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Calculate Posterior p.f. from Poisson Distribution

  1. Jan 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose that the number of defects on a roll of magnetic recording tape has a Poisson distribution for which the mean λ is either 1.0 or 1.5, and the prior of λ is the following:

    L(1.0)=0.4 and L(1.5)=0.6

    If the roll of tape selected at random is found to have 3 defects, what is the posterior p.f. of λ?

    2. Relevant equations

    p(x|λ) = [e^(-λ)]*(λ^x)/x!

    L(1.0|X=3) α f(x|λ)L(λ)

    3. The attempt at a solution

    I think this is all I have to do: (1) calculate the probability of X=3 with parameter λ=1 for a poisson distribution and then multiply this by 0.4. Then do the same thing for λ=1.5 and a prior of 0.6. Right?

    So .4*(1/e)*(1/3!)=0.0245253 and .6*(e^(-1.5))*(1.5^3)/3!=0.0753064

    According to the answers from the book, this is wrong.

    I know the likelihood for the poisson distribution above is

    e^(-nμ)*μ^(Ʃx)/(x1!*...*xn!)

    but I don't think I have to calculate this the same way I'd calculate the posterior using a gamma distribution, for example.

    Any help would be great.
     
  2. jcsd
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