DIJET STATUS

1. DIJET STATUS Kazim Gumus 30 Aug. 2005

2. Signal and QCD Crossection Our signal is spread over many bins, and the background varies widely over the bins. If we were to simply sum up all the bins in our QCD background and compare with all the bins in our Z' signal, then the lowest mass bin where the QCD background is huge would dominate, and we would get a ridiculously wrong result.    That is why, we use the binned maximum likelihood method, which takes into account correctly the relative probabilities in each of the bins.

3. Binned maximum likelihood method Ni: Observed events (Assume that it comes from a mass resonance sinal sitting on a large QCD background) Nsignali: Predicted signal events in the i'th mass bin Nqcdi: Predicted qcd background events in the i'th mass bin Let the signal to be multiplied by an unknown parameter alpha and to be add in the background to obtain the predicted number of events in the i'th mass bin Mui=alpha * Nsignali + Nqcdi The prob, of Pi observing ni events when Mui are predicted is given by Poisson statistics Pi= Mui * exp( - Mui ) / factorial( ni ) The product of Pi over all bins in the mass spectrum is the likelihood function L for seeing the observation given the prediction L = P1 * P2 * .....Pn Minimize the -ln (L) to find the best fit value of the signal normalization parameter alpha.

4. Fit to data and Residual Plot This plot shows the goodness of the fit

5. Two approaches to calculate 95 C.L and 5 Sigma Cross sections 1. Data=qcd (no signal) or Data=qcdfit (no signal) 95 C.L. exclusion 5 sigma exclusion 2. Data=qcd+signal or Data=qcdfit+signal 5 sigma discovery

6. Approach I Likelihoods for 95 C.L and 5 Sigma Exclusion

7. 700 GeV Likelihood (no signal) Data=qcd Data=qcdfit We are constructing the likelihood for our smoothed data sample resulting from a QCD background and a resonance signal as a function of the resonance signal cross section. 

8. Interpretation of Likelihood Distributions If the peak position of the likelihood plot is zero, then we have a perfect fit to the data. If the peak position of likelihood plot for a given mass is different than zero, it means that the fit is not "perfect" at the related mass.  The fit may still be within expected values, since data fluctuates up and down with statistics.  We shouldn't expect the fit to be perfect with real data. Because we should expect some bins to fluctuate high, and others to fluctuate low, with a Gaussian distribution of fluctuations with width determined by the number of events in the data.

9. 2TeV Likelihood (no signal) Data=qcd Data=qcdfit

10. 5TeV Likelihood (no signal) Data=qcdfit Data=qcd

11. Why We Should Use Perfect Fit Likelihoods  It is conceptually more correct,  because we are finding what our sensitivity is "on average" by getting rid of bin-to-bin statistical fluctuations.  It is like repeating the default likelihood for a thousand different experiments and taking the average of them all: That is what we want to know, how sensitive we are on average, not how sensitive we are for this particular data sample with its own statistics.

12. Method for Calculating 95 CL Exclusion To find the 95% C.L cross section for a dijet resonance, find the point on the likelihood distribution, alpha95, such that the integral of the likelihood from alpha95 to infinity is 0.05.  Integrate the likelihood out to some suitably large alpha value, alpha_end, which is large enough for our purposes to be considered infinite: for example one for which the likelihood has dropped to 10^-9 of its peak value.  Use that integral to normalize your likelihood to unit area. Loop over increasing alpha, integrate the normalized likelihood from alpha to alpha_end. Find the first integral that is less than 0.05(closest alpha value). Multiply alpha95 with the total cross section of signal.

13. 95 C.L Sigma Exclusion Results alpha_95 If the data fluctuates beneath the fit, then alpha95(data) < alpha95 (qcd=qcdfit) If the data fluctuates above the fit, then alpha95(data) > alpha95 (qcd=qcdfit)

14. Comments on 95% C.L. Exclusion The likelihood works fine for 95% CL exclusions of a signal, because we are estimating the probability that a high signal could have fluctuated down into the data we are observing, giving no observed signal. 95% CL exclusion says that “In 19 out of 20 cases the signal would not produce data like we observe, but that in 1 out of 20 cases it would accidentally fluctuate down to give an observation compatible with our data”.

15. 5 Sigma Exclusion Cross Section  To find the 5 sigma exclusion cross section for a dijet resonance, find the point on the likelihood distribution, alpha_5sigma, such that the integral of the likelihood from alpha_5sigma to infinity is 2.86 x 10^-7. The area of a Gaussian probability distribution from +5 sigma to infinity is 2.86 x10^-7. Integrate the likelihood out to some suitably large alpha value, alpha_end, which is large enough for our purposes to be considered infinite: for example one for which the likelihood has dropped to 10^-9 of its peak value.  Use that integral to normalize your likelihood to unit area. Loop over increasing alpha, integrate the normalized likelihood from alpha to alpha_end. Find the first integral that is less than 2.86 x 10^-7 (closest alpha value). We are integrating the tail of the likelihood and searching for a probability < 2.86 X 10^-7, rather then integrating the whole likelihood and searching for probability > 1 - 2.86 x 10^-7 = 0.999999714. This avoids problems with knowing the integral to an accuracy of 1 part in 10 million.

16. 5 Sigma Exclusion Results alpha_sigma5

17. 95% C.L. and 5 Sigma Exclusions

18. Approach II Likelihoods for 5 Sigma Discovery

19. What is a 5 sigma discovery? In the simplest scenario, we would be just counting events and a 5 sigma discovery is an observation that is 5 sigma above the background level.  nb= number of background events ns= number of signal events Here the sqrt(nb) is 1 sigma on the expected number of events, and we observe a 5 sigma fluctuation,

20. 5 Sigma Discovery Cross Section Calculation Method Form a new distribution of "signal + background" using this 5 sigma signal. The background should be from the perfect QCD fit.Set data = "signal + background", and find the likelihood distribution for this data. The likelihood distribution will show a Gaussian peak at some value of alpha well separated from zero.Measure the mean and the RMS of this likelihood distribution and the significance of this signal, s = mean/RMS. s should be very close to 5.If s is 5, then you have found the 5 sigma discovery cross section, if s is not 5 then re-estimate the 5 sigma cross section as "new signal cross section = old signal cross section * 5/s and repeat steps 2) - 4). 

21. 700 GeV Likelihood (data contains signal) alpha_5sigma= 33.4 mean= 1.002 rms= 0.1951 mean/rms= 5.1358 alpha_5sigma= 33.4 * 5 / 5.1358 = 32.5166 mean= 1.002 rms= 0.2002 mean/rms=5.0049

22. 2TeV Likelihood (data contains signal) alpha_5sigma= 7.4 mean= 1.002 rms= 0.1953 mean/rms= 5.1305 alpha_5sigma= 7.4 * 5 / 5.1305 = 7.2116 mean= 1.002 rms= 0.2003 mean/rms= 5.0024

23. 5TeV Likelihood (data contains signal) alpha_5sigma= 376.0 * 5 / 4.7453 = 396.1756 mean= 1.023 rms= 0.2079 mean/rms= 4.9206 alpha_5sigma= 376.0 mean= 1.025 rms= 0.216 mean/rms= 4.7453

24. 5 Sigma Discovery Results

25. 95% C.L. and 5 Sigma Discovery

26. Conclusions When there is a lot of data; the 5 sigma exclusions, from the NO SIGNAL likelihood, are at about the same cross section as the 5 sigma discovery cross section from the SIGNAL likelihood. This demonstrates that we have a fairly good understanding of the likelihooddistributions. We can make the cross section limits for new particles.