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 MME of $\theta$ using the first sample moment.

My attempt:

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 have that $E(\log(x)^2)$ = $\theta$ and $Var(\log(x)^2) = 2\theta^2$.

Therefore, I have that $$\mu_1 = E(\log(x)^2) = \theta$$ $$\mu_1-\mu_2 = E((\log(x)^2)^2) = 2\theta^2$$

I solved for $\mu_1 = \sqrt{\mu_2 - 2\theta^2}$ and plugged that into the first equation to get $$\sqrt{\mu_2 - 2\theta^2} = \theta$$ Solving for $\theta$ I got $$\hat\theta_{MME} = \sqrt{\frac{\mu_2}{3}}$$

I just want to check that the way I went about doing this problem was correct. To me, the answer does not seem right.

  • $\begingroup$ MME means what? $\endgroup$
    – AdamO
    Mar 8, 2018 at 15:00
  • $\begingroup$ method of moments estimator $\endgroup$
    – statrat403
    Mar 8, 2018 at 15:02
  • $\begingroup$ and are you using bracket notation [, ] to replace parentheses ? $\endgroup$
    – AdamO
    Mar 8, 2018 at 15:04
  • $\begingroup$ 1. You are mixing up sample and population moments in your notation, it appears to me. 2. You are supposed to use the first sample moment in your solution, but (and I'm guessing here as you never defined $\mu_2$) it appears that you actually are using the second moment of a transformation of the data, not the first sample moment of the data itself which would be $\bar{x}$. $\endgroup$
    – jbowman
    Mar 8, 2018 at 15:31
  • $\begingroup$ Basically, what you want to do is find $\text{E}x$, which will be a function of $\theta$, then find the inverse function so that $\theta$ is a function of $\text{E}x$, and then substitute $\bar{x}$ for $\text{E}x$ in that function, and you're done. $\endgroup$
    – jbowman
    Mar 8, 2018 at 16:09

1 Answer 1


Method of moments

You should calculate ($^\prime$ for raw moments): $$\mu^\prime_2 = E(Y^2) = E((\log(X)^2)^2) = \mu_2 + {\mu^\prime_1}^2 = \text{Var}(\log(X)^2)+ \text{Mean}(\log(X)^2)$$ to obtain eventually $$\begin{align} \mu^\prime_1 &= \theta \\ \mu^\prime_2 &= 3 \theta^2 \end{align}$$

  • You can use an estimate of either these two moments, $\hat\mu_1^\prime$ or $\hat\mu_2^\prime$, to get an estimate $\hat\theta$ of $\theta$. But I guess that the first one would be more efficient (less variance/error)

  • You only need both or more moments together when there are more unknown parameters. E.g. to estimate the two parameters $\mu$ and $\sigma$ in $N(\mu,\sigma^2)$ you use two equations $\mu^\prime_1=\mu$ and $\mu^\prime_2 = \mu^2+\sigma^2$

  • The first moments of the sufficient statistic (yours is $T(X)=log(X)^2$) are sufficient to estimate the parameters. If you would write the derivative of the likelihood function (to calculate the maximum likelihood estimate) then you would get some factorization into a part with $\theta$ and a part with $\sum T(x)$

Your algebra

You happen to get the same $\mu_2=3\theta^2$ expression, but that is because you magically changed your wrong $$\mu_1 - \mu_2 = 2 \theta^2 $$ into the correct (but with different meaning of $\mu_2$) $$\mu_1 = \sqrt{-2\theta+\mu_2} $$ Where a square root and minus sign turn up out of nothing.


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