0

Given that

$$f(x;\theta) = \frac{\pi x}{2 \theta^2} e^{-\frac{\pi x^2}{4 \theta^2}} \text{ for } x>0$$

I wish to find the method of moments estimators for $\theta$ and $\theta^2$.

I have calculated the $k$-th raw moment as $$E(X^k)=\Gamma\left(\frac{k}{2}+1\right) \frac{2^k \theta^k}{\pi^{\frac{k}{2}}}$$

Hence, substituting $k=1$, we obtain that $\hat \theta_{\text{mom}}=\bar X$. Now substituting $k=2$ $$\theta^2=\frac{\pi}{4}E(X^2)=\frac{\pi}{4}(Var(X)+E(X)^2)\Rightarrow\hat\theta^{2}_{\text{mom}}=\frac{\pi}{4}(S_X^2+\bar X^2 )$$

I believe this is how one should find the method of moments estimator for $\theta^2$ however I have come across a solution where they simply say that $$\hat \theta^2_{\text{mom}}=\bar X^2$$

Can anyone shed some light on this?

An Invisible Carrot
  • 420

1 Answers1

1

In a method of moments approach to parameter estimation, the idea is to use as many moments as are necessary to uniquely determine the parameter estimates. Once you found that the first moment was enough to estimate $\theta$, it becomes unnecessary to use higher moments. To use the second moment would mean to ignore the first.

As another example, consider the exponential distribution parametrized by scale; i.e., $$f_X(x) = \frac{1}{\theta} e^{-x/\theta}, \quad x > 0, \quad \theta > 0.$$ The method of moments estimator is easily $\hat \theta_{\text{mom}} = \bar X$. But if you want to estimate $\theta^2$, you could either choose $\widehat{\theta^2}_{\text{mom}} = (\bar X)^2$ or you could do it from the second sample moment; $$\widehat{\theta^2}_{\text{mom}} = \frac{1}{2}\overline{X^2}.$$ They're clearly not equivalent.

Had there been two parameters to the distribution, you would then need to solve a system of equations to estimate both parameters, in which case you might need to use higher order moments. But since you don't, you just use the first one. That's what the solution did.

However, that's not to say that your estimator is invalid: your estimator is a method of moments estimator; it's just not the one that the solution stated. As an exercise, consider the performance of the two estimators: are they biased or unbiased? What is the mean square error of each? Are they consistent estimators? What is the relative efficiency of one to the other?

heropup
  • 135,869