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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

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  • $\begingroup$ You are probably missing a minus sign in the exponential. Otherwise, it's not necessarily a density. $\endgroup$
    – johnny
    Sep 26, 2013 at 16:04
  • $\begingroup$ Yes ok I forgot it when I wrote the problem here but it's fixed now. $\endgroup$
    – Raxel
    Sep 26, 2013 at 16:07
  • $\begingroup$ But can u see if me expected value is right I mean I want to show that it is $\theta$, else the estimator is biased. $\endgroup$
    – Raxel
    Sep 26, 2013 at 16:09
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    $\begingroup$ To compute the MLE, you need a sample ${x_1, \ldots, x_n}$. Assuming iid, the likelihood, that has to be maximised over $\theta$, is the product of $f_\theta(x_i)$. $\endgroup$
    – ocram
    Sep 26, 2013 at 16:15
  • $\begingroup$ Your differentiation looks a bit suspect too - I've fixed the relevant line, but I'm pretty sure the ML estimator is actually $\sqrt{x/2}$ and not $x^2/4$. If I edit the whole thing though other answers will not make sense. $\endgroup$
    – Pat
    Sep 26, 2013 at 16:36

1 Answer 1

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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.

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