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Lesson 0: Preliminaries

Le Thi Xuan Mai

The university of natural sciences

February 24, 2013

T.X.M. Le Bayesian statistics
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Discrete random variables

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The binomial distribution

As its name suggests, the binomial distribution refers to randomvariables with two outcomes.

Example:

smoking status: a person does or does not smoke

healthy insurance coverage: a person does or does not havehealth insurance.

We now consider the first example to demonstrate the calculationof binomial probabilities.

Suppose that four adults have been randomly selected and askedwhether or not they currently smoke. The random variable ofinterest in this example is the number of persons who respondyes to the question about smoking (denoted by X).

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Suppose that in the population, the proportion of people who

would respond yes to this question is , and the probability of aresponse of no is then 1 . Since each person is independentof all the other persons, the binomial probability can be calculatedas follows

P(X = 4) = 4 (the number of 4 responses is 4)

P(X = 3) = 43(1 ) (since there are four occurrences ofthree yes responses)

P(X = 2) = 62(1 )2

P(X = 1) = 4(1 )3

P(X = 0) = (1 )4


C di ib i
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Suppose is 0.25. Then the probability of each outcome is as

follows

P(X = 4) = C04 (0.25)4 (0.75)0 = 0.0039 = P{0 no responses}

P(X = 3) = C14

(0.25)3

(0.75)1 = 0.0469 = P

{1 no response

}P(X = 2) = C24 (0.25)2 (0.75)2 = 0.2109 = P{2 no response}P(X = 1) = C34 (0.25)1 (0.75)3 = 0.4219 = P{3 no response}P(X = 0) = C44 (0.25)0 (0.75)4 = 0.3164 = P{4 no responses}This lead us to define a binomial random variable and calculate itsprobability distribution.


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A random variable Y {0, 1, . . . , n} has a binomial (n, )distribution if [0, 1] and

P(Y = y|, n) = Cyn y(1 )ny for y {0, 1, . . . , n}.For this distribution,

E[Y|] = n;Var[Y|] = n(1 );mode[Y|] = [(n + 1)];p(y|, n) = dbinom(y, n, ).

IfY1 binomial(n1, ) and Y2 binomial(n2, ) areindependent, then Y = Y1 + Y2 binomial(n1 + n2, ). When n = 1 this distribution is called the Bernoulli distribution.

The binomial(n, ) model assume that Y is a sum ofindependent binary() random variables.


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Figure : Binomial distribution with n = 10 and = 0.2.


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The negative binomial distribution

A random variable Y {0, 1, . . .} has a negative binomial(r, )distribution for any positive integer r if 0 1 and

P(y|r, ) = Cyy+r1y(1 )r.

For this distribution,

E[Y|r, ] = r/(1 );Var[Y|r, ] = r/(1 )2;mode[Y

|r, ] = [(1

)(r

1)/] if r > 1, mode[Y

|r, ] = 0 ifr

1

p(y|) = dnbinom(y, ).


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Figure : Negative binomial distribution.


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The name negative binomial can be replaced by inversebinomial because it arises from a series of Bernoulli trials wherethe number of successes is fixed at the outset. In this case, Y isthe number of failure preceding the rth success.

This distribution is sometimes used to model countably infiniterandom variables having variance substantially larger than themean (so that the Poisson model would be inappropriate).

The geometric is the special case of the negative binomialhaving r = 1. It is the nuber of failures preceding the first successin a series of Bernoulli trials.


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The Poisson distribution

A random variable Y {0, 1, . . .} has a Poisson() distribution if > 0 and

P(Y = y|) = ye

y!for y {0, 1, . . .}.


E[Y|] = ;Var[Y|] = ;mode[Y|] = [];p(y|) = dpois(y, ).

IfY1 Poisson(1) and Y2 Poisson(2) are independent,then Y = Y1 + Y2 Poisson(1 + 2). If it is observed that a sample mean is very different than thesample variance, then the Poisson model may not be appropriate.


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Figure : Poisson distribution with = 1, 4, 10.


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Continuous random variables


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The univariate normal distribution

A random variable Y R has a normal(, 2) distribution if2 > 0 and

p(y|, 2) = 122

e12

(y)2/2 for < y < .


E[Y|, 2] = ;Var[Y|, 2] = 2;mode[Y|, 2] = ;p(y

|, 2) = dnorm(y, theta, sigma).

IfX1 normal(1, 21) and X2 normal(2, 22) areindependent, then

aX1 + bX2 + c

normal(a1 + b2 + c, a

221 + b222).


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Figure : Some univariate normal distributions.


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The t distribution

A random variable Y R has a t distribution() if > 0, and

p(y|) = [(+ 1)/2](/2)

1 + y

2

](+1)/2.


E[Y

|] = 0 if > 1;

Var[Y|] = /( 2) if > 2, Var[Y|] = if 1 < 2;mode = 0median = 0p(y|) = dt(y, nu).

The parameter is referred to as the degrees of freedom and isusually taken to be a positive integer, though the distribution isproper for any positive real number .

The t is a common heavy-tailed (but still symmetric and

unimodal) alternative to the normal distribution.T.X.M. Le Bayesian statistics

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Figure : Some t distributions.


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The continuous uniform probability distribution

If a continuous r.v Y can assume any value in the intervala

y

b and only these values, and if its probability density

function p(y|a, b) is constant over that interval and equal to zeroelsewhere, then Y is said to be uniformly distributed and itsdistribution is called a continuous uniform probability distributionor continuous rectangular distribution. Its density distribution is of

the formp(y|a, b) =

1

bafor a y b

0 else where(1)


E[Y] = a+b2

Var[Y] = 112 (b a)2median = a+b2

mode = any value in[a, b].T.X.M. Le Bayesian statistics

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Figure : Uniform distribution.


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h b d b
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The beta distribution

A random variable Y [0, 1] has a beta(a, b) distribution ifa > 0,b> 0 and

p(y|a, b) = (a + b)(a)(b)

ya1(1 y)b1 for 0 y 1.

E[Y|a, b] = a

a+b

;

Var[Y|a, b] = ab(a+b+1)(a+b)2 = E[Y] E[1 Y] 1a+b+1 ;mode[Y|a, b] = a1(a1)+(b1) ifa > 1 and b> 1;p(y

|a, b) = dbeta(y, a, b).

The beta distribution is closely related to the gammadistribution.

A multivariate version of the beta distribution is the Dirichletdistribution.


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Figure : Beta distribution.


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The gamma and inverse-gamma distributions

A random variable Y (0,) has a gamma(a, b) distribution if

p(y|a, b) = ba

(a)ya1eby for y > 0, a > 0, b> 0.

E[Y|a, b] = ab

; Var[Y|a, b] = ab2

;

mode[Y|a, b] = a1b

ifa 1, mode[Y|a, b] = 0 if 0 < a < 1;p(y|a, b) = dgamma(y, a, b). IfY1 gamma(a1, b) and Y2 gamma(a2, b) are independent,then Y1 + Y2 gamma(a1 + a2, b) and Y1/(Y1 + Y2) beta(a1, a2). IfY normal(0, 2) then Y2 gamma(1/2, 1/22). The chi-square distribution with degrees of freedom is thesame as a gamma(/2, 1/2) distribution.

IfY

normal(0, 1) then Y2

chi-square(1).


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Figure : Gamma distributions.


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A random variable Y (0,) has an inverse-gamma(a, b)distribution if 1/Y has a gamma(a, b) distribution. In other words,ifX gamma(a, b) and Y = 1/X, then

Y inverse-gamma(a, b). The density ofY isp(y|a, b) = b

a

(a)ya1eb/y for y > 0.


E[Y|a, b] = ba1 ifa 1,E[Y|a, b] = if 0 < a < 1;

Var[Y|a, b] = b2(a1)2(a2)

ifa 2, Var[Y|a, b] = if 0 < a < 2;mode[X|a, b] = b

a+1 .

Note that the inverse-gamma density is not simply the gammadensity with y replaced by 1/y.

The inverse-2() is a special case of the inverse-gammadistribution, with a = /2 and b= 1/2.


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The scaled in erse chisq are distrib tion
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The scaled inverse chisquare distribution

The family of scaled inverse chi-squared distributions is closelyrelated to two other distribution families: the inverse-chi-squared

distribution and the inverse gamma distribution. Compared to the inverse-chi-squared distribution, the scaleddistribution has an extra parameter 2, which scales thedistribution horizontally and vertically. The relationship between

the inverse-chi-squared distribution the scaled distribution is that if

X Scale-inv-2(, 2) then X2

inv-2()The probability density function of the scaled inverse chi-squareddistribution extends over the domain x> 0 and is

f(x, , 2) =(2/2)/2

(/2)x/21 exp(

2

2x)

where is the degrees of freedom parameter and 2 is the scale

parameter. T.X.M. Le Bayesian statistics

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EX =2

2 for > 2Var(X) =

224

( 2)2( 4) for > 4

mode(X) =2

+ 2 IfX Scale-inv-2(, 2) then kX Scale-inv-2(, k2)

IfX Scale-inv-2(, 2) then X2

inv-2()

IfX Scale-inv-2(, 2) then X Inv-Gamma

2

, 2

2


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Exponential distributions
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Exponential distributions

A random variable Y > 0 has an exponential distribution() if > 0 and

p(y|) = exp(y).For this distribution,

E[Y|] = 1/;Var[Y|] = 1/2;mode[Y|] = 0;

p(y|) = dexp(y, lambda).The exponential distribution is a Gamma(1, ) distribution.


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Figure : Exponential distributions.


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The double exponential distribution (Laplace distribution)
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The double exponential distribution (Laplace distribution)

A random variable Y R has a Laplace distribution(, 2) if

< 0 and

p(y|, 2) = 12

exp(|y |

).


E[Y|, 2] = ;Var[Y|, 2] = 22;mode[Y|, 2] = ;p(y|, 2) = dlaplace(y, mu, sigma).

This distribution is symmetric and unimodal, but has heavier tailsand a somewhat different shape, being strictly concave up on bothsides of .


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Figure : Laplace distributions.


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The logistic distribution
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The logistic distribution

A random variable Y R has a Logistic distribution(, 2) if

< 0 and

p(y|, 2) = exp(y )

[1 + exp( y )]2.


E[Y|, 2] = ;Var[Y|, 2] = (2/3)2;mode[Y

|, 2] = ;

p(y|, 2) = dlogis(y, mu, sigma).

The logistic is another symmetric unimodal distribution, moresimilar to the normal in appearance than the double exponential,

but with even heavier tails.T.X.M. Le Bayesian statistics

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Figure : Some logistic distributions.


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Cauchy distribution
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Cauchy distribution

A random variable Y R has a Cauchy distribution (, 2) if > 0, < < , 2 > 0 and

p(y

|, 2) =

1

[1 + (x

)2

]

.

For this distribution, E[Y] and Var[Y] do not exist, though isthe median of this distribution.

This is a special case of the t distribution t(1, , 2) which has theheaviest possible tails.


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Figure : Some Cauchy distributions.


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The multivariate normal distribution
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A random vector Y Rp

has a multivariate normal distributionnormal(, ) distribution if is a positive definite p pmatrix and

p(y|, ) = (2)p/2||1/2 exp{12

(y)T(y)} for y Rp.


E[Y|, ] = ;Var[Y|, ] = ;mode[Y|, ] = .


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IfY1 normal(1, 1) and Y2 normal(2, 2) areindependent, then

aY1 + bY2 + c normal(a1 + b2 + c, a21 + b22).

IfZ is the vector with elements Z1, . . . ,Zp i.i.d. normal(0, 1)then

Y = + AZ

multivariate normal(, ),

where and AAT = (A is the Choleski factorization of ).

The following R-code will generate an n pmatrix such thatthe rows are i.i.d samples from a multivariate normal distribution:

Z = matrix(rnorm(n p), nrow = n, ncol = p)Y = t(t(Z% %chol(Sigma)) + c(theta)).


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The Wishart and inverse-Wishart disttributions
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A random p p symmetric positive definite matrix Y has aWishart(,M) distribution if the integer

p, M is a p

p

symmetric positive definite matrix and

p(Y|,M) = [2p/2p(/2)|M|/2]1|Y|(p1)/2 exp{tr(M1Y/2)}where

p(/2) = p(p

1)/4

pj=1

[(+ 1 j)/2], and

exp{tr(A)} = exp(

aj,j).


E[Y|,M] = M;Var[Yi,j|,M] = (m2i,j + mi,imj,j);mode[Y|,M] = ( p 1)M.


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The Wishart distribution is a multivariate version of the gammadistribution.

IfX1, . . . ,X i.i.d. multivariate normal(0,Mpp), then

i=1

XiXTi Wishart (,Mpp).

The following R-code is used to generate a Wishart distributedrandom matrix:

X = matrix(rnorm( p),nrow=nu,ncol=p) #standard normal matZ = X%

%chol(M) #

Y = t(Z)% %Z #Wishart matrix


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A random p p symmetric positive definite matrix Y has an
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inverse-Wishart(,M) distribution ifX1 has a Wishart(,M)distribution. In other words, ifX Wishart(,M) and X = Y1,then Y

inverse-Wishart(,M). The density ofY is

p(Y|,M) =[2p/2p(/2)|M|/2]1 |Y|(+p+1)/2 exp{tr(M1Y1/2)}.


E[Y|,M] = ( p 1)1M1;mode[Y|,M] = (+ p+ 1)1M1.

If inverse-Wishart(, S1

), we havemode[Y|, S] = (+ p+ 1)1S. If 0 were the most probable value of a priori, then we couldset S= (0 + p+ 1)0, so that

inverse-Wishart(, [(+ p

1)0]

1) and mode[

|, S] = 0.

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