I'm studying statistics and came across a problem that I'm having some issues wrapping my head around.
I'm given the density function of a Weibull distribution $$ f(y;\lambda,k) = \begin{cases} \frac k \lambda \left(\frac y \lambda\right)^{k-1}e^{-(y/\lambda)^k},& y\geq 0\\ 0, & y<0 \end{cases} $$
I'm supposed to show that the Weibull distributions with fixed $k$ belongs to the exponential family with the form:
$$f_θ(y) = \exp(a(y)b(θ) + c(θ) + d(y))$$
I've taken the logarithm of the function and got:
$$\log f(y;λ,k) = \log (k/λ) + (k-1)\log(y/λ) - k(y/λ)$$
I'm unsure how to derive the canonical parameter and determine where the terms belong. Any tips on how to arrange the logs are welcome.
EDIT:
I've tried rearranging a bit and ended up with $$\log f(y;λ,k) = -y(k/λ) + log(k/λ)-(k-1)*log(y/λ)$$
With a(y) = -y, b(θ)= k/λ, c(θ) = log(k/λ), d(y) = (k-1)*log(y/λ)
With the canonic parameter (θ) = k/λ. Though I'm pretty sure this is incorrect.
First, there is a small mistake on your $\log(f)$ formula
$$\log f(y;λ,k) = \log (k/\color{red}{\lambda}) + (k-1)\log(y/λ) - (y/λ)^{\color{red}k}.$$
Second, I did some digging and it looks like that the natural parameter for Weibull distribution is $\lambda$ and not $k$, meaning that (to write it analogously with your $f_\theta(y)$ in your exponential family) we need write it as
$$f_\lambda(y;k) = \exp\left(a(y;k)b(\lambda;k) + c(\lambda;k) + d(y;k) \right).$$
So, let's look at $\log(f)$
$$ \begin{align} \log f(y;λ,k) &= \log (k/\lambda) + (k-1)\log(y/λ) - (y/λ)^{k}\\ &= \log(k) - \log(\lambda) + (k-1)(\log(y)-\log(λ)) - (y/λ)^{k}\\ &= \underbrace{\log(k) - k \log(\lambda)}_{c(\lambda;k)} + \underbrace{(k-1)\log(y)}_{d(y;k)}+\underbrace{\left(-λ^{-k}\right)}_{b(\lambda;k)}\underbrace{y^k}_{a(y;k)} \end{align} $$
Also, see the Wikipedia article on exponential family which explicitly says that