Download PowerPoint Presentation - Experimental Elementary Particle Physics

Probability and Statistics
What is probability?
What is statistics?
1
Probability and Statistics
 Probability


Formally defined using a set of axioms
Seeks to determine the likelihood that a given
event or observation or measurement will or has
happened
 What is the probability of throwing a 7 using two dice?
 Statistics


Used to analyze the frequency of past events
Uses a given sample of data to assess a
probabilistic model’s validity or determine values
of its parameters
 After observing several throws of two dice, can I
determine whether or not they are loaded

Also depends on what we mean by probability
2
Probability and Statistics
We perform an experiment to collect a
number of top quarks





How do we extract the best value for its
mass?
What is the uncertainty of our best value?
Is our experiment internally consistent?
Is this value consistent with a given theory,
which itself may contain uncertainties?
Is this value consistent with other
measurements of the top quark mass?
3
Probability and Statistics
CDF “discovery” announced 4/11/2011
4
Probability and Statistics
5
Probability and Statistics
Pentaquark search - how can this occur?
2003 – 6.8s effect
2005 – no effect
6
Probability
 Let the sample space S be the space of all
possible outcomes of an experiment
 Let x be a possible outcome



Then P(x found in [x,x+dx]) = f(x)dx
f(x) is called the probability density function (pdf)
It may be called f(x;q) since the pdf could depend
on one or more parameters q
 Often we will want to determine q from a set of
measurements

Of course x must be somewhere so

 f x dx  1

7
Probability
 Definitions of mean and variance are given in
terms of expectation values
Ex   xf x dx  

  
V x  E x  Ex  E x    s
2
2
2
2
8
Probability
 Definitions of covariance and correlation
coefficient


Vxy  covx, y   E x   x  y   y   E xy   x  y
 xy 
covx, y 
s xs y
if x, y independen t then f x,y  f x x  f y  y 
and then E x,y   xyf x,ydxdy  x  y
and so covx, y   0
9
Probability
 Error propagation


Consider y  x  with vari ables x   x1 , x2 ,...xn 


and   E x  and Vij  cov xi , x j


n
 y 


then TS expanding y  x   y        xi  i 
i 1  xi  x  
we find


E  y  x   y  
n 

y y 
2 
E y  x   y     
 Vij
i , j 1 
 xi x j  x  

2

10
Probability
 This gives the familiar error propagation
formulas for sums (or differences) and
products (or ratio)
2
2
2
Using V  y   s y  E y  E  y 
 
we find for
y  x1  x2
s y2  s 12  s 22  2 covx1 , x2 
and for
y  x1  x2
covx1 , x2 
 2  2 2
2
y
x1
x2
x1 x2
s y2
s 12
s 22
11
Uniform Distribution
 Let
 1
for α  x  β

f ( x;  ,  )     
otherwise
 0

x
1
E x    xf  x dx  
dx     
2
  

V x   E  x  E x 

2

2
1
1
1


2
V x     x     
dx     
2
12
  


What is the position resolution of a silicon or
multiwire proportional chamber with detection
elements of space x?
12
Binomial Distribution
 Consider N independent experiments (Bernoulli
trials)
 Let the outcome of each be pass or fail
 Let the probability of pass = p
Probabilit y of n successes  p n 1  p 
N!
But there are
permuation s for
n! N  n !
N distinguis hable objects, grouping them n at a time
N!
N n
n


f (n; N , p ) 
p 1 p
n! N  n !
N n
13
Permutations
 Quick review
Number of permutatio ns for N elements  N !
This considers each element distinguis hable
But n elements of the first type are indistingu ishable so
n! of the elements lead to the same situation
Ditto for the remaining N-n  elements of the second type
Thus accounting for these irrelevant permutatio ns leads to
N
N!
number of unique permuation s is
  
N-n !n!  n 
14
Binomial Distribution
 For the mean and variance we obtain
(using small tricks)
N
En   nf n; N , p   Np
n 0
 
V n  E n  En  Np1  p 
2
2
 And note with the binomial theorem that
N n
N n
N
f (n; N , p)    p 1  p    p  1  p   1

n 0
n 0  n 
N
N
15
Binomial Distribution
Binomial pdf
16
Binomial Distribution
Examples





Coin flip (p=1/2)
Dice throw (p=1/6)
Branching ratio of nuclear and particle
decays (p=Br)
Detector or trigger efficiencies (pass or not
pass)
Blood group B or not blood group B
17
Binomial Distribution
 It’s baseball season! What is the probability of
a 0.300 hitter getting 4 hits in one game?
N  4, p  0.3
4!
4
0
f 4;4,0.3 
 0.3  0.7  0.0081
4!0!
E n  4  0.3  1.2
V n  4  0.3  0.7  0.84
Expect 1.2  0.84 hits for a 0.300 hitter
18
Poisson Distribution
 Consider when
N 
p0
E n   Np  v
The Poisson pdf is
n
v v
f n; v   e
n!
and one finds
E n   v
V n   s 2  v
19
Poisson Distribution
N!
N n
f n; N , p  
p n 1  p 
n! N  n !
N!
 N n for N large
N  n !
p 2 N  n N  n  1
1  p   1  pN  n  
 ...
2!
N 2 p2
N n
1  p   1  Np 
 ...  e  Np  e v
2!
n
n v
N n v v e
f n; N , p  
pe 
for large N and small p
n!
n!
N n
20
Poisson Distribution
 Poisson pdf
21
Poisson Distribution
 Examples

Particles detected from radioactive decays
 Sum of two Poisson processes is a Poisson process






Particles detected from scattering of a beam on
target with cross section s
Cosmic rays observed in a time interval t
Number of entries in a histogram bin when data is
accumulated over a fixed time interval
Number of Prussian soldiers kicked to death by
horses
Infant mortality
QC/failure rate predictions
22
Poisson Distribution
 Let
Let N~1020 atoms
1
1
Let p   12 ~ 2  1020 /s
τ 10 y
Then v  Np  2 / s
and
20 e 2
P (0;2) 
 0.135
0!
21 e 2
P (1;2) 
 0.271
1!
22 e 2
P ( 2;2) 
 0.271
2!
23
Gaussian Distribution
 Gaussian distribution

Important because of the central limit
theorem
 For n independent variables x1,x2,…,xN that are
distributed according to any pdf, then the sum
y=∑xi will have a pdf that approaches a Gaussian
for large N
 Examples are almost any measurement error
(energy resolution, position resolution, …)
E  y    i
i
V y   s i
i
24
Gaussian Distribution
 The familiar Gaussian pdf is
2

1
 x    

f  x;  , s  
exp 
2

2
s
2s 2


E x   
V x   s 2
25
Gaussian Distribution
 Some useful properties of the Gaussian distribution are
in range ±s) = 0.683
in range ±2s) = 0.9555
in range ±3s) = 0.9973
outside range ±3s) = 0.0027
outside range ±5s) = 5.7x10-7

P(x
P(x
P(x
P(x
P(x

P(x in range ±0.6745s) = 0.5




26
c2 Distribution
 Chi-square distribution
1
f  z; n   n / 2
z n / 21e  z / 2 z  0 
2 n / 2 
n  1,2,... is the number of degrees of freedom
Ez   n
V z   2n
The usefulness of this pdf is that for
n independen t xi , i , s i2
n
z
i 1
xi  i 
2
s i2
follows the c 2 distributi on with n d.o.f.
27
c2 Distribution
28
Probability
29
Probability
 Probability can be defined in terms of
Kolmogorov axioms

The probability is a real-valued function defined on
subsets A,B,… in sample space S
For every subset A in S, P A  0
If A  B  0, P A  B   P A  PB 
PS   1

This means the probability is a measure in which
the measure of the entire sample space is 1
30
Probability
 We further define the conditional probability
P(A|B) read P(A) given B
P A  B 
P A | B  
P B 
 Bayes’ theorem
Using P A  B   PB  A
PB | AP A
P A | B  
P B 
31
Probability
 For disjoint Ai
PB    PB | Ai P Ai 
i
then
PB | AP A
P A | B  
 PB | Ai P Ai 
i
 Usually one treats the Ai as outcomes of a
repeatable experiment
32
Probability
 Usually one treats the Ai as outcomes of a
repeatable experiment

Then P(A) is usually assigned a value equal to the
limiting frequency of occurrence of A
A
P  A  lim
n  n

Called frequentist statistics
 But Ai could also be interpreted as hypotheses,
each of which is true or false


Then P(A) represents the degree of belief that
hypothesis A is true
Called Bayesian statistics
33
Bayes’ Theorem
 Suppose in the general population


P(disease) = 0.001
P(no disease) = 0.999
 Suppose there is a test to check for the
disease


P(+, disease) = 0.98
P(-, disease) = 0.02
 But also


P(+, no disease) = 0.03
P(-, no disease) = 0.97
 You are tested for the disease and it comes
back +. Should you be worried?
34
Bayes’ Theorem
 Apply Bayes’ theorem
P(, disease ) P(disease )
P(disease ,) 
P(, disease ) P(disease )  P(, no disease ) P(no disease )
0.98  0.001
P(disease ,) 
 0.032
0.98  0.001  0.03  0.999
 3.2% of people testing positive have the
disease
 Your degree of belief about having the disease
is 3.2%
35
Bayes’ Theorem
 Is athlete A guilty of drug doping?
 Assume a population of athletes in this sport


P(drug) = 0.005
P(no drug) = 0.995
 Suppose there is a test to check for the drug


P(+, drug) = 0.99
P(-, drug) = 0.01
 But also


P(+, no drug) = 0.004
P(-, no drug) = 0.996
 The athlete is tested positive. Is he/she
involved in drug doping?
36
Bayes’ Theorem
 Apply Bayes’ theorem
P(, drug ) P(drug )
P(drug ,) 
P(, drug ) P(drug )  P(, no drug ) P(no drug )
0.99  0.005
P(drug ,) 
 0.55
0.99  0.005  0.004  0.995
and
P(, no drug ) P(no drug )
P(no drug ,) 
P(, no drug ) P(no drug )  P(, drug ) P(drug )
0.004  0.995
P(no drug ,) 
 0.45
0.004  0.995  0.99  0.005
37
 ???
Binomial Distribution
 Calculating efficiencies

Usually use e instead of p
E n  n
E n   Ne  e 
  e
N
N
V n   s 2  Ne 1  e 
We don' t know the true e but we can use our estimate e 
s
e 1  e 
1
e  

n 1  n / N 
N
N
N
This is the error on the efficiency
38
Binomial Distribution
 But there is a problem


If n=0, (e’) = 0
If n=N, (e’) = 0
 Actually we went wrong in assuming the best
estimate for e is n/N

We should really have used the most probable
value of e given n and N
 A proper treatment uses Bayes’ theorem but
lucky for us (in HEP) the solution is
implemented in ROOT



h_num->Sumw2()
h_den->Sumw2()
h_eff->Divide(h_num,h_den,1.0,1.0,”B”)
39