Statistical Errors

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Transcript Statistical Errors

Statistics

David Forrest University of Glasgow May 5 th 2009 1

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

2

highly correlated between two trackers.

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

s e

versus 1/sqrt(N), identifying K (this work)

3

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 e .

• I calculate the fractional change in emittance De .

• I repeat ~500 times and plot distribution of all De for each beam.

• Carried out about 15,000 simulations on Grid (8 beams x 1700 simulations/beam plus repeats) 4

8pi – N=1000 events

De/e 5

before

Checks X, X’

after 0.2pi

2.5pi

6

before

Checks X, X’

after 8.0pi

10.0pi

7

Checks – beta function

Expected beta in absorbers ~420mm, solenoid 330 mm after matching

0.2

p

2.5

p

4.0

p

10.0

p 8

Ds

Results

Events

0.2

p s 1000 2000 10000

1.5

p 1000 2000 10000

2.5

p 1000 2000 10000 0.0897

0.0613

0.0329

0.0168

0.0106

0.0054

0.0117

0.0083

0.0040

0.00417

0.00330

0.00183

0.00065

0.00037

0.00018

0.00051

0.00033

0.00025

de/e 1.726

1.735

1.731

0.084

0.0802

0.0803

-0.0022

-0.0034

-0.0050

rms Sims

0.1102

0.0818

0.0340

0.0169

0.0110

0.0054

0.0124

0.0092

0.0040

450 261 242 545 545 545 421 426 320 9

Events

3.0

p 1000 2000 10000

4.0

p 1000 2000 10000

6.0

p 1000 2000 10000 s 0.0114

0.0079

0.0036

0.0095

0.0066

0.0032

0.0073

0.0064

0.0026

Ds

Results-2

de/e

rms

0.00048

0.00031

0.00016

0.00037

0.00020

0.00015

0.00034

0.00023

0.00017

-0.022

-0.025

-0.026

-0.046

-0.050

-0.051

-0.071

-0.072

-0.072

0.0117

0.0081

0.0036

0.0097

0.0068

0.0031

0.0079

0.0067

0.0028

Sims

513 437 323 440 545 340 358 500 176 10

Events

8.0

p 1000 2000 10000

10.0

p 1000 2000 10000 s 0.0091

0.0071

0.0031

0.0097

0.0069

0.0033

Results-3

Ds de/e 0.00031

0.00035

0.00012

-0.081

-0.083

-0.081

0.00037

0.00025

0.00016

-0.081

-0.085

-0.085

rms

0.0093

0.0070

0.0032

0.0102

0.0068

0.0034

Sims

549 426 547 540 541 359 11

0.2

p 12

1.5

p 13

2.5

p 14

3.0

p 15

4.0

p 16

6.0

p 17

8.0

p 18

10.0

p 19

Beam 0.2

1.5

2.5

3.0

4.0

6.0

8.0

10.0

K

2.533

0.481

0.351

0.356

0.282

0.247

0.287

0.293

d

K

0.188

0.024

0.023

0.019

0.015

0.016

0.014

0.016

K values

C

0.00727

0.00042

0.00052

0.00051

0.00038

0.00024

0.00025

0.00041

s 

K

d

C

0.00325

1

N

C

0.00036

0.00043

0.00031

0.00027

0.00029

0.00022

0.00028

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s De/e =K/sqrt(N) 21

Sans pencil beam

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Physical Meaning (J Cobb)

• There is a physical meaning to this K value • By usual error formula, assuming no correlations:

f

 s

f

2 D e e

in

    e e

out in

e

out

e 

in

e

in

  2 s e e

o u t out

• So without correlations, we have this factor, normally >1, eg if f=-0.08, we get a factor of 1.29

 s e

in

e

in

s

f

2 s

f

 2    e e

out in

 ( 1    s e

o u t

e

out f

)   2 2 2

N

 2 1

N

  1 e e

out in N

2    s e e

in in

 ( 1   1

f

) 2   2 2

N

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Physical Meaning

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

s

sim

k corr

s

nocorr

k corr

K

e

in

e

out

2 2 ( 1 

f

)

k corr

1

N

24

Correlation factor

Preliminary 25

How many muons do we need?

• We want to measure to an error of 0.1% Beam (pi mrad) No correlations (10 6 events) 15 With correlations (10 5 events)

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0.2

1.5

2.5

3.0

4.0

6.0

8.0

10.0

2.4

2.0

2.0

1.8

1.7

1.7

1.7

2.3

1.2

1.3

0.8

0.6

0.8

0.8

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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 27