Estimating Theta for Beta Distribution: Method of Moments vs. MLE?

[mrkb80]

Homework Statement

Let $X_1,...,X_n$ be iid with pdf $f(x;\theta) = \theta x^{\theta-1} , 0 \le x \le 1 , 0 < \theta < \infty$

Find an estimator for $\theta$ by method of moments

Homework Equations

The Attempt at a Solution

I know I need to align the first moment of the beta distribution with the first moment of the sample ($\bar{x}$ or $\dfrac{\Sigma_{i=1}^n x_i}{n}$)

The beta distribution has a first moment of $\dfrac{\alpha}{\alpha + \beta}$I guess my problem is figuring out what my should be alpha and beta from the given pdf, from there it is simply just $\bar{x}=\dfrac{\alpha}{\alpha + \beta}$ and then solving for $\theta$ Any advice?

I did try to take the expected value of the pdf and set it equal to x bar, but I think that is not the correct answer ( I got something like $\hat{\theta}=\dfrac{\bar{x}}{1-\bar{x}}$ )

I also attempted this using MLE and got something like $\hat{\theta}=\dfrac{-n}{\Sigma_{i=1}^n \ln{x_i}}$ but I would also like to solve this problem with method of moments.

Thanks in advance.

 

[haruspex]

I know I need to align the first moment of the beta distribution with the first moment of the sample ($\bar{x}$ or $\dfrac{\Sigma_{i=1}^n x_i}{n}$)

The beta distribution has a first moment of $\dfrac{\alpha}{\alpha + \beta}$


I guess my problem is figuring out what my should be alpha and beta from the given pdf, from there it is simply just $\bar{x}=\dfrac{\alpha}{\alpha + \beta}$ and then solving for $\theta$ Any advice?

I did try to take the expected value of the pdf and set it equal to x bar, but I think that is not the correct answer ( I got something like $\hat{\theta}=\dfrac{\bar{x}}{1-\bar{x}}$ )

If you represent it as a case of a Beta distribution, you get α=θ, β=1, yes? And that gives you the same result as you obtained directly. What makes you think it is wrong?

 

[mrkb80]

Thanks for the reply.

It just felt wrong to me because it was so far from the MLE estimate, but I know that can happen.