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Given the sample mean $\bar{x}$ and the sample variance $s^2$ of a random variable $X$, is it possible to estimate the shape $\sigma^2$ and log-scale $\mu$ of the log-normal distribution, with probability density function $$f_X(x;\mu,\sigma^2) =\frac{1}{x \sqrt{2 \pi \sigma^2}}\, \mathrm{exp}\left( {-\frac{(\ln x - \mu)^2}{2\sigma^2}}\right), \text{ if } x>0,$$ and if so, how are any resulting formulae derived?

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Suppose that $X = \ln(Y)$ follows a Normal distribuion with mean $\mu$ and variance $\sigma^2$ ($N(\mu,\sigma^2)$).

Using $$ E(g(X)) = \int_{-\infty}^{+\infty}g(x)p(x)dx$$ (where $p(x)$ is the pdf of the Normal distribution), we have that

$$E(Y) = E(\exp(X)) = \int_{-\infty}^{+\infty}\exp(x)p(x)dx=\exp(\sigma^2/2+\mu)$$ $$E(Y^2) = E(\exp(X)^2) = \int_{-\infty}^{+\infty}\exp(x)^2p(x)dx=\exp(2\sigma^2+2\mu)$$

from which we conclude that the variance is

$$V(Y) = \exp(2\mu+2\sigma^2)-\exp(2\mu+\sigma^2)$$

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    $\begingroup$ Perhaps you would like to relate your answer with the method of moments or MLE for completeness. $\endgroup$ – user10525 Apr 17 '12 at 15:55
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    $\begingroup$ ...And it's worth performing a quick analysis of the properties of these estimators (the MM estimate of $\sigma^2$ has a strong low bias and the MM estimates of both are more variable than one might expect). $\endgroup$ – whuber Apr 17 '12 at 16:54
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This is not directly an answer to your question:

But if you can instead get the mean and standard deviation of log X then you should be able to reuse the existing estimators for the normal distribution.

That seems to be the most simple method, actually, unless there is additionally a location parameter.

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