Thread: Poisson Distribution Proof and Exercise

1. Poisson Distribution Proof and Exercise

Two related questions here....

Consider a random sample X1,X2, . . . ,Xn from a Poisson(μ) distribution. Using the log–likelihood function, show that the maximum likelihood estimator
for μ is given by ˆμ = ¯X . (That is meant to say "moo hat = X bar"

Suppose the number of buses arriving each rush–hour (8am – 9am) at a bus–
stop in Chicago is thought to follow a Poisson(μ) distribution. In five consecutive daily rush–hours, the numbers of buses arriving at the bus–stop are:

22, 14, 23, 15, 16.

Find the maximum likelihood estimate for μ. What assumption have you made

2. The likelihood function is

${e^{-n\mu}\mu^{\sum x_i}\over \prod (x_i !)}$

so take the log of that, the log likelihood function is then

$-n\mu +\sum x_i\lg \mu -\lg(\prod (x_i !))$

Differentiate this with respect MOO and set it equal to 0.

The solution/point estimator is the sample mean.

You should take the second derivative to prove it's a max, or use the first derivative test.
Not many people do that, but you should.

As for traffic flow in chicago, there are plenty of jokes I can make.
I called a few years ago complaining about how the red lights are not in phase.
The guy in charge is a former student from my school.
The light one block from my apartment is still is out of phase.