Find deviance and canonical link function $g(\mu)=\eta$ where $Z$ is the response variable of Pareto distribution $f(y,\theta)=\theta y^{-\theta-1}$ Find the deviance and the canonical link function $g(\mu)=\eta$ where $Z$ is the response variable of Pareto distribution $f(y,\theta)=\theta y^{-\theta-1}$, where $z=\log y$


I brought the distribution into the exponential family form:

$\mid dy/dz \mid f(y,\theta)=f(z,\theta)$

Hence $f(z,\theta)=\exp \{-\theta z + \log \theta +d(z) \}$

$E(Z)=\mu=\frac{1}{\theta}$

Link function $g(\mu)$:

My notes suggest to use $g(\mu)=-\theta=\eta$ hence $g(\mu)=-1/\mu$

As for Deviance,
$D=-2\log \frac {L(\omega)}{L(\Omega)}$ where $w$ is the reduced model.

For the full model I have: $\hat \mu_i=z_i$ and the reduced model has $\hat \mu_i=\hat \mu_i$

$$D=-2(l(\hat \mu_i)-l(z_i))=-2(\{-\theta z + \log \hat \mu_i +d(z) \}-\{-\theta z + \log z_i +d(z) \})=-2\log \frac{\hat \mu_i}{z_i}$$
Is this correct?
The answer is incorrect because $\theta$ depends on $\hat \mu_i$ in each likelihood.

Thus instead of $\theta z$ you need $z/\mu$ (which in the case of the full model is -1).

Similarly you have jumped straight from $\log\theta$ in the likelihood to $\log\mu$ in the deviance. This should instead be $\log(1/\mu)$ or $-\log\mu$.

Also, as mentioned in the comments, the sum is forgotten.
The answer is incorrect because $\theta$ depends on $\hat \mu_i$ in each likelihood.

Thus instead of $\theta z$ you need $z/\mu$ (which in the case of the full model is -1).

Similarly you have jumped straight from $\log\theta$ in the likelihood to $\log\mu$ in the deviance. This should instead be $\log(1/\mu)$ or $-\log\mu$.

Also, as mentioned in the comments, the sum is forgotten.