(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Calculate Posterior p.f. from Poisson Distribution

Can you offer guidance or do you also need help?

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