Show $E(T^r) = e^{r\mu +\frac{r^2 \sigma^2}{2}}$ in a lognormal distribution.

Question

I'm asked to prove that, if $T$ follows the $LN(\mu, \sigma^2)$, then

$$E(T^r) = \exp\left(r\mu+ \frac{r^2\sigma^2}{2}\right)$$

Can anyone help?

Answer 1

If $T$ follows a log normal distribution, then $\log(T)$ is normally distributed with mean $\mu$ and variance $\sigma^{2}$.

So $E[T^{r}] = E[e^{\log(T^{r})}] = E[e^{r\log(T)}] = \frac{1}{\sqrt{2\pi }\sigma}\int \limits_{-\infty}^{\infty} e^{rx}e^{-(x- \mu)^{2}/\sigma^{2}}\,dx$.

A standard trick is now to work with the integral's exponent and complete the square on it:

\begin{split} \frac{1}{\sqrt{2\pi }\sigma}\int \limits_{-\infty}^{\infty} e^{rx}e^{-(x- \mu)^{2}/\sigma^{2}}\,dx &= \frac{1}{\sqrt{2\pi }\sigma}\int \limits_{-\infty}^{\infty} e^{(-(x- \mu)^{2}/\sigma^{2}) + rx}\,dx\\ &= \frac{1}{\sqrt{2\pi }\sigma}\int \limits_{-\infty}^{\infty} e^{-\frac{1}{\sigma^{2}}((x- \mu)^{2} - \sigma^{2}rx)} \,dx \\ &= \frac{1}{\sqrt{2\pi }\sigma}\int \limits_{-\infty}^{\infty} e^{-\frac{1}{\sigma^{2}}((x^{2} -2\mu x + \mu^{2} - \sigma^{2}rx)} \,dx \\ &= \frac{1}{\sqrt{2\pi }\sigma}\int \limits_{-\infty}^{\infty} e^{-\frac{1}{\sigma^{2}}((x^{2}+ (-2\mu - \sigma^{2}r)x + \mu^{2})} \,dx \\ &= \frac{1}{\sqrt{2\pi }\sigma}\int \limits_{-\infty}^{\infty} e^{-\frac{1}{\sigma^{2}}((\underbrace{x^{2}+ (-2\mu - \sigma^{2}r)x + (\frac{(-2\mu - \sigma^{2}r)}{2})^{2}}_{(x - (2\mu + \sigma^{2}r)/2)^{2} } - (\frac{(-2\mu - \sigma^{2}r)}{2})^{2} + \mu^{2})} \,dx \\ &= \frac{1}{\sqrt{2\pi }\sigma}\int \limits_{-\infty}^{\infty} e^{-\frac{1}{\sigma^{2}}((x - (2\mu + \sigma^{2}r)/2)^{2})} \underbrace{e^{\frac{1}{\sigma^{2}}((\frac{(-2\mu - \sigma^{2}r)}{2})^{2} - \mu^{2})}}_{e^{[(-\mu/\sigma)+ (- \sigma r/2)]^{2}- \frac{\mu^{2}}{\sigma^{2}}}} \,dx \end{split}

Since $[(-\mu/\sigma)+ (- \sigma r/2)]^{2} - \frac{\mu^{2}}{\sigma^{2}} = \frac{\mu^{2}}{\sigma^{2}} + \frac{\sigma^{2}r^{2}}{4} + \mu r - \frac{\mu^{2}}{\sigma^{2}} = \frac{\sigma^{2}r^{2}}{4} + \mu r $ the last line becomes:

$$ e^{\frac{\sigma^{2}r^{2}}{4} + \mu r} \underbrace{ \frac{1}{\sqrt{2\pi }\sigma}\int \limits_{-\infty}^{\infty} e^{-\frac{1}{\sigma^{2}}((x - (2\mu + \sigma^{2}r)/2)^{2})} \,dx}_{1} = e^{\frac{\sigma^{2}r^{2}}{4} + \mu r}$$

So, $E[T^{r}] = e^{\frac{\sigma^{2}r^{2}}{4} + \mu r}$.

Answer 2

You will find an answer here (How to compute moments of log normal distribution?).

The keyword you should have used is "$r$-th moment".

N.B.: lognormal distribution has no moment generating function (Proving that the lognormal distribution has no moment generating function).