Suppose that $X$ follows a distribution with p.d.f. given by \begin{equation} f(x;\theta)=\dfrac{2xe^{-(x/\theta)^2}}{\theta^2} \end{equation}
for $x>0$ Assume that $X_1,X_2,...,X_n$ is a random sample of size $n$ from the distribution. Find the maximum likelihood estimator, $\tilde{\theta}_{ML}$, of $\theta$

I'm having issues with how this simplifies. Can someone check my work?

\begin{equation} L(\theta)=\prod^n_{i=1}f(x;\theta)=\frac{2^n}{\theta^{2n}}\prod^n_{i=1}x_ie^{-\left(\dfrac{\sum^n_{i=z}x_i^2}{\theta^2}\right)} \end{equation}
\begin{equation} log(L(\theta))=nlog(2)-2nlog(\theta)+\sum^n_{i=1}log(x_i)-\frac{\sum^n_{i=1}x_i^2}{\theta^2} \end{equation} \begin{equation} L'(\theta)=-\frac{2n}{\theta}+\frac{2\sum^n_{i=1}x_i^2}{\theta^3} \end{equation}

I think I have done this incorrectly because I don't know how I would isolate $\theta$. Any help would be appreciated.

  • 2
    $\begingroup$ (1) Should be $\frac{1}{\theta^2} \sum_i x_i^2$ not $\frac{1}{\theta^2} \sum_i x_i$ and (2) What do you mean by isolate $\theta$? $\endgroup$ Commented Feb 26, 2017 at 6:11
  • $\begingroup$ When $\theta^2$ moves outside the product doesn't it become $\theta^{2n}$? When I use the log rules should I keep $\theta^2$? As far as isolating $\theta$ is concerned, when I set the last equation equal to $0$ I need to solve for $\theta$ to find the likelihood estimator right? $\endgroup$
    – Lanous
    Commented Feb 26, 2017 at 6:20
  • 2
    $\begingroup$ In your final equation it is legitimate to multiply both sides by $\theta^3$ to clear the denominators. The resulting equation is trivial to solve. Incidentally, in the penultimate formula for $\log(L(\theta))$ the terms in the first summation should be $\log(x_i)$: no squares are involved. (This will not affect the likelihood equations, though.) $\endgroup$
    – whuber
    Commented Feb 26, 2017 at 17:43

1 Answer 1


You can do an easy check of your answer by writing the density as an exponential family. It has sufficient statistic $x^2$, which means that the maximum likelihood estimate must be $\sqrt{\frac{\sum_i x_i^2}{n}}$ as you found.


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.