Im given the question:
Given that the Gamma distribution likelihood $$p(y|β) = Gamma(y; α, β) = \frac{\beta^\alpha}{Γ(\alpha)} y^{α-1} e^{(-βy)}$$
where α is a positive constant and β > 0 is unknown. We place a Gamma prior over β, Gamma(β; α0, β0) as the conjugate prior. Prove the posterior distribution can be written as Gamma(β; α0 + α, β0 + y).
So I do the following:
$$\text{posterior} : p(β|y) \propto p(y|β) \cdot p(β)$$
$$ p(y|β) \cdot p(β) =\frac{\beta^\alpha}{Γ(\alpha)} y^{α-1} e^{(-βy)} \cdot \frac{\beta_0^{\alpha_0}}{Γ(\alpha_0)} β^{α_0-1} e^{-β_0 \cdot β} $$
Simplifies to:
$$ \frac{β_0^{α_0}\cdot y^{α-1}}{Γ(\alpha_0 + \alpha)} β^{α+α_0-1} e^{-β(β_0+y)}$$
And then Im stuck... anybody have a tip or two to share?