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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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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