Let $X_1, X_2,...,X_n$ be iid random variables having pdf $$f(x|\theta) = \frac{1}{x \sqrt{2\pi\theta}}e^{(-[\log x]^2/[2\theta])} I_{(0,\inf)}(x)$$ where $\theta > 0$.

Determine the MLE of $\theta$. Since this is an exponential family distribution, this pdf gets factored into: $h(x) = \frac{1}{x} I_{(0,inf)}$, $c(\theta) = \frac{1}{\sqrt{2\pi\theta}}$, $w(\theta) = -1/2\theta$, and $t(x) = (\log x)^2$

I solved this by using the natural parameterization, so $\eta = \frac{-1}{2\theta}$ and therefore $\theta = \frac{-1}{2\eta}$.

We have that $c(\eta) = \frac{1}{\sqrt{-\pi/\eta}}$ and therefore, $\log c(\eta) = \log\frac{1}{\sqrt{-\pi/\eta}}$.

The log likelihood function is $$L(\eta|x) = \sum\log\frac{1}{x_i} + n\frac{1}{\sqrt{-\pi/\eta}} + \eta\sum(\log x_i)^2$$.

After taking the derivative with respect to $\eta$, I got that $$\frac{d}{d\eta}L(\eta|x)=n\frac{-1}{2\eta} + \sum(\log x_i)^2$$

Setting the derivative equal to zero, we get $$n\frac{-1}{2\eta} + \sum(\log x_i)^2 = 0$$ $$\frac{-1}{2\eta} = -\frac{\sum(\log x_i)^2}{n}$$ Subbing $\theta$ back in we get $$\hat\theta_{MLE} = \frac{\sum(\log x_i)^2}{n}$$

I just want to make sure that my work and rationale is correct in solving this.

  • $\begingroup$ I think the middle term in your L(eta|x) formula should be a logarithm. But your derivative is ok. $\endgroup$ Mar 5, 2018 at 21:19
  • $\begingroup$ Except maybe the minus sign (in the derivative), which I cant reproduce. But in the final line you make again a change of a minus sign (when changing from eta to theta) and end up, I believe, correct. $\endgroup$ Mar 5, 2018 at 21:26
  • $\begingroup$ This is an lognormal distribution you are dealing with (with $\mu = 0$ and $\sigma^2 = \theta$) so I have changed the title of the post to reflect this. $\endgroup$
    – Ben
    Mar 5, 2018 at 22:09
  • $\begingroup$ $\theta$ is not the variance of this distribution: it is the variance of the logarithms. Which, then, is the question you want answered: how to find the MLE of the variance or how to find the MLE of $\theta$? (The latter is something I'm sure you've already done, because it asks how to find the MLE for the variance of a Normal distribution. In this light your final formula ought to look familiar.) $\endgroup$
    – whuber
    Mar 5, 2018 at 23:21

1 Answer 1

  • You can use a computation with a few random examples as a sanity check.

  • Also, you can find the answer by looking up some of the more well known distributions. Hints: You are looking for something with a logarithm and you don't need to worry if the distribution has two parameters (you can set one of them to zero).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.