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I'm trying to solve the following problem:

Let $X_1,...,X_n$ be iid from $\Gamma(\alpha,\beta)$ distribution with density $f(x)=\frac{1}{\Gamma(\alpha)\beta^\alpha}x^{\alpha-1}e^{-\frac{x}{\beta}}$.

Write the density in terms of the parameters $(\alpha,\mu)=(\alpha,\frac{\alpha}{\beta})$. Calculate the information matrix for the $(\alpha,\mu)$ parametrization and show that it is diagonal.

My attempt to solve the problem was writing $\beta$ as function of $\mu$:

$\mu = \frac{\alpha}{\beta} \Rightarrow \beta = \alpha\mu^{-1}$

And then plug into the density function:

$f(x)=\frac{1}{\Gamma(\alpha)(\alpha\mu^{-1})^\alpha}x^{\alpha-1}e^{-\frac{x}{\alpha\mu^{-1}}}$

The problem is, when calculating the Fisher Information, I got the following matrix:

$I(\alpha,\mu) = \left[ {\begin{array}{cc} \psi^{'}(\alpha)+\frac{1}{\alpha}+\frac{1}{\alpha^2} & -\frac{2}{\mu} \\ -\frac{2}{\mu} & \frac{\alpha}{\mu^2} \\ \end{array} } \right]$ , where $\psi(\alpha)=\frac{d\,log\Gamma(\alpha)}{d \alpha}$, the digamma function

Which is not diagonal, because of the $\frac{2}{\mu}$, which is obtained from:

$logf(x) = -log\Gamma(\alpha) - \alpha \, log\alpha +\alpha \, log\mu + (\alpha-1)logx - x\frac{\mu}{\alpha}$

$\frac{d\,logf(x)}{d \mu} = \frac{\alpha}{\mu} - \frac{x}{\alpha}$

$\frac{d^2\,logf(x)}{d \mu \, d \alpha} = \frac{1}{\mu} + \frac{x}{\alpha^2}$

$-E\left[ \frac{d^2\,logf(x)}{d \mu \, d \alpha} \right] = -\frac{1}{\mu} - \frac{\alpha \times \alpha\mu^{-1}}{\alpha^2} =- \frac{2}{\mu}$

What mistake am I committing on writing the reparametrization of Gamma?

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    $\begingroup$ Everything you have done looks fine, but give it a go with the alternative parameterisation $f(x) \propto x^{\alpha-1}e^{-\beta x}$ $\endgroup$ – Nadiels May 8 '18 at 15:24
  • $\begingroup$ When using this parametrization I got: $ I(\alpha,\mu) = \left[ {\begin{array}{cc} \psi^{'}(\alpha) - \alpha^{-1} & 0 \\ 0 & \alpha \mu^{-2} \\ \end{array} } \right]$ which is exactly the expected result. I believe there was a mistake in the book on this question. $\endgroup$ – A.T May 8 '18 at 15:41
  • $\begingroup$ Since the book specifies to use the other parametrization $f(x) \propto x^{\alpha-1}e^{-\frac{x}{\beta}}$ $\endgroup$ – A.T May 8 '18 at 15:47
  • $\begingroup$ I got the same result for $I(\alpha,\mu)$ when using $(\alpha,\mu)=(\alpha,\alpha \beta)$, which makes $\mu$ the mean of the specified distribution. Indeed it is a books mistake. $\endgroup$ – A.T May 8 '18 at 15:55
  • $\begingroup$ Yeah sounds like a mistake - the $(\alpha, \beta)$-parameterisation is often taken to be the $x^{\alpha-1}e^{-\beta x}$ but obviously that isn't set in stone. Anyway might be worth checking if there is an existing errata or else sending a note to the author/publisher for future editions. $\endgroup$ – Nadiels May 8 '18 at 15:57

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