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