Poisson distribution, likelihood ratios

[Auron87]

Homework Statement

An independent, identically distributed sample, x = (x1, ... , xn) of size n, is drawn from a Poisson distribution, parameter A. We want to test the null hypothesis H0 : A = A1 against the alternative hypothesis H1 : A = A2 where A1 < A2.

Write down the likelihood ratio for the data, and show that all likelihood ratio tests of H0 against H1 are of the form: - Reject H0 if the sum from one to n of xi > c.

Homework Equations

LR(x) = f1(x)/f2(x)

The Attempt at a Solution

f0(xi) = (e^(-A1)*A1^xi) / xi!

f1(xi) = (e^(-A2)*A2^xi) / xi!

Then I'm not sure if this is right but I thought each f(x) is equal to the product of its f(xi)

so I have f0(x) = (e^(-A1)*A1^(sum of xi's from 1 to n)) / x1! ... xn!

Sorry don't know if I can put it sigma signs but don't know how and then obviously f1(x) is the same but with A2 instead.

So that gave me after some simplifying

LR(x) = e^(A1 - A2)*(A2/A1)^(sum of xi's)

Now I'm not sure if this is quite right or if I can simplify it further? Any help is much appreciated, thankyou.

 

[mighty2000]

Thank you for your post. Your approach to finding the likelihood ratio for this problem is correct. However, there are a few things that can be clarified and simplified.

Firstly, when finding the likelihood ratio, it is common to use the natural logarithm of the ratio instead of the ratio itself. This is because the natural logarithm is a monotonic function, meaning that it preserves the order of the values. This allows us to work with simpler expressions and still make the same decisions about whether to reject the null hypothesis or not.

Secondly, instead of finding the likelihood ratio for each individual xi, we can find the likelihood ratio for the entire data set x at once. This is because the data is independent and identically distributed, so the same likelihood ratio applies to each xi in the data set.

With these things in mind, we can simplify the expression for the likelihood ratio as follows:

LR(x) = ln(f1(x)/f0(x))

= ln((e^-A2*A2^Σxi)/Πxi!) / (e^-A1*A1^Σxi)/Πxi!)

= ln(e^(A1-A2)*(A2/A1)^(Σxi)/Πxi!) / e^-A1*A1^Σxi

= ln(e^(A1-A2)*(A2/A1)^(Σxi)) - ln(e^-A1*A1^Σxi)

= (A1-A2) + Σxi*ln(A2/A1) - (-A1*Σxi)

= (A1-A2) + Σxi*ln(A2/A1) + A1*Σxi

= A1 + Σxi*ln(A2/A1)

= A1 + n*ln(A2/A1)

So now we have a simplified expression for the likelihood ratio that only depends on the parameter values A1 and A2, and the sample size n. This makes it easier to see that the likelihood ratio test is of the form:

- Reject H0 if A1 + Σxi*ln(A2/A1) > c

or equivalently,

- Reject H0 if Σxi > c' where c' is some constant determined by the values of A1 and A2.

I hope this helps clarify the solution for you. If you have any further questions, please don't hesitate to