# Probability for estimate/poisson distribution

• May 6th 2008, 10:46 AM
iljitj
Probability for estimate/poisson distribution
Hi

I've searched the forum for topic but I can't seem find one with the same problem, so here goes:

I have Xi given as the number of an occurrence pr quarter.

X1...Xn is independent and Xi~Poisson(a)

I have maximized a for the likelihoodfunction to be $\displaystyle a*=\frac{1}{n}\sum{Xi}$ as an estimation of a.

Now to my question:
For a=1.25 and n=10, what is the probability that a* equals respectively 1.1 and 1.25?

My paper is to be done in a couple of hours, so help would be very much appreciated.. (Smile)

Edit: Even though the couple of hours has gone by, I actually still need som help, so if anyone have a hint or two..?
• May 6th 2008, 01:34 PM
CaptainBlack
Quote:

Originally Posted by iljitj
Hi

I've searched the forum for topic but I can't seem find one with the same problem, so here goes:

I have Xi given as the number of an occurrence pr quarter.

X1...Xn is independent and Xi~Poisson(a)

I have maximized a for the likelihoodfunction to be $\displaystyle a*=\frac{1}{n}\sum{Xi}$ as an estimation of a.

Now to my question:
For a=1.25 and n=10, what is the probability that a* equals respectively 1.1 and 1.25?

My paper is to be done in a couple of hours, so help would be very much appreciated.. (Smile)

Edit: Even though the couple of hours has gone by, I actually still need som help, so if anyone have a hint or two..?

The sum of n Poisson iid RV's with parameter a, is a Poission RV with parameter na.

RonL
• May 6th 2008, 02:09 PM
iljitj

Correct me if I'm wrong, but does that mean that I should do it like this:

$\displaystyle P(X=1.1)=\frac{12.5^{11}}{11!}e^-12.5=0.109$

??
• May 6th 2008, 07:51 PM
CaptainBlack
Quote:

Originally Posted by iljitj
$\displaystyle P(X=1.1)=\frac{12.5^{11}}{11!}e^{-12.5}=0.109$