Originally Posted by

**mr fantastic** If X is a random variable with normal distribution then the random variable $\displaystyle Y = e^X$ has a lognormal distribution.

I'll assume a standard normal distribution to make things simple. Then:

$\displaystyle E\left[ e^{tY} \right] = E\left[ e^{te^X} \right]$

$\displaystyle = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{t \, e^x} e^{-\frac{x^2}{2}} \, dx$

$\displaystyle = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{t \, e^x -\frac{x^2}{2}} \, dx$

$\displaystyle \geq \frac{1}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{t \, \left( 1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right) -\frac{x^2}{2}} \, dx$.

Since the exponent is a cubic for which the $\displaystyle x^3$ term is positive (if t > 0), the exponential must approach $\displaystyle + \infty$ as $\displaystyle x \rightarrow \infty$.

Therefore $\displaystyle E\left[ e^{tY} \right] = + \infty$.