Consider the Poisson distribution with parameter lambda. Find the maximum likeli

Question

Consider the Poisson distribution with parameter lambda. Find the maximum likelihood estimator of lambda based on a sample of size in the customer arrivals to a grocery store follow a Poisson distribution, but the rate is unknown The arrivals are observed for 10 hours and the number of customers arriving at the store within each hour is recorded as follows: 12, 8, 7. 8,9,11,14, 22,18, 15. What is the maximum likelihood estimate for the customer arrival rate (arrivals per hour)? Find two different moment estimators for lambda using the facts that for a Poisson random variable X, E(X) = lambda and V(X) = lambda

Explanation / Answer

Solution:

a) To estimate parameter of Poisson() distribution, we recall that µ1 = E(X) = . There is only one unknown parameter, hence we write one equation, µ1 = = m1 = X. ”Solving” it for , we obtain ˆ = X, ¯ the method of moments estimator of . Maximum likelihood. The p.m.f. of Poisson distribution is P(x) = e x x! , and its logarithm is ln P(x) = + x ln ln(x!). Thus, we need to maximize ln P(X1, ..., Xn) = Xn i=1 ( + Xi ln ) + C = n + ln Xn i=1 Xi , where C = Pln(x!) is a constant that does not contain the unknown parameter . Find the critical point(s) of this log-likelihood. Differentiating it and equating its derivative to 0, we get ln P(X1, ..., Xn) = n + 1 Xn i=1 Xi = 0. This equation has only one solution ˆ = 1 n Xn i=1 Xi = X. ¯ Since this is the only critical point, and since the likelihood vanishes (converges to 0) as 0 or , we conclude that ˆ is the maximizer. Therefore, it is the maximum likelihood estimator of . For the Poisson distribution, the method of moments and the method of maximum likelihood returned the same estimator, ˆ = X

b)

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