Hi, I was unsure of how they have calculated the expected numbers based on the estimates: p=0.93295 q=0.06705 k=1.8569 in part (iii) I believe they have used the probability function of the negative binomial distributon - type 2 but I am unsure how they have changed the gamma part into values in the way they have. A break down with explanation would be a great help. Thanks in advance Salil
You should be able to get P(X=0). The rest use the recursive relationship: \( P(X=x) = \frac{k+x-1}{x} q P(X=x-1) \)
Hello everyone, In this same question, for part (i): The solution on page 21 (2013 Q&A bank part 3) says: The expected numbers, based on this estimate, are (using the iterative formula): x =0:......so on and so forth. What did they use and how did they get those answers? I am having a hard time figuring this out, any helps is much appreciated! Cheers!
So I understood what they are using, they are using the recurrence relationship mentioned in chapter 4 under poisson distribution. What I would like to know is why don't these people mention these things in the answer book? Or at least say please see chapter 4 for more explanation. Does my head in, when I am studying and I have to hunt through the index to get this.
I am one of "these people" Apologies. If something is unclear then let us know and we can change it for future generations of students. You can pass suggestions to us using the appropriate subject email address. I'm responsible for updating the CT3 notes this year - so I will include mention of this in the solution. Kind regards
Hi John or anyone else who reads this thread, I am back to studying, had to take a break. Same question 3.21 part (ii): How did they get the L(lambda)? Can anyone please explain the first 3 lines of the equation? Also in this same thread, when John replied to user ssaini, where is the iterative formula for type 2 negative formula given? I just calculated it manually and I get the right answer but just getting P(X=x)/P(X=x-1) is such a hassle. Should I be memorizing it?
Likelihood is the probability of observing our sample since we had \(n_0\) zeros and \(n_1\) 1's, etc we want: \(L(\lambda)=P(X=0)^{n_0}*P(X=1)^{n_1}* ...\) It's in chapter 4. It's not often that you get an exam question where you have to calculate lots of probabilities from the same distribution so just using the formula is usually fine.
