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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.

CuriousQuestioner
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1 Answers1

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The extra "1" in the $\beta$ term comes from the factor $1-\theta$ in $f$.

Prior: $f(\theta)=2(1-\theta)$.

Likelihood: $\binom{8}{3}\theta^3 (1-\theta)^5$.

$\text{Posterior} \propto \text{likelihood}\times \text{prior} \propto \theta^{4-1}(1-\theta)^{7-1}$

angryavian
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  • Ah, so it is likelihood × prior? I think I did likelihood ∝ prior, so I did not account for the other (1-θ), it seems like. – CuriousQuestioner May 20 '14 at 23:41
  • I made a small typo that I have fixed, but yes, posterior is proportional to likelihood times prior. Wikipedia page – angryavian May 20 '14 at 23:59
  • Alright, thanks for the help!

    I am also even more in the dark on another problem, perhaps I could get some assistance also:

    http://math.stackexchange.com/questions/803438/probability-given-x-y-are-independent-rand-variables-and-2-p-d-f-s

     – CuriousQuestioner May 21 '14 at 00:01