MHB Posterior distribution question Pixie's question from Yahoo Answers

[CaptainBlack]

Reposted from Yahoo Answers:

Suppose that Yi is the result of a Bernoulli trial, with probability alpha of success (Yi=1).
If we assign a Unif(0,1) prior distribution to alpha, find the posterior distribution of alpha after the observations:
(a) 1
(b) 0, 1, 1, 0, 0

Thank you so much if you can help! :)

 

[CaptainBlack]

Reposted from Yahoo Answers:

Suppose that Yi is the result of a Bernoulli trial, with probability alpha of success (Yi=1).
If we assign a Unif(0,1) prior distribution to alpha, find the posterior distribution of alpha after the observations:
(a) 1
(b) 0, 1, 1, 0, 0

Thank you so much if you can help! :)

From Bayes' theorem we have:\[P(\alpha|data) = \frac{P(data|\alpha)P(\alpha)}{P(data)}\]

For case (a) we have one success in one trial with prob of success \(\alpha\) so:

\(P(data|\alpha)=\alpha \)

\(P(\alpha)=1\) (since the prior for alpha is \(U(0,1)\) its density is \(1\) for \(\alpha\) in \([0,1]\) and zero elswhere)

\( \displaystyle P(data) = \int_{\alpha=0}^1 P(data|\alpha)P(\alpha)\; d \alpha =\int_{\alpha=0}^1 \alpha \; d \alpha =\Bigl[ \alpha^2/2 \Bigr]_0^1=\frac{1}{2} \)

and so:
\[ P(\alpha|data)=2 \alpha\]For case (b) we have 2 successes in 5 trials so:

\( \displaystyle P(data|\alpha)= b(2;5,\alpha)= \frac{5!}{2!\;3! }\alpha^2 (1-\alpha)^3 \)

CB