0
$\begingroup$

Assuming that:

$0 < \nu, \alpha, y < \infty$

$$f_Y(y; \nu, \alpha) = \frac{y^{\nu-1}{\alpha}^{\nu}e^{-y\alpha}}{\Gamma (\nu)} \mathbb{1}_{Y \in (0, \infty)}$$

$$ = \exp \{ -y\alpha + \nu \log \alpha + (\nu-1) \log y - \log \Gamma (\nu) \} \mathbb{1}_{Y \in (0, \infty)}$$

$$ = \exp \{ \frac{y (- \alpha / \nu ) - [- \log \alpha ]}{1/\nu} + (\nu-1) \log y - \log \Gamma (\nu) \} \mathbb{1}_{Y \in (0, \infty)}$$

such as

$$f_Y(y; \theta, \phi) = \exp \{ \frac{y \theta - [- \log (- \theta) ]}{\phi} - \frac{\log(\phi)}{\phi} + (1/\phi -1) \log y - \log \Gamma (1/\phi) \} \mathbb{1}_{Y \in (0, \infty)}$$

where,

$b(\theta) = - \log (- \theta)$,

$c(\theta) = - \frac{\log(\phi)}{\phi} + (1/\phi -1) \log y - \log \Gamma (1/\phi)$,

$\theta \equiv \frac{-\alpha}{\nu}$, e

$\phi \equiv 1/\nu$

$ \mu \equiv E[Y] = B'(\theta) = \frac{d B(\theta)}{d \theta} = - \frac{1}{\theta} = -\theta^{-1} = \frac{\nu}{\alpha}$

$\text{var} [Y] = B''(\theta)\phi = \frac{d B'\theta)}{d \theta} \phi = \frac{1}{\theta^2} \phi = \phi\mu^2 = \frac{\nu}{\alpha^2}$

And then, replacing $\mu$ in $f_Y(y)$:

$$ f_Y(y) = \exp \{ \frac{y (-\frac{1}{\mu}) - \log (\mu) }{\phi} - \frac{\log(\phi)}{\phi} + (1/\phi -1) \log y - \log \Gamma (1/\phi) \} \mathbb{1}_{Y \in (0, \infty)}$$

So, the canonical link is,

$$g(\mu) = -\frac{1}{\mu}$$

and,

$\theta_i = \eta_i = X_i'\beta$

Question: Assuming a linear predictor of the kind $\hat{\eta} = 2 + 3x_1 + 4x_2$, how one could interpret the impact on the mean of the model when increasing 1 unit in $x_1$? (Considering the canonical link)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.