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Let $X_1,\cdots, X_n$ be a random sample with pdf

$$p(x\mid \theta) = \theta x^{-2}, \text{ with } 0<\theta\le x < \infty$$

Use moment method to estimate $\theta$.

This problem is 7.6 from Casella's Statistical Inference. Strangely, I found that the moment does not exist. For example,

$$E (X) = \int_\theta^\infty \theta x^{-1} \,dx = \theta \ln(x)\biggr|_{\theta}^\infty $$

What should I do in this situation?

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3x89g2
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    Note that $$E_\theta(X^{-1})=\int_\theta^\infty\theta x^{-2}x^{-1}dx=\frac1{2\theta}$$ hence a (biased) consistent estimator is $$\hat\theta_n=\frac{n}{2\sum\limits_{k=1}^n\frac1{X_k}}$$ – Did May 17 '17 at 10:06
  • I see that the answers given here work, but they don't seem to fit within the framework of the text, since it defines the process of finding MOM estimators as equating the first $k$ sample moments to corresponding $k$ population moments, where $k$ appears to be an integer. So I'm curious what solution Casella had in mind. – Nick Koprowicz Mar 11 '20 at 17:19

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Transform your random variable to:

$$ Y =X^{\frac{1}{2}},$$

and find its moment:

$$E(Y) = E \left(X^\frac{1}{2}\right) = \int_\theta^\infty \theta x^{-\frac{3}{2}} \,dx = \left.-2\theta x^{-\frac{1}{2}}\right|_\theta^{+\infty} = 2 \sqrt{\theta}.$$

Given $N$ observation of $X$, an estimator for $\alpha = \sqrt{\theta}$ is:

$$\hat\alpha = \frac{1}{2N} \left(\sum_{i=1}^N \sqrt{X_i}\right),$$

which is unbiased since:

$$E(\hat\alpha) = \frac{1}{2N} N 2 \sqrt{\theta} = \sqrt{\theta}.$$

the_candyman
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