Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient...

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B a s i c C o u n t i n g S t a t i s t i c s

Transcript of Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient...

Page 1: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

Basic

Counting

Statistics

Page 2: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Stochastic Nuclear Observables

2 sources of stochastic observables x in nuclear science:

1) Nuclear phenomena are governed by quantal wave functions and inherent statistics2) Detection of processes occurs with imperfect efficiency (ε < 1) and finite resolution distributing sharp events x0 over a range in x. Stochastic observables x have a range of values with frequencies determined by probability distribution P(x).characterize by set of moments of P

<xn> = ∫ xn P (x)dx; n 0, 1, 2,…

Normalization <x0> = 1. First moment (expectation value) of P:

E(x) = <x> = ∫xP (x)dx

second central moment = “variance” of P(x):σx

2 = <x2- <x2> >

Page 3: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

Uncertainty and Statistics

Nuclear systems: quantal wave functions ψi(x,…;t)

(x,…;t) = degrees of freedom of system, time.

Probability density (e.g., for x, integrate over other d.o.f.)

( )

( )

( )12

12

2

2

212 12

2121

( , ) | , | 1,2,....

( , )( ) | , | 1

1 2, | | ( )2

( , ) | | 1λ

ψ

ψ

ρπ

ψ

+∞ +∞

−∞ −∞

Γ− ⋅ − ⋅

= =

= = =

→ Γ ≈

−= ∝

∫ ∫

ii

ii i

t t

dP x t x t idx

NormalizationdP x tP t dx dx x t

dx

Transition between states M E

dP x t x e e state disappearsdx

1

2

Partial probability rates λ12 for disappearance (decay of 12) can vary over many orders of magnitude no certainty statistics

Page 4: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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The Normal Distribution

10 8 6 4 2 0 2 4 6 8 100

0. 0 5

0. 1

0. 1 5

0. 2

G( )x

G( )σ

x

30 38 46 54 62 70

0.02

0.04

0.06

0.08

Normal (Gaussian) Probability1 10 1−⋅

10 8−

G xn( )

7030 xn

( )2

22

1( ) exp22

2 2ln 2.352

σπσ

σ σ

− = ⋅ −

= =Γ ⋅⋅FWHM x

XX

x

x xP x

Continuous function or discrete distribution (over bins)

Normalized probability

( )1

1

2

22

1( ) exp22 σπσ −∞

− < = ⋅ −

∫X

x

X

x xP x x dx

2σx

P(x)

P(x)

x

x

<x>

Page 5: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Experimental Mean Counts and Variance

1

2 2 2

1

22 2

1

:1 ( )

1 ( )1

(" ")

1 ( )( 1)

N

i populationi

N

ii

population

N

n ii

Average count n in a sample

n n n unknownN

Variance of n in the individual samples

s n nN

Variance error of thesample average n n

s n nN N N

σ

σ

=

=

=

= ⋅ ≠

= = ⋅ −−

= = ⋅ −−

Measured by ensemble sampling expectation values + uncertainties

Sample (Ensemble) = Population instant

236U (0.25mg) source, count α particles emitted during N = 10 time intervals (samples @1 min). λ =??

2

1

. : / 5

(35496 59) n

9

i: m

σ σ−

= ≈ < >

≈ = ±

=

pop

n nStd deviation n N

R n nesultSlightly different from sample to sample

Page 6: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Sample Statistics

0 2 4 6 8 10

1

2

3

101 105.50

5

10Normally Distributed Event

10

0

m i,

σ+

σ−

300 304.50

5

10Normally Distributed Event

10

0

i,

σ+

σ−

( )2

22

1( ) exp22

pop

XX

x xP x

vvπ

− = ⋅ −

x

P(x) Assume true population distribution for variable x

with true (“population”) mean <x>pop = 5.0, νx =1.0

40 1 40 5.50

5

10Normally Distributed Events

10

0

m i,

0

0 σ+

0 σ−

Mean arithmetic sample average <x> = (5.11+4.96+4.96)/3 = 5.01

Variance of sample averages s2=σ2 = [(5.11-5.01)2+2(4.96-5.01)2]/2 = 0.01 s = 0.0075

σx2 = 0.0075/3 = 0.0025 σx = 0.05 Result: <x>pop 5.01 ± 0.05

<x>-σ

<x> =5.11 σ =1.11 <x> =4.96 σ =0.94

<x>+σ<x>

<x> =4.96 σ =1.23

3 independent sample measurements (equivalent statistics):

x

Page 7: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Example of Gaussian Population

0 25 500

5

10Normally Distributed Events

10

0

xm i,

x0

x0 σ+

x0 σ−

500 m

0 5 100

10

2050MC Events in 10 x bins

ii FRAME:=

0 5 100

5

10Normally Distributed Events

xm i,

x0

x0 σ+

x0 σ−

m

0 5 100

10

2010MC Events in 10 x bins

xm xm

Sample size makes a difference ( weighted average)

n = 10 n = 50

The larger the sample, the narrower the distribution of x values, the more it approaches the true Gaussian (normal) distribution.

Page 8: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Central-Limit Theorem

Increasing size n of samples: Distribution of sample means Gaussian normal distrib. regardless of form of original (population) distribution.

The means (averages) of different samples in the previous example cluster together closely. general property:

The average of a distribution does not contain information on the shape of the distribution.

The average of any truly random sample of a population is already somewhat close to the true population average.

Many or large samples narrow the choices: smaller Gaussian width Standard error of the mean decreases with incr. sample size

Page 9: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Integer random variable m = number of events, out of N total, of a given type, e.g., decay of m (from a sample of N ) radioactive nuclei, or detection of m (out of N ) photons arriving at detector. p = probability for a (one) success (decay of one nucleus, detection of one photon)

Choose an arbitrary sample of m trials out of N trialspm = probability for at least m successes(1-p)N- m = probability for N-m failures (survivals, escaping detection)Probability for exactly m successes out of a total of N trials

How many ways can m events be chosen out of N ? Binomial coefficient

Total probability (success rate) for any sample of m events:

Binomial Distribution

( )( ) 1 N mmP m p p −∝ −

P mNm

p pbinomialm N m( ) =

FHG

IKJ − −1b g

! ( 1)!( )! 1

N N N m Nm N m mm

− + ⋅ ⋅ ⋅= = − ⋅ ⋅ ⋅

Page 10: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Moments and Limits

( )( ) 1 N mmbinomial

NP m p p

m−

= −

( )0 0

1 ( ) 1 −

= =

= = −

∑ ∑

N NN mm

binomialm m

NormalizationN

P m p pm

Distributions for N=30 and p=0.1 p=0.3Poisson Gaussian

2 (1 )

(1 ) 1

σ

σ

= ⋅ = ⋅ −

⋅ −=

m

m

Mean and variancem N p and N p p

N p pm N p N

0 5 10 15 200

0.1

0.2

0.3Binomial Dis tributions N=30

0.236

0

Pb N m, 0.1,( )

Pb N m, 0.3,( )

200 m

( )222

1lim ( ) exp22

binomialNmm

x mP m

σπσ→∞

− = ⋅ −

Probability for m “successes” out of N trials, individual probability p

Page 11: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Poisson Probability Distribution

0 5 10 15 200

0.2

0.4

0.6

0.8Po iss on Dist ribu tions

0.607

0

Pp 0.5 m,( )

Pp 3 m,( )

Pp 5 m,( )

Pp 10 m,( )

200 m

P m emPoisson

m

( , )!

µ µ µ

=⋅ −

Probability for observing m events when average is <m> = µ

m=0,1,2,…

µ = <m> = N· p and σ2 = µ

is the mean, the average number of successes in N trials.

Observe N counts (events) uncertainty is σ= √µ

Unlike the binomial distribution, the Poisson distribution does not depend explicitly on p or N !

For large N, p:

Poisson Gaussian (Normal Distribution)

Results from binomial distribution in the limit of small p and large N (N·p > 0)

0lim ( , ) ( , )µ

→→∞

=binomial Poissonpand N

P N m P m

Page 12: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Moments of Transition Probabilities23

17

4 114 1

17

14 1

71/ 2

6.022 10: 0.25 0.25 6.38 10236

3.5946 10 min 5.6362 10 min6.38 10

( ) :5.6362 10 min

" " 2.34 10λ

−− −

− −

⋅= ⋅ = ⋅

⋅= = = ⋅

= = ⋅

= ⋅

N mg mgg

np

NProbability for decay decay rate per nucleus

pcorresponds to half life t a

Small probability for process, but many trials (n0 = 6.38·1017)

0< n0·λ < ∞Statistical process follows a Poisson distribution: n=“random” Different statistical distributions: Binomial, Poisson, Gaussian

Page 13: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Radioactive Decay as Poisson Process

Slow radioactive decay of large sample Sample size N » 1, decay probability p « 1, with 0 < N·p <

137Cs unstable isotope decayt1/2 = 27 years p = ln2/27 = 0.026/a = 8.2·10-10s-1 0Sample of 1 µg: N = 1015 nuclei (=trials for decay)

How many will decay(= activity µ) ? µ =< >= N·p = 8.2·10+5 s-1

Count rate estimate < >=d<N>/dt = (8.2·10+5 ± 905) s-1

estimated

Probability for m actual decays P (µ,m) =

55 8.52 10(8.52 10 )( , )! !

m m

Poissone eP m

m m

µµµ− − ⋅⋅ ⋅ ⋅

= =

N

N

Page 14: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Functions of Stochastic Variables

Random independent variable sets {N1}, {N2},….,{Nn } corresponding variances σ1

2, σ22,….,σn

2

Function f(N1, N2,….,Nn) defined for any tuple {N1, N2,….,Nn}Expectation value (mean) Gauss’ law of error propagation:

( ) ( ) ( )

1 22 2 22 2 21 2

1 2

2 2 2

2, 3,.. 1, 3,.. 1, 2,.., 1

....

| | .. |

f nn

N N N N N N Nn

f f fN N N

f f f

σ σ σ σ

∂ ∂ ∂ = + + + ∂ ∂ ∂

∆ + ∆ + + ∆

Further terms if Ni not independent ( correlations)Otherwise, individual component variances (∆f)2 add.

( )1

1 ,...,,...,

nn N N

f f N N=

Page 15: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Example: Spectral Analysis

Background B

Peak Area A

B1 B2

Analyze peak in range channels c1– c2: beginning of background left and right of peak n = c1 – c2 +1.

Total area = N12 =A+BN(c1)=B1, N(c2)=B2,

Linear background <B> = n(B1+B2)/2

Peak Area A:

( )( )

12 1 2

12 12

2

/ 2

/ 4A

A N n B B

N B Bnσ

= ⋅

⋅ +

+

+

=

( )1 2 11 2 22 1:N N N NN N N NN ± ± ±± = = ± +

Adding or subtracting 2 Poisson distributed numbers N1 and N2:Variances always add

1 2σ σ± ±

Page 16: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Confidence Level

( )2

22

2(| | ) exp22

δ

σπδ

σ

< >+

< >

− < > − < > ≈ ⋅ − =

< ∫x

popx

x xP x x dx CL

( 1 ) 68.3% ( 2 ) 95.4%( 3 ) 99.7%δ σ δ σδ σ

= = = == =

CL CLCL

Assume normally distributed observable x:

( )2

22

1( ) exp22π

− = ⋅ −

pop

poppop

x xP x

vv

Sample distribution with data set

observed average <x> and std. error σ approximate population. Confidence level CL (Central Confidence Interval):

With confidence level CL (probability in %), the true value <xpop> differs by less than δ = nσ from measured average. Trustworthy exptl.

results quote ±3σerror bars!

1σ±

2σ±

3σ±

Measured Probability

Page 17: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Setting Confidence Limits

Example: Search for rare decay with decay rate λ, observe no counts within time ∆t. Decay probability law

dP/dt=-dN/Ndt= exp {- λ·t}. P(λ,t) = symmetric in λ and t

( )

0

0

0

00

0 0

0 | 1

( ) 1

1 1ln[1 ( )] ln[1 ] 0

:

λ λ

λλλ

λ λ

λ λ λ

λ λ λ

∞− ⋅∆ − ⋅∆

− ⋅∆− ⋅∆

∆ = ∆ ⋅ ∆ ⋅ =

≤ = ∆ ⋅ = −

− −= ⋅ − ≤ = ⋅ − >

∆ ∆

=

t t

tt

P t e with t e d

P t e d e

P CLt t

CL

normalized P

normalized P

Upper limit

Higher confidence levels CL (0 CL 1) larger upper limits for a given time ∆t inspected. Reduce limit by measuring for longer period.

no counts in ∆t

Page 18: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Maximum LikelihoodMeasurement of correlations between observables y and x: {xi,yi| i=1-N} Hypothesis: y(x) =f(c1,…,cm; x). Parameters defining f: {c1,…,cm}ndof=N-m degrees of freedom for a “fit” of the data with f.

( )21

1 22

( ,.., ; )1( ,.., ; ) exp22 σπσ

− = ⋅ −

i m ii m

ii

y f c c xP c c x for every

data point

( )

1 11

2

2211

( ,.., ) ( ,.., ; )

1 exp22 σπσ

=

==

= =

∆ = ⋅ −

∑∏

N

m i mi

N Ni

ij ij

P c c P c c x

y

Maximize simultaneous probability

Minimize chi-squared by varying {c1,…,cm}: ∂χ2/∂ci = 0

( ) ( ) ( )2 212

1 2 21 1

( ,.., ; ),.., :χ

σ σ= =

∆ −= =∑ ∑

N Ni i m i

mi ii i

y y f c c xc c

When is χ2 as good as can be?

Page 19: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Minimizing χ2

( ) ( ) ( )2 22

2 21 1

, :χσ σ= =

∆ − −= =∑ ∑

N Ni i i

i ii i

y y a bxa b

Example: linear fit f(a,b;x) = a + b·x to data set {xi, yi, σi}

Minimize:

( ) ( ) ( )22

2 21 1

20 ,

N Ni i i i

i ii i

y a bx y a bxa b

a aχ

σ σ= =

− − − −∂ ∂= = = −

∂ ∂∑ ∑

( ) ( )22

1

20 ,

Ni i i

i i

x y a bxa b

σ=

− −∂= = −

∂∑

2 2 21 1 1

1N N Ni i

i i ii i i

bx y

aσ σ σ= = =

+ =∑ ∑ ∑

2

2 2 21 1 1

N N Ni i i i

i i ii i i

ax

bx x yσ σ σ= = =

+ =∑ ∑ ∑

11 12 1

21 22 2

aad d cd db c

b+ =

+ =11 12

21 22

d dD

d d=

1 12 11 1

2 22 21 22

2 22 2

1 1

1 1 1ia b

i i

c d d ca b

c d d cD D

xD D

σ σσ σ

= =

= =∑ ∑

Equivalent to solving system of linear equations

Page 20: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Distribution of Chi-Squareds

Distribution of possible χ2 for data sets distributed normally about a theoretical expectation (function) with ndof degrees of freedom:

( )( )

( )

22 12 2

2 2

2

2

1 2

2

2

2

e( )

2 2

( ) 1 !

2.507 (1 0.0 )

2 1

833 /

d

ndof

n ndof d

of dof dof

ofdof

n n

P d d

n n n

n

n n

e n n

χ

χ

χ

χχ

σ

χ χ

− −

− −

Γ = − =

= +

= =

(Stirling’s formula)

Reduced χ2:2 2 2

2

( 1)

0 1.5 50%

r dof

r

n N mFor

Confidence

χ χ χ

χ

= = − −

≤ < → ≥

2

2( , ) ( )dof ndof

P n P x dxχ

χ∞

= ∫

0 2 4 6 8 100

0.1

0 .2

0 .3

0 .4Chi-Squared Distribution

P u 1,( )

P u 2,( )

P u 3,( )

P u 4,( )

P u 5,( )

u

<χ2>ndof=5

u:=χ2

Should be P 0.5 for a reasonable fit

Page 21: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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CL for χ2-Distributions

1-CL

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Correlations in Data Sets

Correlations within data set. Example: yi small whenever xi small

x

yuncorrel. P(x,y)

( ) ( )2 2

2 22 2

( , ) ( ) ( )

1 exp2 24 σ σπ σ

= ⋅ =

− − = ⋅ − +

unc

x yx

P x y P x P y

x x y y

( ) ( ) ( )( )( )

2 2 2

2 22 2

2 2 2

1( , )2

2exp

2

π σ σ σ

σ σ

σ σ σ

= ⋅−

− + − − − − ⋅ − −

corr

x y xy

x y

x y xy

P x y

x x y y x x y y

( )

( )

22 22 221cot

2 4

1 1

x yx yxy

xy

xy xy x y xyr correlation coefficient r

σ σσ σα σ

σ

σ σ σ

−−

= + +

= − ≤ ≤

x

ycorrelated P(x,y)

α( )( ) ( , )σ = − −∫xy x x y y P x y dxdy covariance

Page 23: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Correlations in Data Sets

( )( )( ) ( )22

: 1 1

;

1 1:1

σσ σ

σ

− −= − ≤ ≤ +

− ⋅ −

=

= − ⋅ −

∑∑ ∑

∑ ∑ ∑

i i j jij ij

i i j j

ijij

i j

ij i j i j

c c c cr r

c c c c

r

c c c c covariancen n

uncertainties of deduced most likely parameters ci (e.g., a, b for linear fit) depend on depth/shallowness and shape of the χ2 surface

ci

cjuncorrelated χ2 surface

ci

cj

correlated χ2

surface

({ }) ( )

({ }) ( )

=

unc i ii

corr i ii

P c P c

P c P c

Page 24: Stochastic Nuclear Observables...How many ways can mevents be chosen out of N ? Binomial coefficient Total probability (success rate) for any sample of mevents: Binomial Distribution

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Multivariate Correlations

ci

cj

initial guess

search path

Smoothed χ2 surface Different search strategies:

Steepest gradient, Newton method w or w/odamping of oscillations,

Biased MC:Metropolis MC algorithms,

Simulated annealing (MC derived from metallurgy),

Various software packagesLINFIT, MINUIT,….