1
$\begingroup$

I would like help on the following question and I will show my work. Here is the question in my notes and I will follow up with my work:

Q: Suppose a forest is segmented into strips, referred to as transects, within which trees occur roughly in a linear sequence. Assuming trees are independent of each other within a transect, and that a tree within a transect is infected with probability $\pi$, the probability within a transect that the $(r+1)th$ tree is the first one observed to be infected is $$(1-\pi)^r\pi$$ for $r=0,1,2...$ and we have the data for $ten$ independent transects within a forest:

$$0,0,0,1,0,4,1,1,0,0 $$

Thus, if the forester wants to apply a Bayesian inference to the above scenario. She takes as her prior for $\pi$ a probability distribution proportional to:

$$\pi^{3.09-1}(1-\pi)^{8.96-1}$$

Using the data above, find the researcher's posterior probability that $\pi > 0.4$

So here is what I have done: To find the posterior distribution, we first have to find the likelihood thus since $n=10$ and each observation is I.I.D we have:

$$L=\pi(1-\pi)^{\sum_{i=1}^{10}r_i^2}$$ and we multiply $L$, the likelihood by the prior which is where I have a problem. Do we just multiply the prior given to the likelihood and solve? Because if I do that I get:

$$\pi^{3.09}(1-\pi)^{14.96}$$ We can forget about the denominator since it wont be a function of $\pi$ correct? Can anyone help please? Thanks so much!

$\endgroup$
1
$\begingroup$

You have ten positive infected cases (one in each transect) so your likelihood function should be $$L=\pi^{10}(1-\pi)^{\sum_{i=1}^{10}r_i}$$ and so your posterior density function should be proportional to $$\pi^{13.09-1}(1-\pi)^{15.96-1}.$$

Both your prior and posterior distributions are Beta distributions.

$\endgroup$
  • $\begingroup$ may I ask how do you get the r squared $\endgroup$ – natsu Mar 16 '14 at 1:16
  • $\begingroup$ @natsu: Thank you - It should not have been there. I had it because I copied and pasted from the question, which should also not have it. I have edited the answer. $\endgroup$ – Henry Mar 16 '14 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.