MLE for a log-normal on log-scale

Question

Daily traded volumes are modeled as log-normal, meaning $\ln X_i$ is normal with mean $\mu$ and variance $\sigma^2$ . The log-normal density is

$f(x \mid \mu, \sigma^2) = \frac{1}{x\sqrt{2\pi\sigma^2}}\, \exp\!\left(-\frac{(\ln x - \mu)^2}{2\sigma^2}\right), \qquad x > 0.$

Taking natural logs of five observations gives $y_i = \ln x_i$ equal to 2.0, 2.5, 3.0, 3.5, 4.0.

Derive the maximum likelihood estimator of $\mu$ and evaluate it.

Accepted Answer

Write yi = \ln xi. The log-likelihood is \ell(\mu, \sigma^2) = \sumi \left[ -\ln xi - 	frac12\ln(2\pi\sigma^2) - \frac{(yi - \mu)^2}{2\sigma^2} ight]. The term -\ln xi does not involve the parameters, so maximizing over \mu and \sigma^2 is exactly the normal MLE problem applied to the logs yi. Differentiating in \mu, \frac{\partial \ell}{\partial \mu} = \frac{1}{\sigma^2}\sumi (yi - \mu) = 0 \;\Longrightarrow\; \hat\mu = \bar y = \frac{1}{n}\sumi yi. With logs 2.0, 2.5, 3.0, 3.5, 4.0 the mean is \bar y = 15/5 = 3, so \hat\mu = 3. (As a bonus, \hat\sigma^2 = \frac{1}{n}\sumi (yi - \bar y)^2…

MLE for a log-normal on log-scale

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