Statistical Analysis - Maximum Likelihood Fit

  • #1
knowlewj01
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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:

[itex]p(k,\lambda) = \frac{\lambda^k e^{-\lambda}}{k!}[/itex]

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

 
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  • #2
Sorry, I don't think this was very clear. I have done some more reading:

My likelihood function L(λ) is poissonian:

[itex]f(k;\lambda)=\frac{e^{-\lambda}\lambda^k}{k!}[/itex]

Log Likelihood function is:

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

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

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

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

[itex]y=-2L(\lambda)[/itex]

with respect to ω and t_0.

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

[itex]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 !)[/itex]

if my data is unbinned, what is my measured value (ki)? 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:

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

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

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

which implies:

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

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
 
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