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?