probability theory 2 tron anders moger september 13th 2006

Probability theory 2

Tron Anders Moger

September 13th 2006

The Binomial distribution• Bernoulli distribution: One experiment with

two possible outcomes, probability of success P.

• If the experiment is repeated n times

• The probability P is constant in all experiments

• The experiments are independent

• Then the number of successes follows a binomial distribution

The Binomial distribution

If X has a Binomial distribution, its PDF is defined as:

xnx PPxnx

nxXP

)1(

)!(!

!)(

)1()(

)(

PnPXVar

nPXE

Example

• Since the early 50s, 10000 UFO’s have been reported in the U.S.

• Assume P(real observation)=1/100000

• Binomial experiments, n=10000, p=1/100000

• X counts the number of real observations

%5.9095.010000

11

10000

1

0

100001

)0(1)1()real isn observatio oneleast At (100000

XPXPP

The Hypergeometric distribution

• Randomly sample n objects from a group of N, S of which are successes. The distribution of the number of successes, X, in the sample, is hypergeometric distributed:

)!(!

!)!()!(

)!(

)!(!

!

)(

nNn

NxnSNxn

SN

xSx

S

n

N

xn

SN

x

S

xXP

Example

• What is the probability of winning the lottery, that is, getting all 7 numbers on your coupon correct out of the total 34?

71086.1

)!734(!7

!34)!77734()!77(

)!734(

)!77(!7

!7

7

34

77

734

7

7

)7(

XP

The distribution of rare events: The Poisson distribution

• Assume successes happen independently, at a rate λ per time unit. The probability of x successes during a time unit is given by the Poisson distribution:

( )!

( )

( )

xeP x

xE X

Var X

Example: AIDS cases in 1991 (47 weeks)

• Cases per week:

1 1 0 1 2 1 3 0 0 0 0 0 0 2 1 2 2 1 3 0 1 0 0 0

1 1 1 1 1 0 2 1 0 2 0 2 1 6 1 0 0 1 0 2 0 0 0

• Mean number of cases per week:

λ=44/47=0.936

• Can model the data as a Poisson process with rate λ=0.936

Example cont’d:No. of No. Expected no. observed

cases observed (from Poisson dist.)

0 20 18.4

1 16 17.2

2 8 8.1

3 2 2.5

4 0 0.6

5 0 0.11

6 1 0.017

• Calculation: P(X=2)=0.9362*e-0.936/2!=0.17

• Multiply by the number of weeks: 0.17*47=8.1

• Poisson distribution fits data fairly well!

The Poisson and the Binomial

• Assume X is Bin(n,P), E(X)=nP• Probability of 0 successes: P(X=0)=(1-p)n • Can write λ =nP, hence P(X=0)=(1- λ/n)n • If n is large and P is small, this converges to e-λ,

the probability of 0 successes in a Poisson distribution!

• Can show that this also applies for other probabilities. Hence, Poisson approximates Binomial when n is large and P is small (n>5, P<0.05).

Bivariate distributions

• If X and Y is a pair of discrete random variables, their joint probability function expresses the probability that they simultaneously take specific values:– – marginal probability: – conditional probability: – X and Y are independent if for all x and y:

( , ) ( )P x y P X x Y y ( ) ( , )

y

P x P x y( , )

( | )( )

P x yP x y

P y

( , ) ( ) ( )P x y P x P y

Example

• The probabilities for – A: Rain tomorrow

– B: Wind tomorrow

are given in the following table:

0.1 0.2 0.05 0.01

0.05 0.1 0.15 0.04

0.05 0.1 0.1 0.05

No rain

Light rain

Heavy rain

No wind Some wind Strong wind Storm

Example cont’d:• Marginal probability of no rain: 0.1+0.2+0.05+0.01=0.36

• Similarily, marg. prob. of light and heavy rain: 0.34 and 0.3. Hence marginal dist. of rain is a PDF!

• Conditional probability of no rain given storm: 0.01/(0.01+0.04+0.05)=0.1

• Similarily, cond. prob. of light and heavy rain given storm: 0.4 and 0.5. Hence conditional dist. of rain given storm is a

PDF!• Are rain and wind independent? Marg. prob. of no wind:

0.1+0.05+0.05=0.2

P(no rain,no wind)=0.36*0.2=0.072≠0.1

Covariance and correlation

• Covariance measures how two variables vary together:

• Correlation is always between -1 and 1:

• If X,Y independent, then• If X,Y independent, then• If Cov(X,Y)=0 then

( , ) ( ( ))( ( )) ( ) ( ) ( )Cov X Y E X E X Y E Y E XY E X E Y

( , ) ( , )( , )

( ) ( )X Y

Cov X Y Cov X YCorr X Y

Var X Var Y

( ) ( ) ( )E XY E X E Y( , ) 0Cov X Y

( ) ( ) ( )Var X Y Var X Var Y

Continuous random variables

• Used when the outcomes can take any number (with decimals) on a scale

• Probabilities are assigned to intervals of numbers; individual numbers generally have probability zero

• Area under a curve: Integrals

Cdf for continuous random variables

• As before, the cumulative distribution function F(x) is equal to the probability of all outcomes less than or equal to x.

• Thus we get • The probability density function is however

now defined so that

• We get that

( ) ( ) ( )P a X b F b F a

( ) ( )b

a

P a X b f x dx 0

0( ) ( )x

F x f x dx

Expected values

• The expectation of a continuous random variable X is defined as

• The variance, standard deviation, covariance, and correlation are defined exactly as before, in terms of the expectation, and thus have the same properties

( ) ( )E X xf x dx

Example: The uniform distribution on the interval [0,1]

• f(x)=1

• F(x)=x

•

•

1 1121 1

2 200 0

( ) ( )E X xf x dx xdx x

2 2

122 1 1 1

3 4 120

( ) ( ) ( )

( ) 0.5

Var X E X E X

x d x

The normal distribution

• The most used continuous probability distribution: – Many observations tend to approximately

follow this distribution– It is easy and nice to do computations with– BUT: Using it can result in wrong conclusions

when it is not appropriate

Histogram of weight with normal curve displayed

Weight (kg)

95.090.085.080.075.070.065.060.055.050.045.040.0

Distribution of weight among 95 students

Nu

mb

er o

f stu

de

nts

25

20

15

10

5

0

The normal distribution• The probability density function is

• where

• Notation

• Standard normal distribution

• Using the normal density is often OK unless the actual distribution is very skewed

• Also: µ±σ covers ca 65% of the distribution

• µ±2σ covers ca 95% of the distribution

2 2( ) / 2

2

1( )

2

xf x e

( )E X 2( )Var X 2( , )N

(0,1)N

The normal distribution with small and large standard deviation σ

x 2018161412108642

0.4

0.3

0.2

0.1

0

Simple method for checking if data are well approximated by a normal

distribution: Explore

• As before, choose Analyze->Descriptive Statistics->Explore in SPSS.

• Move the variable to Dependent List (e.g. weight).

• Under Plots, check Normality Plots with tests.

Histogram of lung function for the students

Average PEF value measured in a sitting position

800

750

700

650

600

550

500

450

400

350

300

Nu

mb

er o

f stu

de

nts

20

16

12

8

4

0

Std. Dev = 120.12

Mean = 503

N = 95.00

Q-Q plot for lung function

Normal Q-Q Plot of PEFSITTM

Observed Value

800700600500400300200

Exp

ecte

d N

orm

al

3

2

1

0

-1

-2

-3

Age – not normal

Age

35.032.530.027.525.022.520.0

Histogram

Fre

qu

en

cy

50

40

30

20

10

0

Std. Dev = 3.11

Mean = 22.4

N = 95.00

Q-Q plot of age

Normal Q-Q Plot of AGE

Observed Value

40302010

Expe

cte

d N

orm

al

3

2

1

0

-1

-2

SKEWED

40

30

20

10

0

Std. Dev = 1.71

Mean = 1.50

N = 106.00

Skewed distribution, with e.g. the observations 0.40, 0.96, 11.0

A trick for data that are skewed to the right: Log-transformation!

Log-transformed data

LNSKEWD

14

12

10

8

6

4

2

0

Std. Dev = 1.05

Mean = -.12

N = 106.00

ln(0.40)=-0.91ln(0.96)=-0.04ln(11) =2.40

Do the analysis on log-transformed data

SPSS: transform- compute

OK, the data follows a normal distribution, so what?

• First lecture, pairs of terms:– Sample – population

– Histogram – distribution

– Mean – Expected value

• In statistics we would like the results from analyzing a small sample to apply for the population

• Has to collect a sample that is representative w.r.t. age, gender, home place etc.

New way of reading tables and histograms:

• Histograms show that data can be described by a normal distribution

• Want to conclude that data in the population are normally distributed

• Mean calculated from the sample is an estimate of the expected value µ of the population normal distribution

• Standard deviation in the sample is an estimate of σ in the population normal distribution

• Mean±2*(standard deviation) as estimated from the sample (hopefully) covers 95% of the population normal distribution

In addition:

• Most standard methods for analyzing continuous data assumes a normal distribution.

• When n is large and P is not too close to 0 or 1, the Binomial distribution can be approximated by the normal distribution

• A similar phenomenon is true for the Poisson distribution

• This is a phenomenon that happens for all distributions that can be seen as a sum of independent observations.

• Means that the normal distribution appears whenever you want to do statistics

The Exponential distribution

• The exponential distribution is a distribution for positive numbers (parameter λ):

• It can be used to model the time until an event, when events arrive randomly at a constant rate

( ) tf t e

( ) 1/E T 2( ) 1/Var T

Next time:

• Sampling and estimation

• Will talk much more in depth about the topics mentioned in the last few slides today