Consider the probability density function
$$f(x)=\frac{1}{\theta^2}xe^{-x/\theta},\;\;\;\;\; 0\le w\lt\infty,\;\;0\lt\theta\lt\infty$$
Find the maximum likelihood estimator for $\theta$.
I'm really struggling with this question. From my understanding in order to find the maximum likelihood estimator for $\theta$, the function needs to be partially differentiated with respect to $\theta$, equated to zero, and solved for $\theta$; however for this question the differentiation is very messy and even more difficult, is solving the derivative for $\theta$.
Best Answer
The likelihood function is given by:
$$ L(\theta)=\prod_{i=1}^nf(\theta|x_i)=\prod_{i=1}^n\theta^{-2}x_ie^{\frac{-x_i}{\theta}}=\theta^{-2n}\left(\prod_{i=1}^nx_i\right) exp\left(\frac{-1}{\theta}\sum_{i=1}^nx_i\right) $$
Taking the log, to find the log-likelihood:
$$ l(\theta)=log(L(\theta))=-2nlog(\theta)+\sum_{i=1}^nlog(x_i)-\frac{1}{\theta}\sum_{i=1}^nx_i $$
Maximizing this is much simpler, find the derivative with respect to $\theta$ and setting equal to zero:
$$ l'(\theta)=-\frac{2n}{\theta}+\frac{1}{\theta ^2}\sum_{i=1}^n x_i=0$$
I'll leave solving for theta to you