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

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SUMMARY

The discussion focuses on estimating the parameter θ for a Beta distribution using two methods: Method of Moments and Maximum Likelihood Estimation (MLE). The participants clarify that the first moment of the Beta distribution, represented as α/(α + β), can be aligned with the sample mean (x̄) to derive an estimator for θ. Specifically, they confirm that setting α = θ and β = 1 leads to the estimator θ = x̄/(1 - x̄), which is consistent with the MLE result θ = -n/(Σ ln(x_i)). The conversation emphasizes the validity of both estimation methods despite their differing results.

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  • Understanding of Beta distribution properties
  • Familiarity with Method of Moments
  • Knowledge of Maximum Likelihood Estimation (MLE)
  • Basic statistical concepts such as sample mean and expected value
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Statisticians, data analysts, and students studying statistical estimation methods, particularly those interested in the Beta distribution and its applications in various fields.

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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.
 
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mrkb80 said:
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?
 
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.
 

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