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You should be familiar with the Metropolis–Hastings Algorithm, introduced here, and elaborated here.
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!
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))
if(runif(1)<A){
x[i] = proposedx # accept move with probabily min(1,A)
} else {
x[i] = currentx # otherwise "reject" move, and stay where we are
}
}
return(x)
}
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)
plot(z1,type="l")
lines(z2,col=2)
lines(z3,col=3)
Version | Author | Date |
---|---|---|
1543692 | Matthew Stephens | 2022-04-26 |
plot(log_exp_target(z1))
lines(log_exp_target(z2),col=2)
lines(log_exp_target(z3),col=3)
par(mfcol=c(3,1)) #rather odd command tells R to put 3 graphs on a single page
maxz=max(c(z1,z2,z3))
hist(z1,breaks=seq(0,maxz,length=20))
hist(z2,breaks=seq(0,maxz,length=20))
hist(z3,breaks=seq(0,maxz,length=20))
Version | Author | Date |
---|---|---|
c6b20e2 | Matthew Stephens | 2022-04-26 |
Use the function easyMCMC
to explore 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)
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"
return(-Inf)}
else{
return(0)}
}
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))
if(runif(1)<A){
p[i] = newp # accept move with probabily min(1,A)
} else {
p[i] = currentp # otherwise "reject" move, and stay where we are
}
}
return(p)
}
Running this sample for \(n_{AA}\) = 50, \(n_{Aa}\) = 21, \(n_{aa}\)=29.
z=psampler(50,21,29,10000,0.5,0.01)
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).
x=seq(0,1,length=1000)
hist(z,prob=T)
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
hist(z[5001:10000])
Version | Author | Date |
---|---|---|
1543692 | Matthew Stephens | 2022-04-26 |
Investigate how the starting point and proposal standard deviation affect the convergence of the algorithm.
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 \(aa\).
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
}
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
sessionInfo()
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
locale:
[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
loaded via a namespace (and not attached):
[1] Rcpp_1.0.10 rstudioapi_0.14 whisker_0.4.1 knitr_1.42
[5] magrittr_2.0.3 workflowr_1.7.0 R6_2.5.1 rlang_1.0.6
[9] fastmap_1.1.1 fansi_1.0.4 highr_0.10 stringr_1.5.0
[13] tools_4.2.1 xfun_0.37 utf8_1.2.3 cli_3.6.0
[17] git2r_0.31.0 jquerylib_0.1.4 htmltools_0.5.4 rprojroot_2.0.3
[21] yaml_2.3.7 digest_0.6.31 tibble_3.1.8 lifecycle_1.0.3
[25] later_1.3.0 sass_0.4.5 vctrs_0.5.2 fs_1.6.1
[29] promises_1.2.0.1 cachem_1.0.7 glue_1.6.2 evaluate_0.20
[33] rmarkdown_2.20 stringi_1.7.12 bslib_0.4.2 compiler_4.2.1
[37] pillar_1.8.1 jsonlite_1.8.4 httpuv_1.6.9 pkgconfig_2.0.3
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