second case corresponds to the standard definition of a mgf as the expectation value of $e^{k t}$. And here the sum diverges.

A somewhat unusual case would be

$$g_{d=k!^k}(t) = \sum_{k=0}^\infty \frac{m(k)}{k!^k} t^k \tag{5c}$$

Here the sum is convergent for any $t$ and the moments can be found from the $k$-th derivative at $t=0$. But this sum hardly admits a closed analytic expression.


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