Ch8 Q8.11

[CHUN LEONG LEE]

Dear all,

Good day to you. I am looking at Question 8.11 in the practice questions of Chapter 8 Point Estimation. Would like to ask 2 questions about these 2 formulas while referring to the solutions:

1) For the Poisson distribution, probabilities can be calculated iteratively using the relationship:
P0(X =x) = (Lamda / x) * P(X = x−1)

2) The likelihood of obtaining 0n 0’s, 1n 1’s etc (making a total of n), assuming the numbers conform to a Poisson distribution, is the multinomial probability:
L(Lamda) = [n! / (n0!*n1!*n2!*...)] * (e^-Lamda)^n0 * (Lamda * e^-Lamda)^n1 * (Lamda^2 * e^*Lamda)^n2 / 2! * ....

Kinda stuck here, would like to ask on the intuition behind these 2 formulas please

Thank you.

 

[John Lee]

1) You just divide \(P(X=x)\) by \(P(X=x-1)\) for the Poisson.
2) So we have 87,889 zero's, 11,000 one's, 1,000 two's and so on. So the probability of observing this is:

\(P(X=0)^{87,889} * P(X=1)^{11,000} * P(X=2)^{1,000} * ...\)

We then use the Poisson distribution to calculate P(X=0), P(X=1), etc.

But we don't know which order these numbers are in - so there's going to be some combinatorial it to account for the different possibilities. For example, if I flip a coin twice and say there's one head, there's 2 ways of getting that - so the probability isn't just P(head)*P(tail) it's 2*P(head)*P(tail).

Frankly, I wouldn't worry about what the combinatorial bit is as it's a constant and will disappear when we log it and differentiate.