Decision making as a model 6. A little bit of Bayesian statistics.

Decision making as a model

6. A little bit of Bayesian statistics

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Helling: .657, sds = 1.52 sdn

Intercept .831

Az = .756

Opdracht 5.1

Opdracht 5.2

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slope .643

Intercept 1.267

Az = .857

A B C D E

Prev. .75 .60 .25 .10 .06

d' 1.28 1.50 1.77 1.90 1.85

c -.44 -.19 .36 .70 .93

β .57 .76 1.89 3.75 5.57

βuneq .37 .49 1.22 2.41 3.58

LRopt .15 .31 1.37 4.11 7.16

N.B. a course on Bayesians statistics is impossible in one lecture

You might study:

-Bolstad, W.M. (2007). Introduction to Bayesian statistics, 2nd ed. Hoboken N.J. : Wiley.

-Gill, J. (2002). Bayesian methods: A social and behavioral approach. Boca Raton Fl: Chapman & Hall.

Or consult Herbert Hoijtink Irene Klugkist (M&T)

Classical statistics vs Bayesian statistics

Probability: limit of long run relative frequency

Fixed unknown parameters (like θ)

Inference based on likelihood: p(data|θ)

Probability: confidence/strength of belief based on all available prior and actual evidence

θ Stochastic

Inference based on likelihood: p(data|θ) and prior: p(θ)

Confidence Interval Credible Interval p(….≤ x̅� ≤….|μ = x̅� ) ≠ p(….≤ μ ≤….| x̅� )

Given data D, what can be said about the probability of posible values of some unknown quantity θ ?

p(D|θi)•p(θi) p(θi|D) = ---------------------

Σj p(D|θj)•p(θj)

Continous distributions:

NB.: θ is supposed to be a random variable!

L(θ|D)•pdf(θ) pdf(θ|D) = ∞------------------------ L(θ|D)•pdf(θ) dθ

-∞

∫

Bayesian statistics

From p(Y|X) to continous function:

pdf(y|X) pdf of y for some X

L(x|Y) likelihood function of xfor some Y

(normalization constant)

0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

Probability of “Head”

Three possibilities:Three priors

A fair coin?

We will throw the coin several times

beta(20,20)

beta(1,1)

beta(.5,.5)

Beta(a,b):

K∙xa-1∙(1-x)b-1

Γ(a+b)K = ----------- , 0 ≤ x ≤ 1 Γ(a)Γ(b)

∞Γ(a) = e-y ya-1dy (a>0) 0∫Γ(a) = (a-1)! (a integer)

N HR(1-H)N-R

R

N HR(1-H)N-R

R

N HR(1-H)N-R

R

K1∙H19∙(1-H)19

∙

K2∙H0∙(1-H)0 ∙

K3∙H-.5∙(1-H)-.5 ∙

Prior (beta)Likelihood(binomial)

1x head: N=R=1; Posterior(beta)

= K1∙H20∙(1-H)19

= K2∙H1∙(1-H)0

= K3∙H.5∙(1-H)-.5

L=H

beta(20,20)

beta(.5,.5)

beta(1,1)

0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

0 20 40 60 80 100

0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

0 20 40 60 80 100

1: head

2. tail

Likelihoodfunction

posterior

1: head

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0 20 40 60 80 100

0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

0 20 40 60 80 100

3: tail

4: head

2. tail

1: head

3: tail

2. tail

1: head

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0 20 40 60 80 100

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0 20 40 60 80 100

5: head

10x (6 heads)

4: head

3: tail

2. tail

1: head

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0 20 40 60 80 100

0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

0 20 40 60 80 100

20x (9 heads)

40x (18 heads)

0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 10 20 40 60 80 100

0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

0 20 40 60 80 100

80x (38 heads)

160x (69 heads)

Is that fair?

0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

0 20 40 60 80 100

0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

0 20 40 60 80 100

320x (120 heads)

640x (249 heads)

obviously not!

With lots of data the prior distribution does not make much difference anymore (although narrow priors retain their influence longer)

With fewer data the prior distribution does make a lot of difference, so you have to have good reasons for your prior distribution

This was an untypical case: the blue line prior was quite reasonable for coin-like objects, but this “coin” was a computer simulation (with H=.40)

Narrow priors distributions have strong influence on the posterior distribution

Essential: Suitable likelihoods and priors

Priors must be well founded or reasonable, especially when informative (small sd)

Likelihood function must be good model of data (or data producing process)

Priors and likelihood functions preferably conjugate (of same family, so that product is tractible)

Simple case: infer μ from observation D (from normal distribution known variance σD)

pdf(μ|D) L(μ|D) • pdf(μ) (normal prior: mp,σp)

(or something like: g(μ|D) f(D|μ) • g(μ) )

Lengthy derivation results in normal distribution with:

(D – μ)2 (μ – mp)2

- ---------- - ---------- 1 2σD

2 1 2σp2

--------- e • -------- e σD√2π σp√ 2π

mp D ---- + ----σp

2 σD2 1

mean: -------------- and variance: ------------ 1 1 1 1

---- + ---- ---- + ---- σp

2 σD2 σp

2 σD

2

mp D ---- + ----σp

2 σD2 1

mean: -------------- and variance: ------------ 1 1 1 1

---- + ---- ---- + ---- σp

2 σD2 σp

2 σD

2

Add fractions, multiply numerator and denominator by σp2σD

2 :

σD2mp + σp

2D σp2σD

2 mean: ------------------- and variance: --------------

σD2+ σp

2 σD2 + σp

2

If there are n independent observations D1, D2, D3……Dn

mp nmD ---- + ----σp

2 σD2 1

mean: --------------- and variance: ------------ 1 n 1 n ---- + ---- ---- + ---- σp

2 σD2 σp

2 σD2

Add fractions, multiply numerator and denominator by σp2σD

2 :

σD 2mp + σp

2 nmD σp

2σD2

mean: ----------------------- and variance: --------------- σD

2 + nσp

2 σD2 + nσp

2

mp nmD ---- + ----σp

2 σD2 1

mean: --------------- and variance: --------------- 1 n 1 n

---- + ---- ---- + ---- σp

2 σD2 σp

2 σD

2

-Weighing by inverse of variances -Small variance prior weighs heavily-Large n swamps prior -More data: posterior variance decreases

Example ofBayesian statistics at UU(approach Hoijtink c.s.)

Bayesian “AN(C)OVA”In stead of: “is there a significant difference between groups – and if so which? ”

“how much support from data to specific informative models?”

Global description! For more information see:

Klugkist,I., Laudy,O. & Hoijtink, H. (2005). Inequality constrained analysis of variance: A Bayesian approach. Psychological Methods, 10, 477-493.

p(A|B) p(B|A) p(A) ------------- = ------------- • -------- p(¬A|B) p(B|¬A) p(¬A)

posterior odds = likelihood ratio • prior odds (Bayes Factor)

Model selection according to Bayes Factor:

remember?

BF = posterior odds / prior odds

Bayes Factor in general: extent to which data support one model better/worse than other model:

p(D|M1)BF12 = -----------

p(D|M2)

Example 1: four groups.three models (constraints on values μ’s): M1: ( μ1,μ2 ) > ( μ3,μ4 )

M2: μ1 < μ2 < μ3 , μ4 M3: μ1 < μ2 , μ3 ≈ μ4

against encompassing model without constraints:M0: μ1, μ2 , μ3, μ4

For explanatory purpose simpler example 2: two groups:M1: μ1 > μ2

M2: μ1 ≈ μ2

M0: μ1 , μ2

μ2

μ1

μ2

μ1

μ2

μ1

Compute posterior (≈likelihood) for encompassing model (M0)-given data

specify diffuse prior for M0

For every model estimate proportion of encompassing prior (1/c) and of posterior (1/d) satisfying constrictions implied by that model. Works with simulated sampling.

For M0 : 1/c0 = 1/d0 = 1

For M1 : 1/c1 = .5 (viewed from above:)μ2

μ1

(e.g.) 1/d1 = .99

μ2

μ1

For M2 : 1/c2 = .02 and 1/d2 = .003

Select model by Bayes Factors:

In general: p(D|M)∙p(M) p(M|D)∙p(D)p(M|D) = ----------------- p(D|M) = ----------------

p(D) p(M)

p(M1|D)∙p(D) p(M1|D) 1 ------------------ ---------- ------

p(D|M1) p(M1) p(M0|D) d1----------- = ------------------------ = ----------------- = -------------p(D|M0) p(M0|D)∙p(D) p(M1) 1

----------------- ------- ------ p(M0) p(M0) c1

1/dmBFm0 = ----------

1/cm

Bayes factor takes complexity (size of parameter space) in to account of model by denominator 1/c

Our example:

BF10= .99/.5 = 1.98

BF20= .003/.02 = .15

M1 is clearly superior.

PMP1 = 1.98/(1.98 +.15 + 1) = .63

PMP2 = .15/(1.98 + .15 + 1) = .05

PMP0 = 1 /(1.98 + .15 + 1) = .32

Example 1 (four groups) would require four-dimensional drawings,

but can be computed by means of software by Hoijtink &co. (2- 8 groups and 0-2 covariates!)

Decision making as a model 6. A little bit of Bayesian statistics.

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