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I need to prove that the lognormal distribution has no mgf, that is,

$\int_0^\infty \frac{e^{tx}}{\sqrt{2\pi}x}e^\frac{{-(ln(x))^2}}{2} = \infty$.

What is the best way to start this off?

3 Answers3

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For every $t\gt0$, $tx-\frac12(\ln x)^2-\log x\gt0$ for every $x$ large enough, say every $x\geqslant \xi(t)$, hence $$ \int_0^\infty\frac{\mathrm e^{tx}}{\sqrt{2\pi}x}\mathrm e^{-(\ln x)^2/2}\mathrm dx\gt\int_{\xi(t)}^\infty\frac1{\sqrt{2\pi}}\mathrm dx=+\infty. $$

Did
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    I understand your thinking but how exactly would you show $tx-\frac{1}{2} (\ln x)^2-\ln x > 0$? This is the most crucial part of the proof. –  Apr 13 '14 at 17:33
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    I would note that the derivative is $t+o(1)$, hence $\geqslant\frac12t$ for every $x$ large enough, which implies that the limit of the function itself when $x\to\infty$ is $+\infty$. – Did Apr 13 '14 at 18:40
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    @Anonymous: For $u\gt\sqrt{e}$, $\frac{\mathrm{d}}{\mathrm{d}u}\left(\frac{\log(u)}{u}+\frac1{2u}\right) =\frac{\frac12-\log(u)}{u^2}\lt0$. That is, $\frac{\log(u)}{u}+\frac1{2u}$ is monotonically decreasing and by L'Hospital, $\lim\limits_{u\to\infty}\left(\frac{\log(u)}{u}+\frac1{2u}\right)=0$. Thus, for any $t\gt0$, we can find a $u_t$ so that if $u\gt u_t$, then $$\begin{align} \sqrt{t/2}&\gt\frac{\log(u)}{u}+\frac1{2u}\ \sqrt{t/2},u&\gt\log(u)+\frac12\ tu^2&\gt2\log(u)^2+2\log(u)\ tx&\gt\frac12\log(x)^2+\log(x) \end{align}$$ where $x=u^2$. – robjohn Apr 13 '14 at 19:11
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    @Did, could you please elaborate why $t\gt0$, $tx-\frac12(\ln x)^2-\log x\gt0$ (although you mentioned about $t + o(1)$ but I don't understand that implication either), and why that inequality would lead to the second inequality? Thanks in advance.$$ \int_0^\infty\frac{\mathrm e^{tx}}{\sqrt{2\pi}x}\mathrm e^{-(\ln x)^2/2}\mathrm dx\gt\int_{\xi(t)}^\infty\frac1{\sqrt{2\pi}}\mathrm dx=+\infty. $$ – Nemo Jan 28 '20 at 02:16
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There's another cool way to prove this, let's see....

With $c\in\mathbb{R^+},\:X$ ~ LN$(\mu,\sigma^2)\:$ we have $\:Y=\ln(X)$ ~ N$(\mu,\sigma^2)\:$

$M_X(t)=E\large[$$e^{tX}\large]$$=E\large[$$e^{te^Y}\large]=$ $\large\frac{1}{\sigma \sqrt{2\pi}}\LARGE\int_{_{\Large\mathbb{R}}}\normalsize e^{\large{^{te^{^{y}}}}}e^{^{-\large\frac{1}{2}(\frac{y-\mu}{\sigma})^2}}$d$y$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\large\frac{1}{\sigma \sqrt{2\pi}}\LARGE\int_{_{\Large\mathbb{R}}}\normalsize e^{\large{^{te^{^{y}}}}}e^{^{-\large\frac{1}{2\sigma^2}(y^2-2\mu y+\mu^2)}}$d$y$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:>\large\frac{1}{\sigma \sqrt{2\pi}}\LARGE\int_{_{\Large\mathbb{R^+}}}\normalsize e^{\large{{t{(1+y+y^2/2+y^3/6)-\frac{1}{2\sigma^2}(y^2-2\mu y+\mu^2)}}}}$d$y$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sim c\LARGE\int_{_{\Large\mathbb{R^+}}}\normalsize e^{\large{^{y^3}}}$d$y\:\longrightarrow \:\infty$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare$

User 123732
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Hint: when $t > 0$ and $x$ is large, the most important part of the integrand is the $e^{tx}$, which blows up. You might, for example, show that, for any fixed $t > 0$, when $x$ is large enough, $e^{tx/2}/ x > 1$ and $e^{tx/2} e^{-(\ln x)^2/2} > 1$.

Robert Israel
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