1. 1 Statistics David Forrest University of Glasgow May 5 th 2009

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 0.1% 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 to determine error on emittance 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 (8 beams x 1700 simulations/beam plus repeats)

5. 5 8pi – N=1000 events 

6. 6 Checks - X, X’ beforeafter 2.5pi 0.2pi

7. 7 Checks - X, X’ beforeafter 10.0pi 8.0pi

8. 8 Checks – beta function Expected beta in absorbers ~420mm, solenoid 330 mm after matching 0.2  4.0  2.5  10.0 

9. 9 Results Events  rmsSims 0.2  10000.08970.004171.7260.1102450 20000.06130.003301.7350.0818261 100000.03290.001831.7310.0340242 1.5  10000.01680.000650.0840.0169545 20000.01060.000370.08020.0110545 100000.00540.000180.08030.0054545 2.5  10000.01170.00051-0.00220.0124421 20000.00830.00033-0.00340.0092426 100000.00400.00025-0.00500.0040320

10. 10 Results-2 Events  rmsSims 3.0  10000.01140.00048-0.0220.0117513 20000.00790.00031-0.0250.0081437 100000.00360.00016-0.0260.0036323 4.0  10000.00950.00037-0.0460.0097440 20000.00660.00020-0.0500.0068545 100000.00320.00015-0.0510.0031340 6.0  10000.00730.00034-0.0710.0079358 20000.00640.00023-0.0720.0067500 100000.00260.00017-0.0720.0028176

11. 11 Results-3 Events  rmsSims 8.0  10000.00910.00031-0.081 0.0093 549 20000.00710.00035-0.083 0.0070 426 100000.00310.00012-0.081 0.0032 547 10.0  10000.00970.00037-0.081 0.0102 540 20000.00690.00025-0.085 0.0068 541 100000.00330.00016-0.085 0.0034 359

12. 12 0.2 

13. 13 1.5 

14. 14 2.5 

15. 15 3.0 

16. 16 4.0 

17. 17 6.0 

18. 18 8.0 

19. 19 10.0 

20. 20 K values BeamK KK C CC 0.22.5330.1880.007270.00325 1.50.4810.0240.000420.00036 2.50.3510.0230.000520.00043 3.00.3560.0190.000510.00031 4.00.2820.0150.000380.00027 6.00.2470.0160.000240.00029 8.00.2870.0140.000250.00022 10.00.2930.0160.000410.00028

21. 21   =K/sqrt(N)

22. 22 Sans pencil beam

23. 23 Physical Meaning (J Cobb) There is a physical meaning to this K value By usual error formula, assuming no correlations: So without correlations, we have this factor, normally >1, eg if f=-0.08, we get a factor of 1.29

24. 24 Physical Meaning However, there are correlations between input emittance and output emittance, so we include a correlation factor, k corr. The  sim I measure includes this also.

25. 25 Correlation factor Preliminary

26. 26 How many muons do we need? We want to measure to an error of 0.1% Beam (pi mrad)No correlations (10 6 events) With correlations (10 5 events) 0.2 1564 1.5 2.42.3 2.5 2.01.2 3.0 2.01.3 4.0 1.80.8 6.0 1.70.6 8.0 1.70.8 10.0 1.70.8

27. 27 Conclusions We need of order 10 5 muons to achieve 0.1% error on fractional change in emittance Simulations in place for doing a toy Monte Carlo study, to propagate errors from elements of covariance matrix