Let $X_1,...,X_n$ be IID from a $GAMMA(\alpha, \beta)$ distribution, where $E(X_1)=\alpha\beta$. Derive the asymptotic distribution of the method of moment estimator $\tilde{\theta}$ of $\theta=(\alpha,\beta),$ that is:


and give the expression of $W$.

In the above problem, both $\tilde{\theta}$ and $\theta$ should be bold to represent vectors. I can calculate the methods of moments estimators, easily; they are:




where $\bar{x^2}$ represents $\sum x_i^2/n$ and $\bar{x}^2$ represents $(\sum x_i/n)^2$. However, I have no idea how to procde from here. How can I possibly find that limiting distribution with this information?


You can use multivariate Delta-method. Look here (paragraph 5.4) for similar calculations.

By multivariate CLT we can derive the asymptotic distribution of vector of sample moments: $$ \sqrt{n}\left[\pmatrix{\bar x\cr \bar{x^2}}-\pmatrix{\mathbb E[x_1]\cr\mathbb E[x_1^2]}\right]=\sqrt{n}\left[\pmatrix{\bar x\cr \bar{x^2}}-\pmatrix{\alpha\beta\cr\beta^2\alpha(\alpha+1)}\right]\xrightarrow{d} \mathcal N\left(\pmatrix{0\cr 0},\Sigma\right) $$ where $\Sigma$ is a covariance matrix of $\pmatrix{x_1\cr x_1^2}$. Please check all the calculations. $$ \Sigma =\pmatrix{\text{Var}(x_1) & \text{Cov}(x_1,x_1^2)\cr \text{Cov}(x_1,x_1^2) & \text{Var}(x_1^2)} =\pmatrix{\alpha\beta^2 & 2\alpha\beta^3(\alpha+1)\cr 2\alpha\beta^3(\alpha+1) & 2\alpha\beta^4(\alpha+1)(2\alpha+3)}. $$ Next, we have pair of estimates $$ \pmatrix{\tilde\alpha\cr\tilde\beta} = g\left(\bar x,\bar{x^2}\right), $$ where the vector-function $g:\mathbb R^2\to\mathbb R^2$ looks as $$ g(x,y)=\pmatrix{g_1(x,y)\cr g_2(x,y)} = \pmatrix{\frac{x^2}{y-x^2} \cr \frac{y-x^2}{x}}. $$ By multivariate Delta-method, $$ \sqrt{n}(\tilde{\theta}-\theta)=\sqrt{n}\left[g\left(\bar x,\bar{x^2}\right) - g\left(\mathbb E[x_1], \mathbb E[x_1^2]\right)\right] \xrightarrow{d} \mathcal N\left(\pmatrix{0\cr 0},\Sigma^*\right), $$ where $$ \Sigma^*=\left(\nabla_g\left(\mathbb E[x_1], \mathbb E[x_1^2]\right)\right)^T \cdot \Sigma \cdot \left(\nabla_g\left(\mathbb E[x_1], \mathbb E[x_1^2]\right)\right). $$ Here $$ \nabla_g\left(x, y\right) = \pmatrix{\frac{\partial g_1(x,y)}{\partial x} & \frac{\partial g_2(x,y)}{\partial x}\cr \frac{\partial g_1(x,y)}{\partial y} & \frac{\partial g_2(x,y)}{\partial y}} = \pmatrix{\frac{2xy}{(y-x^2)^2}& -\frac{y}{x^2}-1\cr -\frac{x^2}{(y-x^2)^2} & \frac1x }. $$

Substitute expectations instead of variables and get $$ \nabla_g\left(\mathbb E[x_1], \mathbb E[x_1^2]\right) = \pmatrix{ \frac{2(1+\alpha)}{\beta} & -\frac{2\alpha+1}{\alpha} \cr -\frac1{\beta^2} & \frac1{\alpha\beta}} $$

And finally get the covariance matrix of limiting two-dimensional normal distribution: $$ \Sigma^* = \pmatrix{2\alpha(\alpha+1) & -2\beta(\alpha+1) \cr -2\beta(\alpha+1) & \frac{\beta^2(2\alpha+3)}{\alpha}}. $$

At the first sight, this is a valid answer: it is a valid covariance matrix since it is symmetric and positively definite, and the variance of $\tilde \alpha$ does not depend on scaling parameter $\beta$. Note that the distribution of $\tilde \alpha$ does not depend on $\beta$ too since $\frac{x_1}{\beta}$ follows $GAMMA(\alpha,1)$ distribution and $\tilde\alpha$ does not change if we replace $x_i$ with $\frac{x_i}{\beta}$.

  • $\begingroup$ How were you able to calculate the covariance between $x_i$ and $x_i^2$ in the original $\Sigma$ matrix? Similarly, how did you calculate $Var(x_i^2)?$ $\endgroup$ – jippyjoe4 Apr 6 '18 at 7:23
  • $\begingroup$ By definition. $\text{Cov}(x_1, x_1^2)=\mathbb E[x_1^3]-\mathbb E[x_1]\mathbb E[x_1^2]$. $\text{Var}(x_1^2)=\mathbb E[x_1^4]-(\mathbb E[x_1^2])^2$. $\endgroup$ – NCh Apr 6 '18 at 12:52
  • $\begingroup$ Wow. Great work. Lifesaver. It's so nice to finally see an example worked out in as much detail as this one. $\endgroup$ – jippyjoe4 Apr 7 '18 at 6:56

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.