derive MLE for $\theta$

Question

I want to derive the MLE for theta given this equation

$f(x,\theta) = \frac{x}{\theta} \exp \left( - \frac{x^2}{2 \theta} \right)$

do i first take the log and then the derivative of theta?? I think im supposed to get something like this:

$\frac{x^{2}-2\theta}{2\theta^{2}}$

Answer 1

Usually the MLE is based on a random sample, i.e. $X_1,...,X_n$

Thus the loglikelihood is

$$l(\theta)=-nlog\theta-\frac{1}{2\theta}\Sigma_x x^2$$

derive respect to $\theta$, set =0, solve in $\theta$ getting

$$\hat{\theta}=\frac{\Sigma_x x^2}{2n}$$