The posterior density function will be proportional to the likelihood
function, so in this case, the posterior p.d.f. will look like:
g(p) = C*(Pi/2)*sin(Pi*p/2)*(1-p); where C must be such that the integral
of g(p) from 0 to 1 equals 1.
Write the integrand as: C*[ (Pi/2)*sin(Pi*p/2) - (Pi*p/2)*sin(Pi*p/2) ] =
C*[ f(p) - p*f(p) ]; f(p) as in problem 7
Using the hint from problem 7 and the fact the f(p) is a p.d.f., the
integral from 0 to 1 will be:
C*[ 1 - 2/Pi ]
Setting this equal to 1 and solving from C gives C = Pi/(Pi - 2).
Substituting this back into g(p) gives the final result.
----------
> From: DREZEK, STEFVAN S <SDREZEK@amica.com>
> To: 'studygroup4b@lists.casact.org'
> Subject: spring 1998
> Date: Friday, October 02, 1998 11:09 AM
>
> Can someone show me how to arrive at the constant for problem 8 of the
> spring '98 exam?
>
> --Stefvan
>