I have an assignment in which I have to show that Weibull Pareto belongs to exponential family and then mean and variance of term $a(x)$.
$g(x)=\frac{\beta c}{x}\{\beta \log (\frac{x}{\theta})\}^{c-1}\exp \{-(\beta \log (\frac{x}{\theta})^c\}$
For this I have used the structure given by Dobson, as $g(y;\tau)=\exp\{a(x)b(\tau)+c(\tau)d(x)\}$and got $a(x)=\log(x)-\log (\lambda)$, $c(\tau) =-(\alpha \log (\frac{x}{\lambda}))$, $b(\tau) = (c-1)\log (\alpha)$ and $d(x)= -\log (x)$. where $\lambda=\beta c$. Kindly check is this correct.
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1$\begingroup$ @Glen_b kindly see my edit $\endgroup$– J AKCommented Nov 18 at 9:41
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$\begingroup$ What are $\tau$ and $\lambda$? They don't appear in your definition of $g$. $\endgroup$– Glen_bCommented Nov 18 at 10:52
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$\begingroup$ actually $\tau$ represents all parameters. $\endgroup$– J AKCommented Nov 19 at 6:03
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$\begingroup$ Ah. $τ= (β, θ, c)^\top$? $\endgroup$– Glen_bCommented Nov 19 at 10:03
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$\begingroup$ @Glen yes. $g(y; \tau)$ is a formula $\endgroup$– J AKCommented Nov 19 at 12:03
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