Predictive distribution of poisson

[SirEllwood]

1. The distribution of flaws along the length of an artificial fibre follows a Poisson
process, and the number of flaws in a length L is Po(L[tex]\theta[/tex] ). Very little is known about [tex]\theta[/tex]. The number of flaws in five fibres of lengths 10, 15, 25, 30 and 40 metres were found to be 3, 2, 7, 6, 10 respectively. Find the predictive distribution for the number of flaws in another piece of length 60 metres.

Attempt: Well I think i need to find the maximum likelihood estimator first? which is:

l(L[tex]\theta[/tex]) = [tex]\prod[/tex] Po(n_i, L_iL[tex]\theta[/tex])

But i get a little bit confused when doing this, as i am taking the product of the 5 terms because L changes every time.

So am i right in thinking that i would multiply them together so the exponential term would be exp{-120}

Then i get confused with the next term, because i am used to working with an L value that would not change because L has 1 value. So now i have:

(10)^3 * (15)^2... etc

Have i gone about this right?

 

[lippyka]

And if so, how do i continue?Answer:Yes, you have gone about this correctly. To find the maximum likelihood estimator, you need to take the product of the Poisson distributions for each of the lengths, L. The formula for the maximum likelihood estimator is given by:l(L\theta) = \prod_{i=1}^n Po(ni, Li*L\theta)where n is the number of lengths, and Li is the length of the ith fibre. For your example, you would have:l(L\theta) = Po(3, 10*L\theta)*Po(2, 15*L\theta)*Po(7, 25*L\theta)*Po(6, 30*L\theta)*Po(10, 40*L\theta)To find the maximum likelihood estimator, you need to take the partial derivative of l(L\theta) with respect to L\theta and set it equal to 0. You can then solve for L\theta to find the maximum likelihood estimator. Once you have found the maximum likelihood estimator for L\theta, you can use it to find the predictive distribution for the number of flaws in a piece of length 60 metres. The predictive distribution is given by the Poisson distribution with mean L\theta*60.