Last updated: 2023-07-18

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You should be familiar with the Metropolis–Hastings Algorithm, introduced here, and elaborated here.

Caveat on code

Note: the code here is designed to be readable by a beginner, rather than “efficient”. The idea is that you can use this code to learn about the basics of MCMC, but not as a model for how to program well in R!

Example 1: sampling from an exponential distribution using MCMC

Any MCMC scheme aims to produce (dependent) samples from a ``target” distribution. In this case we are going to use the exponential distribution with mean 1 as our target distribution. Here we define this function (on log scale):

log_exp_target = function(x){
  return(dexp(x,rate=1, log=TRUE))

The following code implements a simple MH algorithm. (Note that the parameter log_target is a function which computes the log of the target distribution; you may be unfamiliar with the idea of passing a function as a parameter, but it works just like any other type of parameter…):

easyMCMC = function(log_target, niter, startval, proposalsd){
  x = rep(0,niter)
  x[1] = startval     
  for(i in 2:niter){
    currentx = x[i-1]
    proposedx = rnorm(1,mean=currentx,sd=proposalsd) 
    A = exp(log_target(proposedx) - log_target(currentx))
      x[i] = proposedx       # accept move with probabily min(1,A)
    } else {
      x[i] = currentx        # otherwise "reject" move, and stay where we are

Now we run the MCMC three times from different starting points and compare results:

z1=easyMCMC(log_exp_target, 1000,3,1)
z2=easyMCMC(log_exp_target, 1000,1,1)
z3=easyMCMC(log_exp_target, 1000,5,1)


Version Author Date
1543692 Matthew Stephens 2022-04-26

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c6b20e2 Matthew Stephens 2022-04-26
1543692 Matthew Stephens 2022-04-26
par(mfcol=c(3,1)) #rather odd command tells R to put 3 graphs on a single page

Version Author Date
c6b20e2 Matthew Stephens 2022-04-26


Use the function easyMCMC to explore the following:

  1. how do different starting values affect the MCMC scheme? (try some extreme starting points)
  2. what is the effect of having a bigger/smaller proposal standard deviation? (again, try some extreme values)
  3. try changing the (log-)target function to the following
log_target_bimodal = function(x){
  log(0.8* dnorm(x,-4,1) + 0.2 * dnorm(x, 4, 1)) 

What does this target distribution look like? What happens if the proposal sd is too small here? (try e.g. 1 and 0.1)

Example 2: Estimating an allele frequency

A standard assumption when modelling genotypes of bi-allelic loci (e.g. loci with alleles \(A\) and \(a\)) is that the population is “randomly mating”. From this assumption it follows that the population will be in “Hardy Weinberg Equilibrium” (HWE), which means that if \(p\) is the frequency of the allele \(A\) then the genotypes \(AA\), \(Aa\) and \(aa\) will have frequencies \(p^2, 2p(1-p)\) and \((1-p)^2\) respectively.

A simple prior for \(p\) is to assume it is uniform on \([0,1]\). Suppose that we sample \(n\) individuals, and observe \(n_{AA}\) with genotype \(AA\), \(n_{Aa}\) with genotype \(Aa\) and \(n_{aa}\) with genotype \(aa\).

The following R code gives a short MCMC routine to sample from the posterior distribution of \(p\). Try to go through the code to see how it works.

log_prior = function(p){
  if((p<0) || (p>1)){  # || here means "or"

log_likelihood_hwe = function(p, nAA, nAa, naa){
  return((2*nAA)*log(p)  + nAa * log (2*p*(1-p)) + (2*naa)*log(1-p))

psampler = function(nAA, nAa, naa, niter, pstartval, pproposalsd){
  p = rep(0,niter)
  p[1] = pstartval
  for(i in 2:niter){
    currentp = p[i-1]
    newp = currentp + rnorm(1,0,pproposalsd)
    A = exp(log_prior(newp) + log_likelihood_hwe(newp,nAA,nAa,naa) - log_prior(currentp) - log_likelihood_hwe(currentp,nAA,nAa,naa))
      p[i] = newp       # accept move with probabily min(1,A)
    } else {
      p[i] = currentp        # otherwise "reject" move, and stay where we are

Running this sample for \(n_{AA}\) = 50, \(n_{Aa}\) = 21, \(n_{aa}\)=29.


Now some R code to compare the sample from the posterior with the theoretical posterior (which in this case is available analytically; since we observed 121 \(A\)s, and 79 \(a\)s, out of 200, the posterior for \(p\) is Beta(121+1,79+1).

lines(x,dbeta(x,122, 80))  # overlays beta density on histogram

Version Author Date
1543692 Matthew Stephens 2022-04-26

You might also like to discard the first 5000 z’s as “burnin”. Here’s one way in R to select only the last 5000 z’s


Version Author Date
1543692 Matthew Stephens 2022-04-26


Investigate how the starting point and proposal standard deviation affect the convergence of the algorithm.

Example 3: Estimating an allele frequency and inbreeding coefficient

A slightly more complex alternative than HWE is to assume that there is a tendency for people to mate with others who are slightly more closely-related than “random” (as might happen in a geographically-structured population, for example). This will result in an excess of homozygotes compared with HWE. A simple way to capture this is to introduce an extra parameter, the “inbreeding coefficient” \(f\), and assume that the genotypes \(AA\), \(Aa\) and \(aa\) have frequencies \(fp + (1-f)p*p, (1-f) 2p(1-p)\), and \(f(1-p) + (1-f)(1-p)(1-p)\).

In most cases it would be natural to treat \(f\) as a feature of the population, and therefore assume \(f\) is constant across loci. For simplicity we will consider just a single locus.

Note that both \(f\) and \(p\) are constrained to lie between 0 and 1 (inclusive). A simple prior for each of these two parameters is to assume that they are independent, uniform on \([0,1]\). Suppose that we sample \(n\) individuals, and observe \(n_{AA}\) with genotype \(AA\), \(n_{Aa}\) with genotype \(Aa\) and \(n_{aa}\) with genotype \(aa\).


  • Write a short MCMC routine to sample from the joint distribution of \(f\) and \(p\).

Hint: here is a start; you’ll need to fill in the …

# The first step is probably to code a log-likelihood function for the inbreeding model....
log_likelihood_inbreeding = function(...){

# then use the log-likelihood within your MCMC scheme
fpsampler = function(nAA, nAa, naa, niter, fstartval, pstartval, fproposalsd, pproposalsd){
  f = rep(0,niter)
  p = rep(0,niter)
  f[1] = fstartval
  p[1] = pstartval
  for(i in 2:niter){
    currentf = f[i-1]
    currentp = p[i-1]
    newf = currentf + ...
    newp = currentp + ...
  return(list(f=f,p=p)) # return a "list" with two elements named f and p
  • Use this sample to obtain point estimates for \(f\) and \(p\) (e.g. using posterior means) and interval estimates for both \(f\) and \(p\) (e.g. 90% posterior credible intervals), when the data are \(n_{AA} = 50, n_{Aa} = 21, n_{aa}=29\).

Addendum: Gibbs Sampling

You could also tackle this problem with a Gibbs Sampler (see vignettes here and here).

To do so you will want to use the following “latent variable” representation of the model: \[z_i \sim Bernoulli(f)\] \[p(g_i=AA | z_i=1) = p; p(g_i=AA | z_i=0) = p^2\] \[p(g_i=Aa | z_i = 1)= 0; p(g_i=Aa | z_i=0) = 2p(1-p)\] \[p(g_i=aa | z_i = 1) = (1-p); p(g_i =aa | z_i=0) = (1-p)^2\]

Summing over \(z_i\) gives the same model as above: \[p(g_i=AA) = fp + (1-f)p^2\]


Using the above, implement a Gibbs Sampler to sample from the joint distribution of \(z,f,\) and \(p\) given genotype data \(g\).

Hint: this requires iterating the following steps

  1. sample \(z\) from \(p(z | g, f, p)\)
  2. sample \(f,p\) from \(p(f, p | g, z)\)

R version 4.2.1 (2022-06-23)
Platform: x86_64-apple-darwin17.0 (64-bit)
Running under: macOS Big Sur ... 10.16

Matrix products: default
BLAS:   /Library/Frameworks/R.framework/Versions/4.2/Resources/lib/libRblas.0.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/4.2/Resources/lib/libRlapack.dylib

[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

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[25] later_1.3.0      sass_0.4.5       vctrs_0.5.2      fs_1.6.1        
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