Second post here, hope I can get some help. This
Suppose that there is a proportion θ of defects in a manufacture lot that is unknown, and the prior pdf for θ is:
f(θ) = 2(1-θ) for 0 < θ < 1, 0 otherwise.
Suppose that in a random sample of eight items exactly three are found to be defective. Determine the posterior distribution of θ.
What I have: I ended up with a beta distribution with alpha = 4, and beta = 6. However my book's answer is this: Beta distribution with parameters α = 4 and β = 7.
Can someone explain to me why beta is 7? A step by step solution would be much appreciated, thanks.
The extra "1" in the $\beta$ term comes from the factor $1-\theta$ in $f$.
Prior: $f(\theta)=2(1-\theta)$.
Likelihood: $\binom{8}{3}\theta^3 (1-\theta)^5$.
$\text{Posterior} \propto \text{likelihood}\times \text{prior} \propto \theta^{4-1}(1-\theta)^{7-1}$