If I only know geometric mean and geometric standard deviation of a log-normal distribution how can I calculate the $n$-th moment of the distribution?

In the Wikipedia article I can only see a relationship for the $n$-th moment if the (arithmetic) mean and standard deviation are known:

$$ E\left[X^n\right]=e^{n\mu+\frac{1}{2}n^2\sigma^2} $$

If only the geometric mean and geometric standard deviation are known, is there still a way to calculate the moments?

The geometric mean and standard deviation are defined by:

$$ GM = e^{\mu_l}\quad\mathrm{where}\quad\mu_l = \frac{\sum_{i=1}^N\ln(x_i)}{N} $$


$$ GSD = e^{\sigma_l}\quad\mathrm{where}\quad\sigma_l=\sqrt{\frac{\sum_{i=1}^N\left[\ln(x_i)-\mu_l\right]^2}{N}} $$

  • $\begingroup$ Hi caverac, thanks for your comment! Ok, but is it possible to calculate a possible set of moments for a given GM and GSD? $\endgroup$ – Axel May 14 '18 at 13:47
  • $\begingroup$ I think I misunderstood the question the first time. You have an estimative of both GSD and GM and you want to know if from this you can calculate $\mu$ and $\sigma$, is that it? $\endgroup$ – caverac May 14 '18 at 14:03
  • $\begingroup$ Maybe I also misunderstood some concept about the moments, that's why probably my question is not phrased that well. What I actually want to calculate are the non-central moments as defined in the table in en.wikipedia.org/wiki/Normal_distribution#Moments for a normal distribution. In en.wikipedia.org/wiki/Log-normal_distribution#Geometric_moments there is no such table, but I thought the provided formula would allow to calculate the moments. If I would have $\mu$ and $\sigma$ I could use the formula of course. I am not sure anymore if those are actually the moments I want. $\endgroup$ – Axel May 14 '18 at 14:10
  • $\begingroup$ :) That's exactly the reason I asked you this, because it definitely seems that you're after a way of finding both $\mu$ and $\sigma$ you can use $\mathbb{E}[X^n]$, which for the lognormal distribution is the same as $\mathbb{E}[|X|^n]$ $\endgroup$ – caverac May 14 '18 at 14:16
  • $\begingroup$ So the phrase "moments" when referring to a log-normal distribution does not mean the same when referring to normal (Gaussian) distribution? Because I am after the equivalent of those moments that are defined for the normal distribution as $m_0 = 1, m_1 = \mu, m_2 = \mu^2 + \sigma^2, m_3 = \mu^3 + 3\mu\sigma^2$ and so on. $\endgroup$ – Axel May 14 '18 at 14:23

For simplicity I will call $\mathcal{N}_k$ the $k$-th moment of the normal distribution with parameters $\mu$ and $\sigma$, so for example $\mathcal{N}_1 = \mu$, $\mathcal{N}_2 = \mu^2 + \sigma^2$, $\cdots$

Now if $X$ follows a lognormal distribution, then

\begin{eqnarray} \mathbb{E}[\ln^k X] &=& \int\frac{{\rm d}x~}{x}\frac{\ln^k x}{\sigma\sqrt{2\pi}} \exp\left[-\left(\frac{\ln x - \mu}{2\sigma^2}\right)^2\right] \\ &\stackrel{y=\ln x}{=}&\int{\rm d}y ~\frac{y^k}{\sigma\sqrt{2\pi}}\exp\left[-\left(\frac{y-\mu}{2\sigma^2}\right)^2 \right] \\ &=& \mathcal{N}_k \tag{1} \end{eqnarray}

In you case, you have then

$$ \mu_l = \ln{\rm GM} = \mathbb{E}[\ln X] = \mathcal{N}_1 = \mu $$


$$ \sigma_l^2 = \mathbb{E}[(\ln X - \mu_l)^2] = (\cdots) = \sigma^2 $$

$\mu_l$ and $\sigma_l$ thus allow you to estimate $\mu$ and $\sigma$

  • $\begingroup$ @Axel This was too long to add in the comments, if this is not what you had in mind I will gladly delete it $\endgroup$ – caverac May 14 '18 at 14:52
  • $\begingroup$ Thanks for your answer! So if $\mu_l = \mu$ and $\sigma^2_l = \sigma^2$ can I then just use the same definition $m_0 = 1$, $m_1 = \mu_l$, $m_2 = \mu_l^2 + \sigma_l^2$ to calculate the moments? Wouldn't that mean that it doesn't make a difference if the distribution is normal or log-normal? $\endgroup$ – Axel May 14 '18 at 15:08
  • $\begingroup$ @Axel You have to be careful there! The answer is no! To give you an example, $m_1^{\rm normal} = \mu$, whereas $m_1^{\rm lognormal} = e^{\mu + \sigma^2/2}$ $\endgroup$ – caverac May 14 '18 at 15:12
  • $\begingroup$ Ok, thanks for clarification. So the general formula is $m_n = e^{n\mu+\frac{1}{2}n^2\sigma^2}$? In this formula is $\mu = \mu_l = \ln \mathrm{GM}$ and $\sigma^2 = \sigma_l ^2 = \ln\left(\mathrm{GSD}^2\right)$? $\endgroup$ – Axel May 14 '18 at 15:43
  • $\begingroup$ @Axel $\sigma_l^2 = \ln^2({\rm GSD})$ $\endgroup$ – caverac May 14 '18 at 15:47

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