Bayesian Inference and computational methods – Practical 1
Solutions including R code for implementing these
1. Simulating random sample of size 1000 from N(0, 1)
x = runif(1000, 0, 1)
y = runif(1000, 0, 1)
z = sqrt(-2.0*log(x))*cos(2.0*pi*y)
hist(z)
2. Converting random sample of N(0, 1) to N(10, 10)
z10 = sqrt(10)*z + 10.0
hist(z10)
Here z refers to the sample generated in question 1.
3. Uniform(-1, 1)
x = 2.0*runif(1000, 0, 1) - 1
4. Simulating Exp(2) by inversion of c.d.f.
z = runif(1000, 0, 1) #random sample from U(0, 1)
x = -0.5*log(1 - z) #transformation by inversion of c.d.f.
[Note x = -0.5*log(z) also gives sample Exp(2) since 1-z and z are both i.i.d. U(0, 1)]
5. Simulating 1000 draws from Gamma(3, 2).
This can be done by adding together 3 independent Exp(2) random variables.
Therefore
z = -0.5*(log(runif(1000, 0, 1)) + log(runif(1000, 0, 1)) + log(runif(1000, 0, 1)))
or
z = -0.5*(log(runif(1000, 0, 1)*runif(1000, 0, 1)*runif(1000, 0, 1)))
do the job.
6. Code for simulating from Poisson() using Poisson process with rate 1.
Algorithm counts number of events occurring within time window of length .
for(i in 1:1000) {
k=0
sum = 0.0
while(sum < lam) {
sum = sum – log(runif(1, 0, 1))
k = k+1
}
z[i] = k-1
}
Code for inversion of cdf for Poisson distribution.
for(i in 1:1000){
k=0
q = runif(1, 0, 1)
while(ppois(k, lam) < q) k = k+1
z[i] = k
}
7. Simulate r.s. of size 1000 from Bin(12, 0.5)
Various possibilities exist. One way to do it is to simply add together independent
Bernoulli(0.5) distributions.
for(i in 1:1000) {
sum = 0
for(j in 1:12) sum = sum + trunc(runif(1, 0, 1)+0.5)
z[i] = sum
}
8. Simulating throwing 10 fair dice
for(i in 1:n) {
sum = 0
for(j in 1:10) sum = sum + trunc(6*runif(1, 0, 1)) + 1
z[i] = sum
}
9. Simply use recipe from question 1 to generate 2 independent i.i.d. samples of
size 1000 from N(0, 1), then use formulae from hint on question sheet.
10. Rejection sampling from Beta(2, 2) using q(x) = 1, p(x) = 6x(1-x). Note that
p/q < 3/2. It follows that acceptance probability is 4x(1-x) (check this).
k=1
while(k<1001) {
u = runif(1, 0, 1)
if(runif(1, 0, 1) < 4*u*(1 - u)) {
z[k] = u
k = k+1
}
}
Inversion of c.d.f. for this case requires solving a cubic equation as described in
solutions to Chapter 4, Q. 3.
To obtain sample from Beta(2.2, 2.2) you can use the z generated above and apply
rejection sampling to it. Now q(x) corresponds to the Beta(2.2) density. It follows
that
B(2, 2)
px 

x 0.2 (1  x) 0.2  K 0.25 0.2  c
q x  B2.2, 2.2
Therefore the acceptance ratio for rejection sampler (p/(cq)) is given by
40.2x0.2(1-x)0.2
We can obtain our random sample as:
newz = c(1:1000)
const = 4^0.2
k=1
for(i in 1:1000) {
if(runif(1, 0, 1) < const*z[i]^0.2*(1 - z[i])^0.2) {
newz[k] = z[i]
k = k+1
}
}
On competing this loop k-1 tells us how many samples were generated. Note that k-1
will (almost certainly) be less than 1000 since some of the z's will be rejected.