# Find the maximum likelihood estimator

Suppose that $X$ follows a distribution with p.d.f. given by $$f(x;\theta)=\dfrac{2xe^{-(x/\theta)^2}}{\theta^2}$$
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?

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

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

• (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 $\theta$? Feb 26, 2017 at 6:11
• 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? Feb 26, 2017 at 6:20
• 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.)
– whuber
Feb 26, 2017 at 17:43

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.