2
$\begingroup$

Let $Y_1, Y_2, \dots, Y_n$ be i.i.d RVs from the following distribution:

$$f(y) = \theta y ^{\theta -1} \qquad 0 < y < 1, \quad\theta > 0$$

Show the method of moment estimator of theta is: $\bar Y/(1-\bar Y)$

I must only be seeing solutions that skip steps, because I cannot come to this answer. I can get up to this part (which I believe is in the right track):

$$E(y) = \int_0^1 y \theta y^{\theta -1}\,dy,$$

which leads to $E(y) = \theta/(\theta + 1)$

But I do not know how to proceed from there....

$\endgroup$
  • $\begingroup$ Got something from the answer below? $\endgroup$ – Did Mar 6 '14 at 7:55
1
$\begingroup$

In the method of moments one sets the sample moments equal to the population moments, and then solves for the parameters to be estimated. In this case there's only one such parameter and one uses only the first moment. Thus: $$ \frac{\theta}{1+\theta} = \bar Y. $$ Consequently $$ \theta=\bar Y(1+\theta) = \bar Y+\theta\bar Y $$ $$ \theta-\theta\bar Y= \bar Y $$ $$ \theta(1-\bar Y) = \bar Y $$ $$ \theta = \frac{\bar Y}{1-\bar Y} $$

So that is the estimate of $\theta$ by the method of moments.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.