# Predictive distribution of poisson

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• May 3rd 2011, 05:54 PM
sirellwood
Predictive distribution of poisson
Hi everyone,

Anyone got any ideas on how to do this?

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 $\theta$ ). Very little is known about $\theta$ . 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.

Thanks!
• May 3rd 2011, 08:13 PM
CaptainBlack
Quote:

Originally Posted by sirellwood
Hi everyone,

Anyone got any ideas on how to do this?

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 $\theta$ ). Very little is known about $\theta$ . 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.

Thanks!

It would help if we knew what you had been studying when this was set.

It looks like a maximum likelihood estimator is wanted, so the first step is to write down the likelihood for the data in terms of the unknown parameter $\theta$.

$\displaystyle \it{l}(\theta)=\prod po(n_i, L_i \theta)$

where $n_i$ is the number of flaws observed in the $i$th sample of length $L_i$

CB
• May 6th 2011, 07:22 AM
sirellwood
ok, im just getting a little bit confused when doing this, as i am taking the product of the 5 terms because L $\theta$ changes every time.

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

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

(10 $\theta$)^3 * (15 $\theta$)^2.... etc

Have i gone about this right? How to i simplify this into one term?
• May 6th 2011, 07:30 AM
sirellwood
If that didnt make sense, what im trying to say is that usually in a likelihood estimator of po( $\lambda$) = -n $\lambda$ + sumof(ki)ln( $\lambda$) - sumof(ln(ki!)), where ki are the observed values.

But in this case, $\lambda$, lambda has 5 different values because L $\theta$ changes so this has thew me?
• May 6th 2011, 11:10 AM
CaptainBlack
Quote:

Originally Posted by sirellwood
If that didnt make sense, what im trying to say is that usually in a likelihood estimator of po( $\lambda$) = -n $\lambda$ + sumof(ki)ln( $\lambda$) - sumof(ln(ki!)), where ki are the observed values.

But in this case, $\lambda$, lambda has 5 different values because L $\theta$ changes so this has thew me?

$\theta$ is the parameter and that is the same in each term.

CB
• May 10th 2011, 07:05 AM
sirellwood
so do i kind of just ignore the L values?

Because currently for the log likelihood:

-n $\lambda$ + sumof(ki)ln( $\lambda$) - sumof(ln(ki!)), where ki are the observed values.

I have -n $\lambda$ = -120 $\theta$

then for the next part sumof(ki)ln( $\lambda$), is it 28ln( $\theta$)? This just seems to ignore the differing values of L?