Question on expected value of an estimator

Question

I have been given the pdf: $$f_{\theta}(x)=\frac{x}{\theta^2}e^{-\frac{x^2}{2\theta^2}}$$ and I want to know if the MLE estimator $\theta$ is unbiased.

Attempt: I want to maximize the function so I take the natural logarithm and then I differentiate with respect to $\theta$ as follows: $$\frac{d}{d\theta}ln(f_\theta (x))=\frac{-2}{\theta}+\frac{x^2}{\theta^3}$$ which equals zero iff $\theta =x^2/4$. Thus $\hat{\theta}=\frac{x^2}{4}$

Now I want to evaluate $E[\hat{\theta}]$ which I'm having trouble to implement. Is it $$E[\hat{\theta}]=E[x^2/4]=\int_0^{\infty}\frac{x^2}{4}f_\theta (x^2/4)dx$$ or am I misunderstanding the definition of $E[\hat{\theta}]$?

Edit:differentiation fixed

Answer 1

Expected value of a function of a random variable $X$ is given by

$$ E[g(X)] = \int g(x) f(x) \operatorname{dx} $$

So in your case it would be as follows

$$ E\left[\frac{x^2}{4}\right] = \int^{\infty}_{0} \frac{x^{2}}{4} \frac{x\exp\left(-\frac{x^2}{2\theta^2}\right)}{\theta^2} \operatorname{dx}$$

In other words, you should not evaluate your pdf at a particular point but let it run through all the domain.

EDIT: Besides, @ocram is right. Assuming you have an i.i.d. sample $(x_{1}, ..., x_{n})$, your likelihood function to be maximized should be

$$ L(\theta) = \prod_{i=1}^{n} f_{\theta}(x_{i}) = \frac{\prod_{i=1}^{n}x_{i}}{\theta^{2n}} \exp\left(- \frac{\sum_{i=1}^{n} x_{i}^{2}}{2\theta^2} \right)$$

Of course, if $n=1$, you would fall back to the previous case.