1. 1 Statistics Update David Forrest University of Glasgow

2. 2 The Problem We calculate 4D emittance from the fourth root of a determinant of a matrix of covariances...We want to measure fractional change in emittance with 1% error. The problem is compounded because our data is highly correlated between two trackers.

3. 3 How We Mean To Proceed W e assume that we will discover a formula that takes the form Sigma=K*(1/sqrt(N)) where K is some constant or parameter to be determined. How do we determine K? 1) First Principles: do full error propagation of cov matrices → difficult calculation 2) Run a large number of G4MICE simulations, using the Grid, to find the standard deviation for every element in the covariance matrix → Toy Monte Carlo 3) Empirical approach: large number of simulations to plot   versus 1/sqrt(N), identifying K (this work)

4. 4 What I’m Doing 3 absorbers (Step VI), G4MICE, 4D Transverse Emittance I plot 4D Transverse Emittance vs Z for some number of events N, for beam with input emittance . I calculate the fractional change in emittance . I repeat ~500 times and plot distribution of all  for each beam. Carried out about 15,000 simulations on Grid (7 beams x 1700 simulations/beam plus repeats)

5. 5 8pi – N=1000 events 

6. 6 8pi – N=2000 events 

7. 7 8pi – N=10000 events 

8. 8 I get sigma for the distribution 8pi Events1/sqrt(N)SigmaError on Sigma 10000.03160.01040.0004 20000.02240.00710.0003 100000.010.00350.0001

9. 9 8pi

10. 10 I get sigma for the distribution 6pi Events1/sqrt(N)SigmaError on Sigma 10000.03160.00850.0004 20000.02240.00740.0003 100000.010.00300.0002

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15. 15 6pi

16. 16 10pi rogue – not satisfied is correct

17. 17   =K/sqrt(N)   mm rad  K 0.20.591 2.50.58 30.389 40.31 60.25 80.416 100.317

18. 18 Discussion Behaviour   =K/sqrt(N) verified for all beams For   =0.001 require N~10 5 not N~10 6 as in the case of K=1 At low  the value of k should tend to 1 as stochastic processes dominate (but might plateau around 0.6). As we increase  it reaches a minimum at 6  and then rises again. This is strange so we think that the minimum at 6pi is because beam tuned for 6pi, so at other emittances, beam not well tuned so not optimum.