880.p20 winter 2006 richard kass probability and statistics many of the process involved with...

880.P20 Winter 2006 Richard Kass

Probability and StatisticsMany of the process involved with detection of particles are statistical in nature: Number of ion pairs created when proton goes through 1 cm of gas

Energy lost by an electron going through 1 mm of leadThe understanding and interpretation of all experimental data depend on statistical and probabilistic concepts:

“The result of the experiment was inconclusive so we had to use statistics” how do we extract the best value of a quantity from a set of measurements? how do we decide if our experiment is consistent/inconsistent with a given theory? how do we decide if our experiment is internally consistent? how do we decide if our experiment is consistent with other experiments? how do we decide if we have a signal (i.e. evidence for a new particle)?Our pentaquark example from the SPring-8 (LEPS) experiment :

signal: 19 eventsSignificance: 4.6Assuming gaussian stats the prob. for a 4.6 effect is ~4x10-6)

What are the authors trying to say here?If this bump is accidental, then the accident rate is 1 in 4 million. OrIf I repeated the experiment 4 million times I would get a bump this big or bigger.What do the authors want you to think?Since the accident rate is so low it must not be an accident, therefore it is physics!

K+n5=dudus


Probability & Statistics & Reality

CLAS 2003/2004 CLAS 2005 g11

Sometimes it is not a question of statistical significance!Again, the pentaquark state +(1540) gives a great example:Consider the CLAS experiment at JLAB: 2003/4 report a 7.8 effect (~6x10-15 according to MATHEMATICA)

2005 report NO Signal! (better experiment)

What size signal should we expect?

Lesson: This is not a statistics issue, but one of experiment design and implementation.


Definition of probability by example (“empirical”):Suppose we have N trials and a specified event occurs r times.

example: the trial could be rolling a dice and the event could be rolling a 6. define probability (P) of an event (E) occurring as:

P(E) = r/N when N examples:

six sided dice: P(6) = 1/6for an honest dice: P(1) = P(2) = P(3) = P(4) =P(5) = P(6) =1/6

coin toss: P(heads) = P(tails) =0.5P(heads) should approach 0.5 the more times you toss the coin.

For a single coin toss we can never get P(heads) = 0.5!

By definition probability (P) is a non-negative real number bounded by 0P 1

if P = 0 then the event never occurs

if P = 1 then the event always occurs

Let A and B be subsets of S then P(A)≥0, P(B)≥0

Events are independent if: P(AB) = P(A)P(B)

Coin tosses are independent events, the result of the next toss does not depend on previous toss.

Events are mutually exclusive (disjoint) if: P(AB) = 0 or P(AB) = P(A) + P(B)

In tossing a coin we either get a head or a tail.

Sum (or integral) of all probabilities if they are mutually exclusive must = 1.

intersection union

How do we define Probability?

{1,3} A {1,2,3,5}

1)(1 dxxporxPi

i p(x) =probability distribution function (pdf)

One can find 4 definitions ofprobability: mathematical,empirical, objective, and subjective.(see STATISTICS by Barlow)

“objective”probability

“mathematical” probability


Probability can be a discrete or a continuous variable.

Discrete probability: P can have certain values only. examples:

tossing a six-sided dice: P(xi) = Pi here xi = 1, 2, 3, 4, 5, 6 and Pi = 1/6 for all xi.

tossing a coin: only 2 choices, heads or tails.

for both of the above discrete examples (and in general)

when we sum over all mutually exclusive possibilities:

Continuous probability: P can be any number between 0 and 1. define a “probability density function”, pdf, f(x):

with a continuous variable Probability for x to be in the range a x b is:

Just like the discrete case the sum of all probabilities must equal 1.

We say that f(x) is normalized to one.Probability for x to be exactly some number is zero since:

P xi i 1

f x dx dP x xdx

P(ax b) f x a

b dx

f x

dx 1

f x xa

xa dx 0

NOTATION

xi is called a

random variable

Probability=“area under the curve”

Note: in the above example the pdf depends on only 1 variable, x. In general, the pdf can depend on manyvariables, i.e. f=f(x,y,z,..). In these cases the probability is calculated from a multi-dimensional integration.

Two Types of Probability


Some Common Probability Distributions

Examples of some common P(x)’s and f(x)’s:

Discrete = P(x) Continuous = f(x)

binomial uniform, i.e. = constant

Poisson Gaussian

exponential

chi square

How do we describe a probability distribution?

mean, mode, median, and variance

For a continuous distribution these quantities are defined by:

For discrete distribution the mean and variance are defined by:

Mean Mode Median Variance average most probable 50% point width of distribution

xf(x)dx

f x x xa

0 0.5 f(x)dx

a

2 f(x) x 2dx

nxn

ii /)( 2

1

2

nxn

ii /

1


Some Continuous Probability Distributions

mode median mean

symmetric distribution (gaussian)Asymmetric distribution showing the mean, median and mode

Chi-square distribution Student t distributionu

v=gaussianv=1 Cauchy (Breit-Wigner)

For a Gaussian pdfthe mean, mode,and median are all at the same x.

For many pdfsthe mean, mode,and median are in different places.

Remember: Probability is the area under these curves!For many pdfs its integral can not be done in closed form, use a table to calculate probability.


Uniform distribution and Random NumbersWhat is a uniform probability distribution: p(x)?

p(x)=constant (c) for a x bp(x)=zero everywhere else

Therefore p(x1)dx1= p(x2)dx2 if dx1=dx2 equal intervals give equal probabilities

For a uniform distribution with a=0, b=1 we have p(x)=1

11)(1

0

1

0

1

0 cdxccdxdxxp

What is a random number generator ?A number picked at random from a uniform distribution with limits [0,1] All major computer languages (FORTRAN, C) come with a random number generator.

FORTRAN: RAN(iseed)The following FORTRAN program generates 5 random numbers: iseed=12345

do I=1,5 y=ran(iseed) type *, y enddo end

0.1985246 0.8978736 0.2382888 0.3679854 0.3817045

If we generate “a lot” of random numbersall equal intervals should contain the sameamount of numbers. For example: generate: 106 random numbers expect: 105 numbers [0.0, 0.1]

105 numbers [0.45, 0.55]

x

p(x)

1

10


Uniform Random Numbers & Monte CarloThe uniform pdf is the basis of all Monte Carlo Calculations!The Monte Carlo method is commonly used to simulate experiments. A google search yielded 3,960,000 hits for “Monte Carlo Method”First reference (http://www.geocities.com/CollegePark/Quad/2435/history.html):“The Monte Carlo method provides approximate solutions to a variety of mathematical problems by performing statistical sampling experiments on a computer. The method applies to problems with no probabilistic content as well as to those with inherent probabilistic structure. Among all numerical methods that rely on N-point evaluations in M-dimensional space to produce an approximate solution, the Monte Carlo method has absolute error of estimate that decreases as N superscript -1/2 whereas, in the absence of exploitable special structure all others have errors that decrease as N superscript -1/M at best.”

“The method is called after the city in the Monaco principality, because of a roulette, a simple random number generator. The name and the systematic development of Monte Carlo methods dates from about 1944.”

“The real use of Monte Carlo methods as a research tool stems from work on the atomic bomb during the second world war. This work involved a direct simulation of the probabilistic problems concerned with

random neutron diffusion in fissile material”

Basically, it is a way to do really complicated integrals!A certain molecule always has a rectangular shape. However, the length of a side varies uniformly between 0.5 and 1Å. Calculate the probability that the area of a molecule is0.5 Å2.Suppose I want the volume to be 0.5 Å3?

1

5.0

1

/5.0

22 dxdyPx


Uniform Random Numbers & Monte CarloGiven the uniform pdf we can generate all other pdfs!For example: if RAN is uniform in (0,1) then a+(b-a)*RAN is uniform in (a,b)

Suppose we want to generate random numbers according to a pdf=p(x) startingfrom our uniform random number pdf (=r). Let r0=random number uniform in (0,1), we want to find the x that satisfies:

xxr

dxxprdxxprdx0

0

00

)()(0

Whether or not we can actually invert this equation depends on whether or notp(x) can be integrated in closed form. Works for exponential pdf: )1ln(1)()( 0

/

0

/

rxedxxpe

xp xxx

Does not for gaussian since integral can not be done in closed form.BUT lots of other clever ways of generating pdfs, e.g. gaussian: g=sin2r1(-2lnr2)1/2

When all else fails, can use “acceptance-rejection”

x

p(x)

1

0

1) normalize p(x): p(x)/pmax Max. of function=12) pick a random number to represent x and calculate p(x)/pmax

3) pick another random number=y4) if y<p(x)/pmax accept otherwise reject5) repeat 2)-4) lots of times…Problem: this algorithm can be very inefficient

“inversion”


Calculation of mean and variance:example: a discrete data set consisting of three numbers: {1, 2, 3}

average () is just:

Complication: suppose some measurements are more precise than others. Let each measurement xi have a weight wi associated with it then:

variance (2) or average squared deviation from the mean is:

is called the standard deviation rewrite the above expression by expanding the summations:

Note: The n in the denominator would be n -1 if we determined the average () from the data itself.

xi

ni1

n 123

32

xii1

n wi / wi

i1

n

2 1n

(xi )2

i1

n

2 1n

xi2 2

i1

n 2 xi

i1

n

i1

n

1n

xi2

i1

n 2 22

1n

xi2

i1

n 2

The variancedescribesthe width of the pdf !

“weighted average”

This is sometimes written as:<x2>-<x>2 with <> average of what ever is in the brackets

Discrete Probability Distributions

Real life example: 3 measurements of branching fraction of B-D0K*-


Using the definition of from above we have for our example of {1,2,3}:

The case where the measurements have different weights is more complicated:

Here is the weighted mean If we calculated from the data, 2 gets multiplied by a factor n/(n1).

Example: a continuous probability distribution, This “pdf” has two modes!

It has same mean and median, but differ from the mode(s).

2 1n

xi2

i1

n 2 4.67 22 0.67

2 w i (x ii1

n )2 / w i

2

i1

n w ix i

2

i1

n / w i

2

i1

n 2

constantc,20for sin)( 2 xxcxf

Discrete & Continuous Probability Distributions

f(x)=sin2x is not a true pdf since it is not normalized!f(x)=(1/) sin2x is a normalized pdf (c=1/).

x sin2 xdx /0

2 sin2 xdx

0

2

modex

sin2 x 0 2

,32

median sin2 xdx /0

sin2 xdx1

20

2

2

0

2sin:Note xdx


Probability, Set Theory and StuffThe relationships and results from set theory are essential to the understanding of probability.Below are some definitions and examples that illustrae the connection between set theory,probability and statistics.

We define an experiment as a process that generates “observations” and a sample space (S)as the set of all possible outcomes from the experiment:

simple event: only one possible outcomecompound event: more than one outcome

As an example of simple and compound events consider particles (e.g. protons, neutrons) madeof u (“up”), d (“down”), and s (“strange”) quarks. The u quark has electric charge (Q) =2/3|e|(e=charge of electron) while the d and s quarks have charge =-1/3|e|. Let the experiment be the ways we combine 3 quarks to make a Q=0, 1, or 2 state.

Event A: Q=0 {ssu, ddu, sdu} note: a neutron is a ddu stateEvent B: Q=1{suu, duu} note: a proton is a duu stateEvent C: Q=2 {uuu}

For this example events A and B are compound while event C is simple.

The following definitions from set theory are used all the time in the discussion of probability.Let A and B be events in a sample space S.Union: The union of A & B (AB) is the event consisting of all outcomes in A or B.Intersection: The intersection of A & B (AB) is the event consisting of all outcomes in A and B.Complement: The complement of A (A´) is the set of outcomes in S not contained in A. Mutually exclusive: If A & B have no outcomes in common they are mutually exclusive.


Probability, Set Theory and StuffReturning to our example of particles containing 3 quarks (“baryons”):The event consisting of charged particles with Q=1,2 is the union of B and C: BCThe events A, B, C are mutually exclusive since they do not have any particles in common.

S

AB

A common and useful way to visualize union, intersection, and mutually exclusive is to use a Venn diagram of sets A and B defined in space S:

S

AB

AB: intersection of A&B

S A

B

AB: union of A&B A & B mutually exclusive

S

AB

Venn diagram of A&B

The axioms of probabilities (P):a) For any event A, P(A)≥0. (no negative probabilities allowed)b) P(S)=1. c) If A1, A2, ….An is a collection of mutually exclusive events then:

n

iin APAAAP

121 )()(

(the collection can be infinite (n=∞))

From the above axioms we can prove the following useful propositions:a) For any event A: P(A)=1-P(A´) b) If A & B are mutually exclusive then P(AB)=0c) For any two events A & B: P(AB)=P(A)+P(B)-P(AB)

items b, c are “obvious”from their Venn diagrams


Probability, Set Theory and StuffExample: Everyone likes pizza. Assume the probability of having pizza for lunch is 40%, the probability of having pizza for dinner is 70%, and the probability of having pizza for lunch and dinner is 30%. Also, this person always skips breakfast. We can recast this example using:

P(A)= probability of having pizza for lunch =40%P(B)= probability of having pizza for dinner = 70%P(AB)=30% (pizza for lunch and dinner)

1) What is the probability that pizza is eaten at least once a day? The key words are “at least once”, this means we want the union of A & B

P(AB)=P(A)+P(B)-P(AB) = .7+.4-.3 =0.82) What is the probability that pizza is not eaten on a given day?

Not eating pizza (Z´) is the complement of eating pizza (Z) so P(Z)+P(Z´)=1 P(Z´)=1-P(Z) =1-0.8 = 0.2

3) What is the probability that pizza is only eaten once a day?This can be visualized by looking at the Venn diagram and realizing we need to exclude the overlap (intersection) region.

P(AB)-P(AB) = 0.8-0.3 =0.5

prop. c)

prop. a)

The non-overlapping blue area is pizza for lunch, no pizza for dinner. The non-overlapping red area is pizza for dinner, no pizza for lunch.

pizza for lunch

pizza for dinner


Conditional Probability

Frequently we must calculate a probability assuming something else has occurred. This is called conditional probability.Here’s an example of conditional probability: Suppose a day of the week is chosen at random. The probability the day is Thursday is 1/7. P(Thursday)=1/7 Suppose we also know the day is a weekday. Now the probability is conditional, =1/5. P(Thursday|weekday)=1/5

the notation is: probability of it being Thursday given that it is a weekday

Formally, we define the conditional probability of A given B has occurred as:P(A|B)=P(AB)/P(B)

We can use this definition to calculate the intersection of A and B:P(AB)=P(A|B)P(B)

For the case where the Ai’s are both mutually exclusive and exhaustive we have:

n

iiinn APABPAPABPAPABPAPABPBP

12211 )()|()()|()()|()()|()(

For our example let B=the day is a Thursday, A1= weekday, A2=weekend, then: P(Thursday)=P(thursday|weekday)P(weekday)+P(Thursday|weekend)P(weekend) P(Thursday)=(1/5)(5/7)+(0)(2/7)=1/7


Bayes’s TheoremBayes’s Theorem relates conditional probabilities. It is widely usedin many areas of the physical and social sciences.Let A1, A2,..Ai be a collection of mutually exclusive and exhaustive events with P(Ai)>0 for all i.Then for any other event B with P(B)>0 we have:

n

jjj

iiii

APABP

APABP

BP

BAPBAP

1

)()|(

)()|(

)(

)()|(

We call: P(Aj) the aprori probability of Aj occurringP(Aj|B) the posterior probability that Aj will occur given that B has

occurredP(B|Aj) the likelihood

Independence has a special meaning in probability:Events A and B are said to be independent if P(A|B)=P(A)Using the definition of conditional probability A and B are independent iff:

P(AB)=P(A)P(B)

Probability it is a weekday given it is Thursday.Probability it is Thursday given it is a weekday.


Independence Example

If the scans are independent then: P(12)=P(1)P(2)

Let’s consider a situation where we are trying to determine the number ofevents (N) we have in our data sample. This is a “classic” problem that comes inmany different situations: ancient times: scanning emulsion or bubble/spark chamber photos modern times: using computer algorithms to find rare events (top, beyond, GKZ, B, etc)

Consider the case where “scan 1” finds N1 events and “scan 2” finds N2 events.The number of events found by both scans is N12. What can we say about the total number of events, N?

N

NP

N

NP

N

NP 1221 )21()2()1(

P(1)

P(2)

P(12)

N

N

N

N

N

NPPP 1221)21()2()1(

12

21

N

NNN


Example of Bayes’s TheoremWhile Bayes’s theorem is very useful in physics, perhaps the bestillustration of its use is in medical statistics, especially drug testing.Assume a certain drug test: gives a positive result 97% of the time when the drug is present:

P(positive test|drug present)=0.97 gives a positive result 0.4% of the time if the drug is not present (“false positive”)

P(positive test|drug not present)=0.004Let’s assume that the drug is present in 0.5% of the population (1 out of 200 people).

P(drug present)=0.005P(drug not present)=1-P(drug present)=0.995

What is the probability that the drug is not present and you have a positive test?P(drug is not present|positive test)=????

Bayes’s Theorem gives:

)presentdrug()presentdrug|positivetest()presentnotdrug()presentnotdrug|positivetest(

)presentnotdrug()presentnotdrug|positivetest(

)positivetest |presentnotdrug(

PPPP

PP

P

45.0)005.0)(97.0()005.01)(004.0(

)005.01)(004.0()positivetest |presentnotdrug(

P

Thus there is a 45% chance that the test comes back positive even if you are drug free!The real life consequence of this large probability is that drug tests are often administered twice!


Another Example of Bayes’s TheoremAssume we are trying to separate pions from kaons using (e.g.) a cerenkov counter (CC)99% of the time when a pion goes through our CC we get a signal: P(signal|pion)=0.995.5% of the time when a kaon goes through our CC we get a signal: P(signal|kaon)=0.055 Suppose we are interested in looking for -- This decay of the -lepton has never been observed, example of a 2nd class current! (wrong “G-parity”)Theory says BF(-- )=1.5x10-5. (ranges from 1.2-1.6x10-5)There is a similar decay --that has been measured with BF(-- )=27x10-5

)()|s()()K|s(

)()|s(

)signal|(

PignalPKPignalP

PignalP

P

Thus there is ~49% chance that an event with a CC signal is --

Suppose we have an event where our CC gives us a signal.What is the probability that the event is --?

95.0105.11027

1027

)()(

)()(

05.0105.11027

105.1

)()(

)()(

55

5

55

5

BFKBF

KBFKP

BFKBF

BFP

49.0)05.0)(99.0()95.0)(055.0(

)05.0)(99.0()signal|(

P


Bayes’s Theorem ContinuedHow can we improve the signal to noise (S/N)?Suppose we add (or use) another independent particle detector where 90% of the time when a pion goes through our CC we get a signal 15% of the time when a kaon goes through our CC we get a signalThis is not such a great detector. But it really helps: P(signal|pion)=(0.99)(0.9)=0.89 P(signal|kaon)=(0.055)(0.15)=0.00825

Thus now is ~85% chance that an event with a CC signal is --

85.0)05.0)(89.0()95.0)(00825.0(

)05.0)(89.0()signal|(

P

Adding particle ID has improved our S/N: No particle ID S/N=1.5/27=0.06 One PID detector S/N=.49/.51=0.96 Two PID detectors S/N=0.85/0.15=5.7

It is usually more practical to build two moderately good PID detectors than onesuper high quality PID detector. In BaBar we do PID with a cerenkov counter and the drift chamber.

)physics)(detector())()K|s(

)()|s(/

KPignalP

PignalPNS


Accuracy: The accuracy of an experiment refers to how close the experimental measurement is to the true value of the quantity being measured. Precision: This refers to how well the experimental result has been determined, without regard to the true value of the quantity being measured.

Just because an experiment is precise it does not mean it is accurate!!example: measurements of the neutron lifetime over the years:

Steady increase in precision of the neutron lifetime but are any of these measurements accurate?

This figure showsvarious measurementsof the neutron lifetimeover the years.

The size of barreflects theprecision ofthe experiment

Accuracy and Precision


Use results from probability and statistics as a way of calculating how “good” a measurement is.

most common quality indicator:

relative precision = [uncertainty of measurement]/measurement

example: we measure a table to be 10 inches with uncertainty of 1 inch.

relative precision = 1/10 = 0.1 or 10% (% relative precision)

Uncertainty in measurement is usually square root of variance:

= standard deviation

is usually calculated using the technique of “propagation of errors”.

However this is not what most people think it is! We will discuss this in more detail soon.

Statistical and Systematic ErrorsResults from experiments are often presented as:

N ± XX ± YY

N: value of quantity measured (or determined) by experiment.

XX: statistical error, usually assumed to be from a Gaussian distribution.

With the assumption of Gaussian statistics we can say (calculate) something about how well our experiment agrees with other experiments and/or theories.

Expect ~ 68% chance that the true value is between N - XX and N + XX.

YY: systematic error. Hard to estimate, distribution of errors usually not known.

Examples: mass of proton = 0.9382769 ± 0.0000027 GeV (only statistical error given)

mass of W boson = 80.8 ± 1.5 ± 2.4 GeV (both statistical and systematic error given)

Measurement Errors (or Uncertainties)


What’s the difference between statistical and systematic errors?Statistical errors are “random” in the sense that if we repeat the measurement enough times:

XX 0

Systematic errors do not 0 with repetition.

examples of sources of systematic errors:

voltmeter not calibrated properly

a ruler not the length we think is (meter stick might really be < meter!)

Because of systematic errors, an experimental result can be precise, but not accurate!

How do we combine systematic and statistical errors to get one estimate of precision?

Can be a problem!

two choices:

tot = XX + YY add them linearly

tot = (XX2 + YY2)1/2 add them in quadrature

We will discuss a detailed averaging procedure next week………...

Some other ways of quoting experimental results

lower limit: “the mass of particle X is > 100 GeV”

upper limit: “the mass of particle X is < 100 GeV”

asymmetric errors: mass of particle X 100 34 GeV

Measurement Errors (or Uncertainties)

Example: the error in the mean m:If we repeat a measurement n times and each measurement has uncertainty then:

nm

880.p20 winter 2006 richard kass probability and statistics many of the process involved with...

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