This is from Casella Berger chapter example 10 (11) part d.

we have the pdf $f(x|\mu,\beta)= \frac{x^{\mu/\beta -1 e^{-x/\beta}}}{\gamma(\mu/\beta) \beta^{\mu/\beta}}$. This pdf was reparameterized to have mean $EX= \mu$ and variance $VX= \mu \beta$.

I would like to compare the MOM estimator, to the MLE variance using relative efficiency in the asymptotic distribution.

consider the previous gamma distribution, and we have the MLE of the variance $\hat{\sigma^2}= 1/n \sum (X_i-\bar{X})^2$, i am trying to find the variance of this estimator $V(\hat{\sigma^2})$.

So for brevity the MOM estimator $\beta = \frac{\widehat{\sigma^2}}{\bar{x}}$.

So in order to find the variance of the MOM estimator $\beta$ and apply the multivariate delta method for the variance of the MOM, i need to find the numerator variance and denominator variance. So $V(\bar{X})= \frac{\mu *\beta}{n}$, and i am struggling to find the $V(\hat{\sigma^2})$

my first attempt is $V(\hat{\sigma^2})= 1/n V(X_1^2- 2X_1*\bar{X} + \bar{X}^2)$, but these higher terms require finding the 4th moment of the gamma distribution.

The solution says $V(\hat{\sigma^2})$ should be approximated $\approx \frac{\mu\beta^3 *2}{n}$. From here i know how to apply the multivariate delta method, but the approximation of the MLE variance i am stuck on.

any thoughts or suggestions to finding the variance estimator in the gamma distribution?

should i assume that the gamma is approximately chi-square?

The goal is to compare the asymptotic relative efficiency of the MOM and the MLE given

$ARE(\hat{\beta},\tilde{\beta})$ =[V$\hat{\sigma^2}/\bar{x}][-\partial^2 / \partial \beta^2 log L(\mu,\beta |X)]$

where hte information of the PDF is $n\mu / \beta^3$


1 Answer 1


The pdf of a $\Gamma(\alpha,\beta)$ distribution is $$g(x|\alpha,\beta)=\frac{\beta^{\alpha}x^{\alpha-1}e^{-\beta x}}{\Gamma(\alpha)}:x\in [0,\infty)$$ According to Wikipedia, the kurtosis of a random variable possessing a $\Gamma(\alpha,\beta)$ distribution equals $\frac{6}{\alpha}+3$. Your pdf $f(x|\mu,\beta)$ corresponds to a $\Gamma(\mu/\beta,1/\beta)$ distribution, so it has kurtosis $\kappa:=\frac{6\beta}{\mu}+3$. This parameter becomes apparent in establishing the variance of your MLE $\hat{\sigma}^2$.

Take $S^2=\frac{1}{n-1}\sum_{j=1}^n(X_j-\bar{X})^2$ to be the sample standard deviation of our simple random sample from a $\Gamma(\mu/\beta,1/\beta)$ population. Wikipedia says $$V(S^2)=\frac{\sigma^4}{n}\Bigg(\kappa-1+\frac{2}{n-1}\Bigg)$$ Using $\hat{\sigma}^2=\frac{(n-1)S^2}{n}$, $\kappa=\frac{6\beta}{\mu}+3$, and $\sigma^2=\mu \beta$, it follows that

$$V\Bigg(\frac{n\hat{\sigma}^2}{n-1}\Bigg)=\frac{(\mu \beta)^2}{n}\Bigg(\frac{6\beta}{\mu}+2+\frac{2}{n-1}\Bigg)$$ Since $V(cX)=c^2V(X)$, $$V(\hat{\sigma}^2)=\frac{(\mu \beta)^2(n-1)^2}{n^3}\Bigg(\frac{6\beta}{\mu}+\frac{2}{n-1}+2\Bigg)$$

  • $\begingroup$ What’s kurtosis? $\endgroup$
    – Jellyfish
    Apr 28, 2021 at 5:54
  • $\begingroup$ kurtosis is the 4th central moment divided by the square of the variance. $\endgroup$ Apr 28, 2021 at 15:48
  • $\begingroup$ could you please include the wiki link, i can't find. thank you. $\endgroup$ Apr 30, 2021 at 3:26
  • $\begingroup$ Here is the wiki link for Kurtosis: en.wikipedia.org/wiki/Kurtosis $\endgroup$
    – Matthew H.
    Apr 30, 2021 at 3:29
  • $\begingroup$ Here is for variance. Look under "distribution of sample variance." en.wikipedia.org/wiki/Variance $\endgroup$
    – Matthew H.
    Apr 30, 2021 at 3:30

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