# Chapter 1 Monte Carlo

By Monte Carlo techniques we refer to computational algorithms that rely on random sampling.

## 1.1 Probability and inference refresher

Consider a random variable $$X$$ and a random sample from it $$X_1,X_2,\ldots,X_n$$, we define:

• The sample mean as $\overline{X}_n=\frac{1}{n}\sum_{i=1}^n X_i$

• The sample variance as $S_n^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline{X}_n)^2$

• The sample proportion as the fraction of sample observations with a given characteristic, $\hat{p}_n=\frac{\#\{X_i\textsf{ with the characteristic}\}}{n}.$

Laws of Large Numbers

If $$\{X_n\}_n$$ is a sequence of independent and identically distributed random variables with $${\mathbb E}[X_i]=\mu$$, then

• WLLN: for any $$\varepsilon>0$$, $$\lim\limits_{n\rightarrow\infty} P(|\overline{X}_n-\mu|\geq\varepsilon)=0\,;$$

• SLLN: $$P\left(\lim\limits_{n\rightarrow\infty}\overline{X}_n=\mu\right)=1\,,$$

where $$\overline{X}_n=\frac{1}{n}\sum_{i=1}^n X_i$$.

Observe that if $$X_i\sim{\rm B}(1,p)$$, then $$\overline{X}_n=\hat{p}_n$$ and $$\mu=p$$.

Central Limit Theorem

• Lyapunov If $$\{X_n\}_n$$ is a sequence of iid r.v.s with mean $$\mu$$ and variance $$\sigma^2<\infty$$, then $\frac{\sum_{i=1}^n X_i-n\mu}{\sigma\sqrt{n}}=\frac{\overline{X}_n-\mu}{\sigma/\sqrt{n}}\xrightarrow{d}Z\,,$ where $$Z\sim{\rm N}(0,1)$$. If needed, we can substitute $$\sigma$$ by is estimate $$S_n$$.

• DeMoivre-Laplace If $$X_n\sim{\rm B}(n,p)$$, then $\frac{X_n-np}{\sqrt{np(1-p)}}\xrightarrow{d}Z\,,$ where, we can divide numerator and denominator by $$n$$ to obatin $\frac{\hat{p}_n-p}{\sqrt{\frac{p(1-p)}{n}}}\xrightarrow{d}Z\,,$ and here we can substitute the $$p$$’s in the denominator by $$\hat{p}_n$$ if needed.

Statistical analysis of simulated data (Conf. Intervals)

• Approximate $$(1-\alpha)100\%$$ CI on $$\mu$$ (the mean) $\left[\overline{x}_n-z_{\alpha/2}\frac{s_n}{\sqrt{n}}\,,\,\overline{x}_n+z_{\alpha/2}\frac{s_n}{\sqrt{n}}\right]$

• Approximate $$(1-\alpha)100\%$$ CI on $$p$$ (a proportion) $\left[\hat{p}_n-z_{\alpha/2}\sqrt{\frac{\hat{p}_n(1-\hat{p}_n)}{n}}\,,\,\hat{p}_n+z_{\alpha/2}\sqrt{\frac{\hat{p}_n(1-\hat{p}_n)}{n}}\right]$

$$z_{\alpha/2}=qnorm(1-\alpha/2)$$

## 1.2 Statistical validation techniques (GoF tests)

• Kolmogorov-Smirnoff Goodness-of-Fit test (continuous data, given $$F_0$$)

Test. $$H_0:\,F=F_0$$ vs. $$H_1:\,F\neq F_0$$

Test statistic. $$D=\max_x|\hat{F}_n(x)-F_0(x)|$$, with $$\hat{F}_n$$ empirical cdf.

The distribution of $$D$$ (under $$H_0$$) does NOT depend on $$F_0$$ (continuous).

data <- c(.56,.35,.89,.81,.36,.19,.65,.42,.12,.05)
ks.test(data,"punif")
##
##  One-sample Kolmogorov-Smirnov test
##
## data:  data
## D = 0.18, p-value = 0.8473
## alternative hypothesis: two-sided

Statistical validation techniques (GoF tests)

plot(ecdf(data))
aux <- seq(-.1,1.1,0.01)
points(aux,punif(aux),type="l")
D <- ks.test(data,"punif")$statistic points(aux,punif(aux)+D,type="l",col="red") points(aux,punif(aux)-D,type="l",col="red") Statistical validation techniques (GoF tests) • Chi-Square Goodness-of-Fit test (discrete data, given $$F_0$$) Test. $$H_0:\,F=F_0$$ vs. $$H_1:\,F\neq F_0$$ Test statistic. $$X^2=\sum^k_{i=1}\frac{(N_i-np_i)^2}{np_i}$$, $$k$$ levels of the variable, $$N_i$$ observed counts, and $$np_i$$ expected counts. Under $$H_0$$ $$X^2\sim\chi^2_{k-1}$$. observed <- table(c(3,2,1,2,2,1,2,1,2)) probs <- rep(1/3,3) chisq.test(x=observed,p=probs) ## ## Chi-squared test for given probabilities ## ## data: observed ## X-squared = 2.6667, df = 2, p-value = 0.2636 n <- sum(observed) par(mfrow=c(1,2)) barplot(observed,main="observed") barplot(probs*n,main="expected") • For specific distributions (e.g. normal), we can use specific tests (e.g. Shapiro-Wilk normality test) or graphical tools (e.g. qq-plots). • When some parameters are unespecified: • Continuous distribution, use KS test and simulate to estimate critical value or p-value. • Discrete distribution, use Chi-square test and substract the number of estimated parameters from the degrees of freedom. Randomness tests • Wald-Wolfowitz runs test of randomness for continuous data Test. $$H_0:\,$$i.i.d. observations vs. $$H_1:\,$$ not i.i.d. observations Test statistic. $$U$$: # runs (group of consecutive values either above or below threshold – median), $$N_1$$ (resp. $$N_2$$): # values above (below) threshold $\frac{U-\left(\frac{2 N_1 N_2}{N_1+N_2}+1\right)}{\sqrt{\frac{2N_1 N_2(2N_1 N_2-N_1-N_2)}{(N_1+N_2)^2(N_1+N_2-1)}}}$ # install.packages("randtests") require(randtests) ; data("iris") runs.test(iris$Sepal.Length[iris$Species=="setosa"])$p.value
##  0.7428503
attach(iris)
plot(Sepal.Length[Species=="setosa"],type="l")
points(Sepal.Length[Species=="setosa"])
abline(h=median(Sepal.Length[Species=="setosa"]),col="red") detach(iris)

## 1.3 Random numbers Prima materia

The first matter in simulation are random numbers in the $$(0,1)$$ interval (indepent random variables with a $${\rm U}(0,1)$$ distribution).

Is it possible to build such random numbers?

If fair coins do exist, it is.

Consider a sequence of tosses of a fair coin $$x_1,x_2,x_3,\ldots$$, where

$x_i=\left\{\begin{array}{ll}1&\text{ if Head on i-th toss}\\ 0 & \text{ if Tail on i-th toss}\end{array}\right.$ We can interpret the sequence as the binary expansion of a number in the $$[0,1]$$ interval,

$\sum_{i=1}^\infty\frac{x_i}{2^i}\,.$

MC <- 200
simul.bin <- vector(length=MC)
set.seed(1)
for(i in 1:MC){
samp <- sample(c(0,1),replace=T,size=10)
simul.bin[i] <- sum(samp/2^((1:10)*samp))
}
ks.test(simul.bin,punif)$p.value ##  0.8532117 runs.test(simul.bin)$p.value
##  0.3949537
layout(matrix(c(1,2),1,2,byrow=T),widths=c(1,3))
hist(simul.bin,probability=T)
plot(simul.bin,type="l")
abline(h=median(simul.bin),col="red") Pseudorandom numbers

The numbers we work with have been determiniscally generated, so they are not truely random.

A multiplicative congruential generator works as follows:

1. Set $$m$$ (prime) large.
2. Set $$a<m$$.
3. Set seed $$x_0$$ (initial value).
4. Set $$x_n\equiv a x_{n-1}\,\,(m)$$, which means $$x_n$$ is the remainder of $$ax_{n-1}/m$$.
5. For every $$n$$ return $$x_n/m$$.
MC <-1000 ; seed <- 1 ; m <- 2^35-31 ; a <- 5^5
simul <- vector(length=MC)
simul <- (a*seed)%%m
for(i in 2:MC) simul[i] <- (a*simul[i-1])%%m
simul <- simul/m
ks.test(simul,"punif")$p.value ##  0.9174393 runs.test(simul)$p.value
##  0.3113298

Replicability and efficiency vs. true randomness

The methods we use do not produce truely random numbers, but they have two properties that are essential in Statistics:

• We can duplicate our results. By fixing a seed (initial value) we always obtain the same sequence of (pseudo)random numbers.

• They are efficient. Fast and simple to implement.

MC<-1000000 ; ptm <- proc.time() ; set.seed(1)
simul.unif <- runif(MC)
proc.time()-ptm
##    user  system elapsed
##    0.03    0.00    0.04
ptm <- proc.time()
seed <- 1 ; m <- 2^35-31 ; a <- 5^5
simul <- vector(length=MC)
simul <- (a*seed)%%m
for(i in 2:MC) simul[i] <- (a*simul[i-1])%%m
simul <- simul/m ; proc.time()-ptm
##    user  system elapsed
##    0.22    0.02    0.23

## 1.4 Approximation of probabilities and volumes

We can approximate the volue of some given compact set $$C\subset{\mathbb R}^d$$ by considering a $$d$$-dimensional box $$B$$ containing it, $$C\subset B$$. We can simulate points from the box at random, the volume of $$C$$ is approximately the fraction of points in $$C$$ times the volume of $$B$$.

Consider $$B=[a_1,b_1]\times[a_2,b_2]\times\ldots\times [a_d,b_d]$$

1. Set $$inside=0$$.
2. Repeat Steps $$3$$ to $$4$$ $$n$$ times.
3. Generate $$U_1$$ random number in $$[a_1,b_1]$$, $$U_2$$ random number in $$[a_2,b_2]$$, $$\ldots,U_d$$ random number in $$[a_d,b_d]$$.
4. If $$(U_1,\ldots,U_d)\in C$$, then $$inside = inside+1$$
5. Set $$vol=(inside/n)*(b_1-a_1)*\cdots*(b_d-a_d)$$ and stop.

Approximate the area of a circle of radius 1

aux <- seq(0,1,.01)
plot(aux,sqrt(1-aux^2),type="l",xlab="",ylab="")
lines(c(1,0,0),c(0,0,1))
MC <- 10 ; set.seed(1)
points(runif(MC),runif(MC),pch=3) n <- 1000 ; set.seed(1)
x <- runif(n)
y <- runif(n)
sum(x^2+y^2<=1)/n*4
##  3.148
prop.test(sum(x^2+y^2<=1),n=n,conf.level=0.95)$conf.int*4 ##  3.040123 3.246918 ## attr(,"conf.level") ##  0.95 Approximation of probabilities by simulation For each random experiment, we can simulate a random number in the interval $$(0,1)$$ and assume that an event with assigned probability $$p$$ occurs if it is less than $$p$$. The expression (runif(1)<p) is a logical value (TRUE or FALSE) that is interpreted as numerical value 1 (TRUE) or 0 (FALSE) when operating. The intesection of two events occurs whenever both of them do simulataneously ocurr, logical operator ‘AND’, product as numbers. The union occurs whenever at least one of them occurs, logical operator ‘OR’, maximum as numbers. Approximation of probabilities by simulation (example) An urn contains 1 white ball and 2 red balls. One ball is taken from the urn at random. If it is a white ball, it is returned into the urn together with another white ball. If it is a red ball, it is returned together with other 2 red balls. In a second stage, another ball is taken from the urn at random. Events • $$W_1\equiv$$’first ball is white’ • $$R_1\equiv$$’first ball is red’ • $$W_2\equiv$$’second ball is white’ • $$R_2\equiv$$’second ball is red’ Probabilities • $$P(W_1)=1/3$$; $$P(R_1)=2/3$$ • $$P(W_2|W_1)=1/2$$; $$P(R_2|W_1)=1/2$$ • $$P(W_2|R_1)=1/5$$; $$P(R_2|R_1)=4/5$$ • Total probability rule What is the probability that the second ball extracted from the urn is red? $P(R_2)=P(W_1)P(R_2|W_1)+P(R_1)P(R_2|R_1)=\frac{1}{3}\cdot\frac{1}{2}+\frac{2}{3}\cdot\frac{4}{5}=0.7\,.$ Bayes’ rule If the second ball extracted from the urn is red, what is the probability that the first one was alse red? $P(R_1|R_2)=\frac{P(R_1)P(R_2|R_1)}{P(R_2)}=\frac{\frac{2}{3}\cdot\frac{4}{5}}{0.7}=0.762\,.$ MC <- 1000 ; set.seed(1) ball.1 <- runif(MC) ball.2 <- runif(MC) R1 <- (ball.1<2/3) R2 <- R1*(ball.2<4/5)+(1-R1)*(ball.2<1/2) p1 <- sum(R2)/MC ; p1 ##  0.703 p2 <- sum(R1*R2)/sum(R2) ; p2 ##  0.7496444 95% CIs on the probabilities p1-sqrt(p1*(1-p1)/MC)*qnorm(.975) ##  0.6746793 p1+sqrt(p1*(1-p1)/MC)*qnorm(.975) ##  0.7313207 n2 <- sum(R2) p2-sqrt(p2*(1-p2)/n2)*qnorm(.975) ##  0.7176203 p2+sqrt(p2*(1-p2)/n2)*qnorm(.975) ##  0.7816685 ## 1.5 Monte Carlo integration Integrals over $$(0,1)$$ For any function $$g$$, the integral $$\int_0^1 g(x)\,{\rm d}x$$ is the expectation of random variable $$U\sim{\rm U}(0,1)$$ After LLN, the sample mean of $$g(U_1),\ldots,g(U_n)$$ can be used to approximate the mean of randon variable $$g(U)$$: 1. Generate $$U_1,\ldots,U_n$$ random numbers in $$(0,1)$$. 2. Transform each $$U_i$$ into $$Y_i=g(U_i)$$. 3. Return $$n^{-1}\sum_{i=1}^n Y_i$$ and stop. Asymptotic confidence inteval of an integral Approximate $$(1-\alpha)100\%$$ CI on $$\mu$$ (the mean) $\left[\overline{x}_n-z_{\alpha/2}\frac{s_n}{\sqrt{n}}\,,\,\overline{x}_n+z_{\alpha/2}\frac{s_n}{\sqrt{n}}\right]$ with $$z_{\alpha/2}=qnorm(1-\alpha/2)$$. Take each $$X_i$$ as $$Y_i=g(U_i)$$, and use is average and sample variance. 1. Generate $$U_1,\ldots,U_n$$ random numbers in $$(0,1)$$. 2. Transform each $$U_i$$ into $$Y_i=g(U_i)$$. 3. Obtain sample mean and variance, $$\overline{y}_n$$ and $$s^2_n$$. 4. Return $$\overline{y}_n\pm z_{\alpha/2}s_n/\sqrt{n}$$ and stop. Area of a quarter circle $\int_0^1 \sqrt{1-x^2}\,{\rm d}x$ MC <-1000 ; set.seed(1) ; y <- sqrt(1-runif(MC)^2) ; mean(y) ##  0.7845459 t.test(y)$conf.int
##  0.7704274 0.7986644
## attr(,"conf.level")
##  0.95
4*t.test(y)$conf.int ##  3.081710 3.194658 ## attr(,"conf.level") ##  0.95 Integrals over $$(a,b)$$ Transform to integral over $$(0,1)$$ as with the change of variable $$t=(x-a)/(b-a)$$. If $$U\sim{\rm U}(0,1)$$, then $$a+(b-a)U\sim{\rm U}(a,b)$$, so alternatively to generating randdom numbers in $$(0,1)$$ and transforming, it is possible to generate random numbers in $$(a,b)$$ as runif(MC,min=a,max=b). Integrals over $$(a,\infty)$$ Transform to integral over $$(0,1)$$ as where the change of variable is $$t=(1+x)^{-1}$$. $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,{\rm d}x=2\int_0^1\frac{t^{-2}}{\sqrt{2\pi}}e^{-(t^{-1}-1)^2/2}\,{\rm d}t$ MC <-10000 ; set.seed(1) ; u <- runif(MC) y <- 2*exp(-(1/u-1)^2/2)/(u^2*sqrt(2*pi)) t.test(y)$conf.int
##  0.9803571 1.0074944
## attr(,"conf.level")
##  0.95

Multiple integrals

Consider now a function of two (or more) variables, $$g:{\mathbb R}^2\mapsto{\mathbb R}$$, the integral $$\int_0^1\int_0^1 g(x_1,x_2)\,{\rm d}x_1{\rm d}x_2$$ is thus the expectation of two independent random variables in the unit interval, $$X_1,X_2\sim{\rm U}(0,1)$$

We can approximate the integral using the LLN over rv $$g(X_1,X_2)$$:

1. Generate $$U_1,\ldots,U_n$$ and $$U'_1,\ldots,U'_n$$ random numbers in $$(0,1)$$.
2. Transform each pair $$(U_i,U'_i)$$ into $$Y_i=g(U_i,U'_i)$$.
3. Return $$n^{-1}\sum_{i=1}^n Y_i$$ and stop.

If the multiple integral is over a region different from the unit square (hypercube) we can transform it.

Alternatively, we can consider a wider region and discard those generated points that do not lie in it. In such a case, we should keep on simulating until we have a fixed number of points inside the region.

A new step should be introduced between Septs 1 and 2 of the previous algorithm.

• If some $$(U_i,U'_i)$$ does not lie inside the region, generate a new pair.