Statistical Analysis - Maximum Likelihood Fit

[knowlewj01]

Homework Statement

I have a set of data from the DAMA experiment in which a detector attempted to measure collisions with 'WIMP's [Weakly Interacting Massive Particles] as a candidate for dark matter. The detector records the time in days of a collision event. After binning the data and performing a Chi sqared test to a sine function I need to perform an 'Unbinned Maximum Likelihood Fit'.

As I understand the maximum likelehood fit is calculated using the probability distribution function (which i think is poissonian) for each data point.
After this I'm at a loss. Could anyone perhaps decribe the steps involved or even point me in the direction of a good guide to this test?

Thanks

Homework Equations

Poissonian PDF:

p(k,\lambda) = \frac{\lambda^k e^{-\lambda}}{k!}

k - observed # of events
λ - expected # of events

(However, Surely the data needs to be binned for a poissonian distribution to apply at all?)

The Attempt at a Solution

 

[knowlewj01]

Sorry, I don't think this was very clear. I have done some more reading:

My likelihood function L(λ) is poissonian:

f(k;\lambda)=\frac{e^{-\lambda}\lambda^k}{k!}

Log Likelihood function is:

L(\lambda)=ln\left(\Pi_{i}^{n} f(k_i;\lambda)\right)

Heres where i get a bit lost, I think my expected value λ should be a periodic function of the form:

\lambda_i(\omega,t_0;t_i)=cos(\omega[t_i-t_0])

The remaining steps (i think) are to substitute λ into the likelihood function and then to minimize the expression:

y=-2L(\lambda)

with respect to ω and t_0.

Does this sound right? Here's the expression i get:

L(\lambda)=-\Sigma_i^n cos(\omega[t_i - t_0]) + \Sigma_i^n k_i ln(cos(\omega[t_i - t_0]) - \Sigma_i^n ln(k_i !)

if my data is unbinned, what is my measured value (k_i)? I don't think the detector records more than one count in a day, so could i make my effective bin size 1 day? this would eliminate the final term (as 0! = 1! = 1, and ln(1) = 0) giving:

L(\lambda)=-\Sigma_i^n cos(\omega[t_i - t_0]) + \Sigma_i^n k_i ln(cos(\omega[t_i - t_0])
(although this would disappear when minimizinag anyway)

after some rearranging and minimising with respect to omega i find:

\Sigma_i^n \frac{k_i}{cos(\omega[t_i-t_0])}=1

which implies:

\Sigma_i^n cos(\omega[t_i-t_0]) = 1

as k_i is only non zero when we observe a detection.

Does any of this look right? I've never done one of these before and examples of this type are difficult to find.

Thanks