computabonal&physicsparticle.korea.ac.kr/class/2016/phys461/cphy_stat-ch02.pdf · 2016. 6. 11. ·...

Korea U/Dept. Physics, Prof. Eunil Won (All rights are reserved)

ComputaBonal PhysicsExamples of Probability FuncBons

Korea University Eunil Won

1


Binomial distribuBon

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• Binomial distribuBon Consider a series of N independent trials with only two possible outcomes. When the desired outcome of one trial is p, the probability of finding n desired outcome from N trials is governed by the binomial distribuBon.

The probability to have n desired outcomes in N events is

f(n;N, p) =N !

n!(N � n)!pn(1� p)N�n

The expectaBon and the variance are

V [n] = E[n2]� (E[n])2 = Np(1� p)

Can you prove them?

E[n] =NX

n=0

nN !

n!(N � n)!pn(1� p)N�n = Np


Binomial distribuBon -‐ proof 1

3

The expectaBon value can be found as

E[n] =NX

n=0

nN !

n!(N � n)!pn(1� p)N�n

= NpNX

n=1

(N � 1)!(n� 1)!

�(N � 1)� (n� 1)

�!p(n�1)(1� p)(N�1)�(n�1)

with m = n� 1, M = N � 1,

= NpMX

m=0

M !

m!(M �m)!pm(1� p)M�m

= Np


Binomial distribuBon -‐ proof 2

4

The variance can be found as

E[n(n� 1)] =NX

n=0

n(n� 1) N !n!(N � n)!p

n(1� p)N�n

=NX

n=2

N !

(n� 2)!(N � n)!pn(1� p)N�n

= N(N � 1)p2NX

n=2

(N � 2)!(n� 2)!

�(N � 2)� (n� 2)

�!pn�2(1� p)(N�2)�(n�2)

= N(N � 1)p2

E[n2] = E[n(n� 1) + n] = E[n(n� 1)] + E[n]

) V [n] = E[n2]� (E[n])2 = N(N � 1)p2 +Np� (Np)2 = Np(1� p)


Binomial distribuBon -‐ example

5

N= 5, p=0.5 N=10, p=0.5

N=20, p=0.2 N=20, p=0.5

Can you write ROOT C++ code that draws this?


Binomial distribuBon

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#include //// Drawing a binomial function//// E. Won ([email protected])

Double_t f1(Int_t n, Int_t N, Double_t p) { Double_t fac = TMath::Power(p,n)*TMath::Power((1.0-p),N-n); for (Int_t i=1;iSetStyle("Plain"); gROOT->ForceStyle();

TCanvas* c = new TCanvas("c","",200,10,600,400); c->Divide(2,2); h1 = new TH1F("h1","",20,0.0,20); h1->SetMaximum(0.4); h2 = new TH1F("h2","",20,0.0,20); h2->SetMaximum(0.4); h3 = new TH1F("h3","",20,0.0,20); h3->SetMaximum(0.4); h4 = new TH1F("h4","",20,0.0,20); h4->SetMaximum(0.4); h1->SetFillColor(3);h2->SetFillColor(3); h3->SetFillColor(3);h4->SetFillColor(3); for (Int_t i=0;iFill(i,f1(i, 5,0.5)); h2->Fill(i,f1(i,10,0.5)); h3->Fill(i,f1(i,20,0.2)); h4->Fill(i,f1(i,20,0.5)); } c->cd(1); h1->Draw(); c->cd(2); h2->Draw(); c->cd(3); h3->Draw(); c->cd(4); h4->Draw();}

- the source code

The funcBon f1 returns the binomial distribuBon

mailto:[email protected]


Binomial distribuBon -‐ example

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• Histogram Suppose we are interested in one parBcular bin of a compound histogram. Then “success” may correspond to geVng an entry in this parBcular bin, and “failure” corresponds to an entry in any other bin of the histogram -‐ so it is related with the binomial distribuBon.

The probability p for a success is p = n/N when n and N are entries in the bin and the total number of events in the histogram.

V [n] = Np(1� p) = N nN

⇣1� n

N

⌘= n

⇣1� n

N

⌘

�[n] =

rn⇣1� n

N

⌘⇡

pn for p ! 0 (or N ! 1)

Half size of the verBcal bar shows the “error of yi”

�yi : error of yi�yi =

pyi

We did this when we discussed histogram errors!


MulBnomial distribuBon

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• MulBnomial distribuBon GeneralizaBon of the binomial distribuBon with m different possible outcomes. For a parBcular outcome has the probability pi, and of course

mX

i=1

pi = 1

f(n1, ..., nm;N, p1, ..., pm) =N !

n1!n2!..nm!pn11 p

n22 ...p

nmm

The expectaBon and the variance areE[ni] = Npi and V [ni] = Npi(1� pi)

ex) For three possible outcomes

f(ni, nj ;N, pi, pj) =N !

ni!nj !(N � ni � nj)!pnii p

njj (1� pi � pj)

N�ni�nj

The covariance becomescov[ni, nj ] = �Npipj


Poisson DistribuBon

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• Poisson distribuBon LimiBng case of the binomial distribuBon when N becomes large, p very small but the product Np (i.e. the expectaBon value) remains a finite value ν

f(n; ⌫) =⌫n

n!e�⌫

E[n] =1X

n=0

n⌫n

n!e�⌫ = ⌫ V [n] =

1X

n=0

(n� ⌫)2 ⌫n

n!e�⌫ = ⌫


ν=2 ν=10


Poisson DistribuBon

10

• Poisson distribuBon LimiBng case of the binomial distribuBon when N becomes large, p very small but the product Np (i.e. the expectaBon value) remains a finite value ν -‐ proof

Np = ⌫ and SBrling’s formula gives N ! ⇡p2⇡NNNe�N

so,N !

n!(N � n)!pn(1� p)N�n = 1

n!

p2⇡NNNe�Np

2⇡(N � n)(N � n)(N�n)e�(N�n)⇣ ⌫N

⌘n⇣1� ⌫

N

⌘N�n

⇡ 1n!

N (N�n)e�N

(N � n)(N�n)e�(N�n)⌫n

⇣1� ⌫

N

⌘N�n

⇡ 1n!

1

(1� nN )(N�n)en⌫n

⇣1� ⌫

N

⌘N�n

⇡ 1n!

1

(1� nN )Nen⌫n

⇣1� ⌫

N

⌘N (1� ⌫N )�n

(1� nN )�n

⇡ e�⌫⌫n

n!

limN!1

⇣1 +

✏

N

⌘N= e✏

is used.


Uniform DistribuBon

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• Uniform distribuBon The uniform p.d.f. for the conBnuous variable is defined by

f(x;↵,�) =

⇢ 1��↵ ↵ x �0 otherwise

E[x] =

Z �

↵

x

� � ↵dx =1

2(↵+ �)

V [x] =

Z �

↵[x� 1

2(↵+ �)]2

1

� � ↵dx =1

12(� � ↵)2



ExponenBal DistribuBon

12

• ExponenBal distribuBon The exponenBal p.d.f. for the conBnuous variable is defined by

f(x, ⇠) =1

⇠

e

�x/⇠ (0 x < 1)


E[x] =1

⇠

Z 1

0xe

�x/⇠dx = ⇠

V [x] =1

⇠

Z 1

0(x� ⇠)2e�x/⇠dx = ⇠2

ξ=1ξ=2ξ=5


ExponenBal DistribuBon

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• ExponenBal distribuBon : the source code

ξ=1ξ=2ξ=5

//// Drawing exponential functions//// E. Won ([email protected])

Double_t f1(Double_t *x, Double_t *xi) { return TMath::Exp(-1.0*x[0]/xi[0])/xi[0]; }

void exp(){ gROOT->SetStyle("Plain"); gROOT->ForceStyle(); TF1 *fun1 = new TF1("fun1",f1,0.0,5.0,1); TF1 *fun2 = new TF1("fun2",f1,0.0,5.0,1); TF1 *fun3 = new TF1("fun3",f1,0.0,5.0,1); fun1->SetParameter(0,1.0); fun1->Draw(); fun2->SetLineColor(kBlue); fun2->SetParameter(0,2.0); fun2->Draw("LSAME"); fun3->SetLineColor(kRed); fun3->SetParameter(0,5.0); fun3->Draw("LSAME");}

ξ=1ξ=2ξ=5



Gaussian DistribuBon

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• Gaussian distribuBon The gaussian p.d.f. for the conBnuous variable is defined by

f(x;µ,�

2) =

1p2⇡�

2exp

⇣� (x� µ)

2

2�

2

⌘

E[x] = µ

V [x] = �2


μ=0,σ=2μ=4,σ=3μ=0,σ=6Can you prove them?


N-‐dim. Gaussian DistribuBon

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• Generalized N-‐dimensional gaussian distribuBon The N-‐dim. gaussian p.d.f. for the conBnuous variables x is defined by

f(x;µ, V ) =1

(2⇡)N/2|V |1/2exp

h� 1

2

(x� µ)TV �1(x� µ)i

where x and μ are column vectors containing x1,...,xn and μ1,...,μn, xT and μT are the corresponding row vectors, and |V| is the determinant of a symmetric NxN matrix V.

The expectaBon and the (co)variance are

E[xi] = µi, V [xi] = Vii, cov[xi, xj ] = Vij

The two dimensional example becomesf(x1, x2;µ1, µ2,�1,�2, ⇢) =

1

2⇡�1�2

p1� ⇢2

⇥ exp(

� 12(1� ⇢2)

h⇣x1 � µ1

�1

⌘2+

⇣x2 � µ2

�2

⌘2� 2⇢

⇣x1 � µ1

�1

⌘⇣x2 � µ2

�2

⌘i)

where is the correlaBon coefficient ⇢ = cov[x1, x2]/(�1�2)


Chi-‐square DistribuBon

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• The χ2(chi-‐square) distribuBon The χ2(chi-‐square) distribuBon of the conBnuous variable z is defined by

f(z;n) =1

2n/2�(n/2)zn/2�1e�z/2, n = 1, 2, ... (0 z < 1)

where the parameter n is called the number of degrees of freedom and the gamma funcBon is defined by

�(x) =

Z 1

0e

�tt

x�1dt

The expectaBon and the variance are found to be

E[z] =

Z 1

0z

1

2n/2�(n/2)zn/2�1e�z/2 dz = n,

V [z] =

Z 1

0(z � n)2 1

2n/2�(n/2)zn/2�1e�z/2 dz = 2n


Chi-‐square DistribuBon

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//// E. Won ([email protected])//// gammln from Numerical Recipes

float gammln(float xx){ double x,y,tmp,ser; static double cof[6]={76.18009172947146,-86.50532032941677, 24.01409824083091,-1.231739572450155, 0.1208650973866179e-2,-0.5395239384953e-5}; int j;

y=x=xx; tmp=x+5.5; tmp -= (x+0.5)*log(tmp); ser=1.000000000190015; for (j=0;jSetStyle("Plain"); gROOT->ForceStyle(); TH2F *h2 = new TH2F("h2","",2,0.0,20.0,2,0.0,0.5); h2->Draw(); TF1 *fun1 = new TF1("fun1",f1,0.0,20.0,1); TF1 *fun2 = new TF1("fun2",f1,0.0,20.0,1); TF1 *fun3 = new TF1("fun3",f1,0.0,20.0,1); TF1 *fun4 = new TF1("fun4",f1,0.0,20.0,1); fun1->SetParameter(0, 1); fun1->Draw("LSAME"); // rest of lines skipped ...

n=1n=2n=5n=10

Can you understand how it is done? (I took the logarithm first!)



Binomial, Poisson, and Gaussian PDF

18

Binomial Poisson

Gaussian

N ! 1, Np = µ

N ! 1 µ ! 1

computabonal&physicsparticle.korea.ac.kr/class/2016/phys461/cphy_stat-ch02.pdf · 2016. 6. 11. ·...

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