Examples using ROOT - JLab Computer …elton/talks/examples_hugs2013.pdfElton S. Smith HUGS Summer...

Elton S. Smith HUGS Summer School June 2013 1

Examples using ROOT http://root.cern.ch

Ex1: Event Generation (Binomial Distribution) Ex2: Generate various random distributions Ex3: Linear fits Ex4: Determination of the area under a Gaussian Ex5: Bayesian inference

Elton S. Smith, Jefferson Lab

Elton S. Smith HUGS Summer School June 2013

Exercise 1 – Generate Binomial Distribution

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Exercise: Use the ROOT random number generator to generate 10 random numbers according to the Binomial distribution. Compare with the parent probability distribution function.


Exercise 1 - output

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Exercise 2 – Various random distributions

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Extra credit: Generate random number according to an arbitrary input function.

Exercise: Use the ROOT random number generator to generate random numbers according to a 1.  Uniform distribution 2.  Exponential distribution 3.  Gaussian distribution 4.  Poisson distribution

Compare with the parent probability distribution function.


Random distributions

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

200

400

600

800

1000

1200

1400

1600

1800

2000

uniform distribution uniformEntries 100000

Mean 0.4997

RMS 0.2886

/ ndf 2! 94.51 / 98

p0 6.3± 1001

p1 10.950± -4.053

uniform distribution

0 5 10 15 20 25 30 35 400

2000

4000

6000

8000

10000

exponential distribution decayEntries 100000

Mean 3.999

RMS 4.002

/ ndf 2

! 88.19 / 91

Constant 0.004± 9.213

Slope 0.0008± -0.2507

exponential distribution

-10 -8 -6 -4 -2 0 2 4 6 8 100

500

1000

1500

2000

2500

3000

3500

4000

bell distribution bellEntries 100000

Mean 3.001

RMS 1.998

/ ndf 2! 68.21 / 75

Constant 15.4± 3988

Mean 0.006± 3.003

Sigma 0.004± 1.999

bell distribution

0 1 2 3 4 5 6 7 8 9 100

2000

4000

6000

8000

10000

12000

14000

16000

18000

20000

22000

Poisson distribution countsEntries 100000

Mean 2.995

RMS 1.71

/ ndf 2

! 1704 / 7

Constant 9.68e+01± 2.25e+04

Mean 0.01± 3.36

Sigma 0.006± 1.793

Poisson distribution


Polynomial function

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Exercise 3 - Linear Fits

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Assume a parent distribution of the form y(x) = a + bx, a=5, b=1 Assume one experiment collects a data set of ten points of the form (xi, yi±σ), i=0,1,2,...9, with the measurements yi following a Gaussian distribution with a fixed width σ=0.5.   Invent the data points yi for one experiment.   Fit the data yi to the form y = a + bx.   Determine y and the uncertainty of y as a function of x from the fit.


Exercise 3 - Linear Fits

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Extra credit   Generate 1000 Monte Carlo experiments   For each experiment fit the data set to the functional form given above   Plot the difference between the fitted function and data   Histogram the difference between the fitted parameters and the true parameter a and b   Use the width of the distribution to determine the uncertainty in the parameters.   Compare with the estimated uncertainties in the fit


Linear Fit – one “experiment”

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Fit for one “experiment” showing the fitted parameters Repeat 1000 times


Fitted Results to 1000 “experiments”

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For each fit, plot the fitted value of the intercept and slope. Fit the distributions to Gaussian functions

Mean = 5.004 ± 0.010 Sigma = 0.303

Mean = 0.9997 ± 0.0018 Sigma = 0.05626 What is the relation

between these two?


Plot difference between fitted and true values

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Fit Gaussian to slices

y(x)-yfit

Uncertainty on σy can be

computed using

σy2=σa

2+x2σb2+2xσab

Correlation term is important

σab=0


Exercise 4 - Fitted Gaussian area

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Many measurements of a variable x have been accumulated in one experiment. The measurements are dominated by experimental resolution, so the distribution of measurements is Gaussian.   Obtain Integral

  Generate 1000 measurements (one experiment) according to a Gaussian distribution   Fit the distribution to a Gaussian and determine the area under the curve and its uncertainty.

  Systematic Study (extra credit)   Generate 100 experiments with 1000 measurements each and empirically determine areas and uncertainties   Use fit defaults and option=‘L’   Compare and discuss


Integral =!

2!A"b

Var(Integral) = 2!!A2#33 + "2#11 + 2A"#13

"b2

b = bins/xunits

Fitted Gaussian area – one “experiment”

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Variable x (xunits)

(969)

(5) (322)

Fitting Option=default


Fitted Gaussian area – 100 “experiments”

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Events Generated Fitted Gaussian Sum

Mean = 1000 ± 3.7 Mean = 976.5 ± 3.7 Sigma = 34 Sigma = 35


Integral =!

2!A"b

Var(Integral) = 2!!A2#33 + "2#11 + 2A"#13

"b2

b = bins/xunits

Fitted Gaussian area – one “experiment”

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Variable x (xunits)

(997)

(5) (322)

Fitting Option=‘L’


Fitted Gaussian area – 100 “experiments”

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Events Generated Fitted Gaussian Sum

Mean = 1000 ± 3.7 Mean = 998 ± 3.5 Sigma = 34 Sigma = 32


Summary of examples of fitting

  Linear fit   Obtained values of the uncertainties of the parameters by

Monte Carlo.   Demonstrated these uncertainties corresponded to the

computed estimate   The correlation term must be included

  Determination of areas under Gaussian distributions   The area and uncertainties were computed   The computation requires inclusion of the correlation

term.   Monte Carlo determination of the area shows that the use

of least-square fits lead to estimates for the area that are systematically low

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Bayesian inference - example

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An experiment is interested in identifying pions in the presence of a large number of muon tracks. A detector is designed to identify pions, but has a 96% efficiency for tagging pions as pions and a 1% chance of misidentifying a muon and tagging it as a pion.   Assume there are 10 times more muons than pions.   Assume that a track has been tagged as a pion.   What is the degree of belief that the track is a pion?   How does this change if there are 100 times more muons than pions?


Bayesian inference

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P (!|id) =P (id|!)P (!)

P (id|!)P (!) + P (id|µ)P (µ)P (id|!) = 0.96

P (!) = 1P (id|µ) = 0.01

P (µ) = 10Z (cm)

760 780 800 820 840 860 880 900

Y (

cm

)

-100

-80

-60

-40

-20

0

20

40

60

80

100

BCAL

BCAL

CDC

CDC

FDC FDC FDC FDC

TO

F

target

FCAL

solenoid

1MWPC

Iron

2MWPC

Muon Detector

π µ

µ

P (!|id) = 0.91

P (!|id) = 0.49 P (µ) = 100


Dependence on background ratio

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! / µRatio of 20 40 60 80 100

| id

)!

P(

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

efficiency=0.96! misidentification=0.01µ

Examples using ROOT - JLab Computer …elton/talks/examples_hugs2013.pdfElton S. Smith HUGS Summer...

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