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I'm trying to solve a question where I have to find $V(\hat{\alpha})$ using the delta-method where $V$ is notation for variance and $\hat{\alpha}$ is the method of moment estimator for a beta distribution. To be more specific, I write down the question below:

Suppose $X_1, ..., X_n$ be a random sample from the following distribution,

$$f(x)=\alpha x^{\alpha -1}, 0<x<1\text{ and }\alpha>0.$$

  1. Let $\hat{\alpha}$ be the method of moment estimator. Obtain the $V(\hat{\alpha})$ using the delta-method.

I know that this function is just Beta distribution where $X\sim\mathrm{Beta}(\alpha,1)$. So the method of moment $E(X^k)=\frac{\Gamma(\alpha +1)\Gamma(\alpha +k)}{\Gamma(\alpha)\Gamma(\alpha +k+1)}$, where $\Gamma(\cdot)$ is a Gamma function.

But how do I get the variance of the method of moment using delta method? I am not sure how the usage of the delta method will get me the variance. Thank you.

Michael Hardy
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kim
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    You're estimating only one parameter, so you need only the first moment, which is $\operatorname E(X) = \dfrac\alpha{\alpha+1}.$

    In estimation by the method of moments, one sets the sample moment equal to the population moment and then solves that equation for the parameter to be estimated. Thus: $$ \frac{X_1 + \cdots + X_n} n = \overline X = \frac\alpha{\alpha+1}, $$ and then solving this for $\alpha,$ we get $$ \alpha = \frac{\overline X}{1 - \overline X}. $$ That is the estimator of $\alpha$ by the method of moments.

     – Michael Hardy Mar 25 '21 at 03:09

1 Answers1

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I believe the delta method only gives you an approximation of the variance, which becomes more accurate as $n$ goes to infinity (because it is based on a Taylor approximation).

The variance I get using the delta method is $\frac{\alpha(\alpha+1)^2}{n(\alpha+2)}.$ Here is how.

The method of moment estimator $\alpha$ solves the equation $$ \frac{\Gamma(\hat\alpha+1)\Gamma(\hat\alpha+1)}{\Gamma(\hat\alpha)\Gamma(\hat\alpha+2)} = \frac{\hat\alpha}{\hat\alpha+1} = \frac{1}{n}\sum_{i=1}^n X_i=: \hat\mu$$ where I am using the fact that $\Gamma(x+1)=x\Gamma(x)$. By inverting this we obtain $\hat\alpha = \frac{1}{1-\hat\mu} - 1$. Therefore we are interested in $\mathrm{Var}(\frac{1}{1-\hat\mu})$. We can use the following Taylor approximation, valid if $\hat\mu\approx\mu = \frac{\alpha}{\alpha+1}$ the true mean (which should hold by LLN): $$ \frac{1}{1-\hat\mu} \approx \frac{1}{1-\mu} + \frac{1}{(1-\mu)^2}(\hat\mu - \mu). $$

Therefore, $$ \mathrm{Var}(\hat\alpha) \approx \mathrm{Var}\left(\frac{1}{(1-\mu)^2}(\hat\mu-\mu)\right) = \frac{1}{(1-\mu)^4}\mathrm{Var}(\hat\mu). $$

Using the fact that $\mathrm{Var}(\hat \mu)$ is $\frac{1}{n}$ times the variance of a single $\mathrm{Beta}(\alpha, 1)$ variable, which is $\frac{\alpha}{(\alpha+1)^2(\alpha+2)}$, and plugging in $1-\mu = \frac{1}{1+\alpha}$, we obtain the final answer for the delta method approximated variance

$$ \mathrm{Var}(\hat\mu) \approx \frac{\alpha(1+\alpha)^2}{n(\alpha+2)} $$

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