Expected value of a lognormal distribution [duplicate]

Solution 1:

Standard method to find expectation(s) of lognormal random variable.

1)

Determine the MGF of $U$ where $U$ has standard normal distribution.

This comes to finding the integral:$$M_U(t)=\mathbb Ee^{tU}=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{tu}e^{-\frac12u^2}du=e^{\frac12t^2}$$

2)

If $Y$ has lognormal distribution with parameters $\mu$ and $\sigma$ then it has the same distribution as $e^{\mu+\sigma U}$ so that: $$\mathbb EY^{\alpha}=\mathbb Ee^{{\alpha}\mu+{\alpha}\sigma U}=e^{{\alpha}\mu}\mathbb Ee^{{\alpha}\sigma U}=e^{{\alpha}\mu}M_U({\alpha}\sigma)=e^{{\alpha}\mu+\frac12{\alpha}^2\sigma^2}$$

Solution 2:

Hint:

By the substitution $y=e^z$, you transform to

$$\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty} e^z\exp\left(-\frac{1}{2}\left(\frac{z-\mu}{\sigma}\right)^2\right)e^z\,\mathrm{d}z=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty} \exp\left(-\frac{1}{2}\left(\frac{z-\mu}{\sigma}\right)^2+2z\right)\,\mathrm{d}z$$ which you can reduce to a standard Gaussian integral by shifing the variable, giving the value $1$.