presentation slides for my simulation course at dauphine

Simulation: a ubiquitous tool for statistical computation

Statistique exploratoire (NOISE)

I introduction to computer language R

I rudiments of simulation (aka Monte Carlo methods)

I illustration by the boostrap method


Organisation

I weekly computer labs using R and Rstudio

I one computer account per student

I midterm and final exams on anonymous account

I exam made of programming problems (e.g., 3 out of 6)

I exam evaluation based on executable R code

I handouts and on-line help available on anonymous account

I no other document


Reference

I documents onhttps://www.ceremade.dauphine.fr/~xian/Noise.html

I R. Drouihlet, P. Lafaye de Micheaux & B. Liquet (2010) Lelogiciel R Springer, Paris

I C. Robert & G. Casella (2007) Introduction to Monte CarloMethods with R Springer, New York

I Manuals on http://cran.r-project.org/manuals.html

(CRAN is the Comprehensive R Archive Network)


Simulation: a ubiquitous tool for statisticalcomputation

Christian P. Robert

Universite Paris-Dauphinehttp://www.ceremade.dauphine.fr/~xian

September 26, 2013

http://www.ceremade.dauphine.fr/~xian


Outline

Simulation, what’s that for?!

Producing randomness by deterministic means

Monte Carlo principles

Simulated annealing



Illustrations

Necessity to “(re)produce chance” on a computer

I Evaluation of the behaviour of a complex system (network,computer program, queue, particle system, atmosphere,epidemics, economic actions, &tc)

[ c© Office of Oceanic and Atmospheric Research]



Illustrations


I Production of changing landscapes, characters, behaviours incomputer games and flight simulators

[ c© guides.ign.com]



Illustrations


I Determine probabilistic properties of a new statisticalprocedure or under an unknown distribution [bootstrap]

(left) Estimation of the cdf F from a normal sample of 100 points;

(right) variation of this estimation over 200 normal samples



Illustrations


I Validation of a probabilistic model

Histogram of 103 variates from a distribution and fit by this distribution density



Illustrations


I Approximation of a integral

[ c© my daughter’s math book]



Illustrations


I Maximisation of a weakly regular function/likelihood

[ c© Dan Rice Sudoku blog]



Illustrations


I Pricing of a complex financial product (exotic options)

Simulation of a Garch(1,1) process and of its volatility (103 time units)



Illustrations


I Handling complex statistical problems by approximateBayesian computation (ABC)

core principle

I Simulate a parameter value (at random) and pseudo-data from thelikelihood until the pseudo-data is “close enough” to the observeddata, then

I keep the corresponding parameter value

[Griffith & al., 1997; Tavare & al., 1999]



Illustrations



demo-genetic inference

Genetic model of evolution from acommon ancestor (MRCA)characterized by a set of parametersthat cover historical, demographic, andgenetic factorsDataset of polymorphism (DNA sample)observed at the present time

97

!""#$%&'()*+,(-*.&(/+0$'"1)()&$/+2!,03 !1/+*%*'"4*+56(""4&7()&$/.+.1#+4*.+8-9':*.+

Différents scénarios possibles, choix de scenario par ABC

Le scenario 1a est largement soutenu par rapport aux

autres ! plaide pour une origine commune des

populations pygmées d’Afrique de l’Ouest Verdu et al. 2009



Illustrations



Pygmies population demo-genetics

Pygmies populations: do theyhave a common origin? whenand how did they split fromnon-pygmies populations? werethere more recent interactionsbetween pygmies andnon-pygmies populations?

94

!""#$%&'()*+,(-*.&(/+0$'"1)()&$/+2!,03 !1/+*%*'"4*+56(""4&7()&$/.+.1#+4*.+8-9':*.+



Pseudo-random generator

Pivotal element/building block of simulation: always requiresavailability of uniform U (0, 1) random variables

[ c© MMP World]





0.13331390.30262990.43429660.23953570.32237230.85311620.39214570.76252590.17019470.2816627...





Definition (Pseudo-random generator)

A pseudo-random generator is a deterministic function f from ]0, 1[to ]0, 1[ such that, for any starting value u0 and any n, thesequence

{u0, f(u0), f(f(u0)), . . . , fn(u0)}

behaves (statistically) like an iid U (0, 1) sequence





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0.0 0.2 0.4 0.6 0.8

0.0

0.2

0.4

0.6

0.8

x t+1

10 steps (ut , ut+1) of a uniform generator



Philosophical foray

¡Paradox!

While avoiding randomness, the deterministic sequence

(u0, u1 = f(u0), . . . , un = f(un−1))

must resemble a random sequence!

Debate on whether or not true

randomness does exist (Laplace’s

demon versus Schroedinger’s

cat), in which case pseudo

random generators are not

random (von Neuman’s state of

sin)



True random generators

Intel circuit producing “truly random” numbers:There is no reason physical generators should be“more” random than congruential (deterministic)pseudo-random generators, as those are validgenerators, i.e. their distribution is exactly known(e.g., uniform) and, in the case of parallelgenerations, completely independent



True random generators

Intel generator satisfies all benchmarks of“randomness” maintained by NIST:Skepticism about physical devices, when comparedwith mathematical functions, because of (a)non-reproducibility and (b) instability of the device,which means that proven uniformity at time t doesnot induce uniformity at time t + 1



A standard uniform generator

The congruencial generator on {1, 2, . . . ,M}

f(x) = (ax + b) mod (M)

has a period equal to M for proper choices of (a, b) and becomes agenerator on ]0, 1[ when dividing by M + 1



A standard uniform generator

The congruencial generator on {1, 2, . . . ,M}

f(x) = (ax + b) mod (M)


Example

Takef(x) = (69069069x + 12345) mod (232)

and produce... 518974515 2498053016 1113825472 1109377984 ...i.e.... 0.1208332 0.5816233 0.2593327 0.2582972 ...



A standard uniform generatorThe congruencial generator on {1, 2, . . . ,M}

f(x) = (ax + b) mod (M)




Approximating π

My daughter’s pseudo-code:

N=1000π = 0for I=1,N do

X=RDN(1), Y=RDN(1)if X2 + Y2 < 1 thenπ = π + 1

end ifend forreturn 4*π/N



Approximating π


N=1000π = 0for I=1,N do



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100 simulations



Approximating π


N=1000π = 0for I=1,N do



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pi = 3.108

1000 simulations



Approximating π


N=1000π = 0for I=1,N do



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pi = 3.136

10,000 simulations



Approximating π


N=1000π = 0for I=1,N do



106 simulations



Distributions that differ from uniform distributions

Proble

Given a probability distribution withdensity f , how can we producerandomness according to f ?!

I implemented algorithms in aresident software only available forcommon distributions

I new distributions may require fastresolution

I no approximation allowed

xf (

x)

an arbitrary density



The Accept-Reject Algorithm (remember π?!)

Given a probability distribution with density f , how can we producerandomness according to f ?!

The uniform distribution on thesub-graph region

Sf = {(x , u); 0 ≤ u ≤ f (x)}

produces a marginal distributionin x with density f .(”Fundamental theorem ofsimulation”)

x

f (x)



The Accept-Reject Algorithm (remember π?!)

Given a probability distribution with density f , how can we producerandomness according to f ?!

In practice, this means we areback to throwing dots on a boxcontaining

Sf = {(x , u); 0 ≤ u ≤ f (x)}

and counting those inside!

x

f (x)

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The Accept-Reject Algorithm

Refinement of the above through construction of a proper box:

1. Find a hat on f , i.e. a density g that can be simulated andsuch that

supx

f (x)/g(x) = M <∞

2. Generate dots on the subgraph of g , i.e. Y1,Y2, . . . ∼ g , andU1,U2, . . . ∼ U ([0, 1])

3. Accept only the Yk ’s such that

Uk ≤ f (Yk)/Mg(Yk)



The Accept-Reject AlgorithmRefinement of the above through construction of a proper box:

x

f (x)

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Properties

I Does not require normalisation constant of f

I Does not require an exact upper bound M

I Requires on average M Yk ’s for one simulated X (efficiencymeasure)



Slice sampler

There exists other ways of exploitating the fundamental idea ofsimulation over the subgraph:



Slice sampler


If direct uniform simulation on

Sf = {(u, x); 0 ≤ u ≤ f (x)}

is too complex [because of unavailable hat] use instead a randomwalk on Sf :



Slice sampler


this means doing random jumps in vertical then horizontaldirection, accounting for the boundaries

I 0 ≤ u ≤ f (x), i.e. U(0, 1)

I f (x) ≥ u, i.e. x ∼ US(u)



Slice sampler

Slice sampler algorithm

For t = 1, . . . ,Twhen at (x (t), ω(t)) simulate

1. ω(t+1) ∼ U[0,f (x(t))]

2. x (t+1) ∼ US(t+1) , where

S(t+1) = {y ; f (y) ≥ ω(t+1)}.



Slice sampler

Slice sampler algorithm

For t = 1, . . . ,Twhen at (x (t), ω(t)) simulate

1. ω(t+1) ∼ U[0,f (x(t))]

2. x (t+1) ∼ US(t+1) , where

S(t+1) = {y ; f (y) ≥ ω(t+1)}.

0.0 0.2 0.4 0.6 0.8 1.0

0.00

00.

002

0.00

40.

006

0.00

80.

010

x

y



The Metropolis–Hastings algorithm

Further generalisation of the fundamental idea to situations whenthe slice sampler cannot be easily implemented





Idea

Create a sequence (Xn) such that, for n ‘large enough’, the densityof Xn is close to f , only using a ‘local’ knowledge of f ...

This is the domain of Markoviansimulation methods: there is a Markovdependence between the Xn’s

Andreı Markov





Idea

Create a sequence (Xn) such that, for n ‘large enough’, the densityof Xn is close to f , only using a ‘local’ knowledge of f ...

This is the domain of Markoviansimulation methods: there is a Markovdependence between the Xn’s

Markov bar, Melbourne




Further generalisation of the fundamental idea to situations whenthe slice sampler cannot be easily implemented‘local’ exploration can produce ‘global’ vision:

[ c© Civilization 5]



The Metropolis–Hastings algorithm (2)

If f is the density of interest, we pick a proposal conditional density

q(y |xn)

such that

I it connects with the current value xnI it is easy to simulate

I it is positive everywhere f is positive



Random walk Metropolis–Hastings

ProposalYt = Xn + εn,

where εn ∼ g , independent from Xn, and g symmetrical




ProposalYt = Xn + εn,


Motivation

local perturbation of Xn / myopic exploration of its neighbourhood

Yn accepted or rejected depending on the relative values of f (Xn)and f (Yn)



Random walk Metropolis–HastingsProposal

Yt = Xn + εn,





Resulting algorithm


Starting from X (t) = x (t)

1. Generate Yt ∼ g(y − x (t))

2. Take

X (t+1) =

Yt with proba. min

{1,

f (Yt)

f (x (t))

},

x (t) otherwise



Properties

I Always accepts higher point and sometimes lower points(similar to gradient algorithms)

I Depends on the dispersion of g

I Average probability of acceptance

% =

∫ ∫min{f (x), f (y)}g(y − x) dxdy

I close to 1 if g has a small variance [Danger!]I far from 1 if g has a large variance [Re-Danger!]



Properties

scale=1



Properties

scale=2



Properties

scale=3



General purpose of Monte Carlo methods

Given a probability density f known up to a normalizing constant,f (x) ∝ f (x), and an integrable function h, compute

I(h) =

∫h(x)f (x)dx =

∫h(x)f (x)dx∫f (x)dx

when∫h(x)f (x)dx is intractable.

[Remember π!!]



Monte Carlo basics

Generate a pseudo-randomsample x1, . . . , xN from f andestimate I(h) by

IMCN (h) = N−1

N∑i=1

h(xi ).

and let N grow to infinity...



Monte Carlo basics


IMCN (h) = N−1

N∑i=1

h(xi ).


A. Doucet, C. Andrieu, X, A. Philippe, J. Rosenthal, E. Moulines



Monte Carlo basics


IMCN (h) = N−1

N∑i=1

h(xi ).




Monte Carlo basics


IMCN (h) = N−1

N∑i=1

h(xi ).


Caveat

Often impossible or inefficient to simulate directly from f



Importance Sampling

For proposal distribution with density q(x), alternativerepresentations

I(h) =

∫h(x){f /q}(x)q(x)dx

Principle

Generate an iid sample x1, . . . , xN ∼ q and estimate I(h) by

I ISq,N(h) = N−1N∑i=1

h(xi ){f /q}(xi ).


Simulated annealing

Optimisation problems

A (genuine) puzzle

During a dinner with 20 couples sitting at four tables with tenseats, everyone wants to share a table with everyone. Theassembly decides to switch seats after each serving towards thisgoal. What is the minimal number of servings needed to ensurethat every couple shared a table with every other couple at somepoint? And what is the optimal switching strategy?

[http://xianblog.wordpress.com/2012/04/12/not-le-monde-puzzle/]

solution to the puzzle


Simulated annealing

Stochastic minimisation

Example (Egg-box function)

Consider the function

h(x , y) = (x sin(20y) + y sin(20x))2 cosh(sin(10x)x)

+ (x cos(10y)− y sin(10x))2 cosh(cos(20y)y) ,

to be minimised.


Simulated annealing


Example (Egg-box function)

Consider the function

h(x , y) = (x sin(20y) + y sin(20x))2 cosh(sin(10x)x)

+ (x cos(10y)− y sin(10x))2 cosh(cos(20y)y) ,

to be minimised. (I knowthat the global minimum is 0for (x , y) = (0, 0).)


Simulated annealing


Example (egg-box function (2))

Instead of solving the first order equations

∂h(x , y)

∂x= 0 ,

∂h(x , y)

∂y= 0

and of checking that the second order conditions are met, we cangenerate a random sequence in R2

θj+1 = θj +αj

2βj∆h(θj , βjζj) ζj


Simulated annealing


Example (egg-box function (2))

we can generate a random sequence in R2

θj+1 = θj +αj

2βj∆h(θj , βjζj) ζj

where

� the ζj ’s are uniform on the unit circle x2 + y2 = 1;

� ∆h(θ, ζ) = h(θ + ζ)− h(θ − ζ);

� (αj) and (βj) converge to 0


Simulated annealing

Calibration parameterschoice 1 2 3 4

αj 1/ log(j + 1) 1/100 log(j + 1) 1/(j + 1) 1/(j + 1)βj 1/ log(j + 1).1 1/ log(j + 1).1 1/(j + 1).5 1/(j + 1).1

I α ↓ 0 slowly,∑

j αj =∞I β ↓ 0 more slowly,∑

j(αj/βj)2 <∞

I Scenarios 1-2: not enoughenergy

I Scenarios 3-4: good

[ c©George Casella 1951–2012]


Simulated annealing

The traveling salesman problem

A classical allocation problem:

I Salesman who needs tovisit n cities in a row

I Traveling costs betweenpairs of cities are known

I Search of the optimalcircuit

Aboriginal art, NGA, Melbourne


Simulated annealing

The traveling salesman problem

A classical allocation problem:

I Salesman who needs tovisit n cities in a row

I Traveling costs betweenpairs of cities are known

I Search of the optimalcircuit

Procter & Gamble competition, 1962


Simulated annealing

An NP-complete problem

The traveling salesmanproblem is an example ofmathematical problems thatrequire explosive resolutiontimes

Aboriginal art, NGA, Melbourne


Simulated annealing


The traveling salesmanproblem is an example ofmathematical problems thatrequire explosive resolutiontimesNumber of possible circuits n!and exact solutions availablein O(2n) time

Exact solution for 15, 112 German cities

found in 2001 in 22.6 CPU years.


Simulated annealing


The traveling salesmanproblem is an example ofmathematical problems thatrequire explosive resolutiontimesNumerous practicalconsequences (networks,integrated circuit design,genomic sequencing, &tc.)

Exact solution for the 24, 978 Sweedish cities

found in 2004 in 84.8 CPU years.


Simulated annealing

Resolution via simulation

The simulated annealing algorithm:

I name is borrowed from metallurgy

I metal manufactured by a slowdecrease of temperature(annealing) stronger than whenmanufactured by fast decrease

[ c©Joachim Robert, ca. 2006]


Simulated annealing


Repeat

I Random modifications of parts of the original circuit with costC0

I Evaluation of the cost C of the new circuit

I Acceptation of the new circuit with probability

min

{exp

{C0 − C

T

}, 1

}[Metropolis et al., 1953]


Simulated annealing


Repeat

I Random modifications of parts of the original circuit with costC0

I Evaluation of the cost C of the new circuit

I Acceptation of the new circuit with probability

min

{exp

{C0 − C

T

}, 1

}T , temperature, is progressively reduced

[Metropolis et al., 1953]


Simulated annealing

A family meeting (1)

Recall the table puzzle

Definition of a target function

Set the story of the 20 couple tables during the 6 courses as a(20, 6) matrix, e.g.,

S =

1 1 3 2 4 22 1 3 3 2 4...

......

......

...3 2 2 2 4 4

and define a penalty function as the number of missed encounters


Simulated annealing


Recall the table puzzle

Definition of a target function

I=sample(rep(1:4,5))

for (i in 2:6)

I=cbind(I,sample(rep(1:4,5)))

meet=function(I){

M=outer(I[,1],I[,1],"==")

for (i in 2:6)

M=M+outer(I[,i],I[,i],"==")

M

}

penalty=function(M){ sum(M==0) }

penat=penalty(meet(I))


Simulated annealing


Random switch of couples

I Pick two couples [among the 20 couples] at random withprobabilities proportional to the number of other couples theyhave not seenprob=apply(meet(I)==0,1,sum)

I switch their respective position during one of the 6 courses

I accept the switch with Metropolis–Hastings probabilitylog(runif(1))<(penalty(old)-penalty(new))/gamma


Simulated annealing



For instance, propose to replace

S =

1 1 3 2 4 22 1 3 3 2 4...

......

......

...3 2 2 2 4 4

with S ′ =

1 1 3 2 2 22 1 3 3 4 4...

......

......

...3 2 2 2 4 4


Simulated annealing



for (t in 1:N){

prop=sample(1:20,2,prob=apply(meet(I)==0,1,sum))

cour=sample(1:6,1)

Ip=I

Ip[prop[1],cour]=I[prop[2],cour]

Ip[prop[2],cour]=I[prop[1],cour]

penatp=penalty(meet(Ip))

if (log(runif(1))<(penat-penatp)/gamma){

I=Ip

penat=penatp}

}


Simulated annealing


Solution

I

[1,] 1 4 3 2 2 3

[2,] 1 2 4 3 4 4

[3,] 3 2 1 4 1 3

[4,] 1 2 3 1 1 1

[5,] 4 2 4 2 3 3

[6,] 2 4 1 2 4 1

[7,] 4 3 1 1 2 4

[8,] 1 3 2 4 3 1

[9,] 3 3 3 3 4 3

[10,] 4 4 2 3 1 1

[11,] 1 1 1 3 3 2

[12,] 3 4 4 1 3 2

[13,] 4 1 3 4 4 2

[14,] 2 4 3 4 3 4

[15,] 2 3 4 2 1 2

[16,] 2 2 2 3 2 2

[17,] 2 1 2 1 4 3

[18,] 4 3 1 1 2 4

[19,] 3 1 4 4 2 1

[20,] 3 1 2 2 1 4[ c© http://www.metrolyrics.com]


Simulated annealing

Solving sudokus

Given a partly filled Sudoku grid (with a single solution),I define a random Sudoku grid by filling the empty slots at

randoms=matrix(0,ncol=9,nrow=9)

s[1,c(1,6,7)]=c(8,1,2)

s[2,c(2:3)]=c(7,5)

s[3,c(5,8,9)]=c(5,6,4)

s[4,c(3,9)]=c(7,6)

s[5,c(1,4)]=c(9,7)

s[6,c(1,2,6,8,9)]=c(5,2,9,4,7)

s[7,c(1:3)]=c(2,3,1)

s[8,c(3,5,7,9)]=c(6,2,1,9)


Simulated annealing

Solving sudokus

Given a partly filled Sudoku grid (with a single solution),

I define a random Sudoku grid by filling the empty slots atrandom

I define a penalty function corresponding to the number ofmissed constraints#local score

scor=function(i,s){

a=((i-1)%%9)+1

b=trunc((i-1)/9)

boxa=3*trunc((a-1)/3)+1

boxb=3*trunc(b/3)+1

return(sum(s[i]==s[9*b+(1:9)])+

sum(s[i]==s[a+9*(0:8)])+

sum(s[i]==s[boxa:(boxa+2),boxb:(boxb+2)])-3)

}


Simulated annealing

Solving sudokus

Given a partly filled Sudoku grid (with a single solution),

I define a random Sudoku grid by filling the empty slots atrandom

I define a penalty function corresponding to the number ofmissed constraints

I fill “deterministic” slotsI make simulated annealing moves

# random moves on the sites

i=sample((1:81)[as.vector(s)==0],sample(1:sum(s==0),1,pro=1/(1:sum(s==0))))

for (r in 1:length(i))

prop[i[r]]=sample((1:9)[pool[i[r]+81*(0:8)]],1)

if (log(runif(1))/lcur<tarcur-target(prop)){

nchange=nchange+(tarcur>target(prop))

cur=prop

points(t,tarcur,col="forestgreen",cex=.3,pch=19)

tarcur=target(cur)

}


Simulated annealing

Solving sudokus

I many possible variants in theproposals

I rather slow and sometimesreally slow

I may get stuck on a penalty of2 (and never reach zero)

I does not compete at all withnonrandom solvers

presentation slides for my simulation course at dauphine

Education

ubiquitous tool

complex statistical

illustrations necessity

computer account

computer evaluation

new statistical procedure

computer program

computer games