Pre-requisites

Be familiar with the concept of joint distribution and a conditional distribution. Ideally also with the concept of a Markov chain and its stationary distribution.

Overview

Gibbs sampling is a very useful way of simulating from distributions that are difficult to simulate from directly. However, in this introduction to the key concept, we will use a Gibbs sampler to simulate from a very simple distribution that could be simulated from in other ways.

Gibbs Sampling

Suppose $X$ and $Y$ are two binary random variables with joint distribution $\Pr(X=x,Y=y) = p_{X,Y}(x,y)$ given by the following table:

That is, for example, $p_{X,Y}(0,0)=0.6$.

The conditional distribution of $X$ given any given value is easy to compute by the usual formula for conditional probability, $\Pr(A | B) = \Pr(A \cap B)/\Pr(B)$. For example: $$\Pr(X=0 | Y=0) = \Pr(X=0 \cap Y=0)/Pr(Y=0) = 0.6/0.75 = 0.8$$ and so $$\Pr(X=1 | Y=0) = 1-0.8 = 0.2.$$ Similarly $$\Pr(X=0 | Y=1) = 0.1/0.25 = 0.4$$ and so $$\Pr(X=1 | Y=1) = 0.6$$ .

We can just as easily compute the conditional distribution of $Y$ for any given value of $X$: $$\Pr(Y=0 | X=0) = 6/7$$ $$\Pr(Y=1 | X=0) = 1/7$$ $$\Pr(Y=0 | X=1) = 1/2$$ $$\Pr(Y=1 | X=1) = 1/2$$

Now the question: what if we start at some value of $X,Y$ and proceed to iterate the following steps: 1. Simulate a new value of $X$ from $\Pr(X | Y=y)$ where $y$ is the current value of $Y$. 2. Simulate a new value of $Y$ from $\Pr(Y | X=x)$ where $x$ is the current value of $X$ (generated in 1.) What happens?

Let’s try it:

#returns 1 with probability p, and 0 with probability 1-p
rbernoulli=function(p){return(1*runif(1)<p)}

# sample from distribution X given Y above
sample_XgivenY = function(y){
 if(y==0){
   x = rbernoulli(0.2) # returns 1 with probability 0.2; otherwise 0
 } else {
   x = rbernoulli(0.6)
 } 
  return(x)
}
#' sample from distribution Y given X above
sample_YgivenX = function(x){
  if(x==0){
    y = rbernoulli(1/7)
  } else {
    y = rbernoulli(0.5)
  }
  return(y)
}
set.seed(100)
niter = 1000
X = rep(0,niter)
Y = rep(0,niter)
X[1]=1
Y[1]=1 # start from (1,1)
for(i in 2:niter){
  X[i] = sample_XgivenY(Y[i-1])
  Y[i] = sample_YgivenX(X[i])
}
res = data.frame(X=X,Y=Y)

Here is what happens for the first 20 iterations:

head(res,20)

   X Y
1  1 1
2  1 1
3  1 1
4  1 1
5  0 0
6  0 0
7  0 0
8  0 0
9  0 0
10 0 0
11 0 0
12 0 0
13 0 0
14 0 0
15 0 0
16 0 0
17 0 0
18 0 0
19 0 0
20 1 0

And here is a summary of what proportion of the rows are of each type:

table(data.frame(X=X,Y=Y))/niter

   Y
X       0     1
  0 0.617 0.092
  1 0.154 0.137

As you can see, the proportion of iterations in which $X=x$ and $Y=y$ is very close to $\Pr(X=x,Y=y)=p_{X,Y}(x,y)$. This is not a coincidence!

Explanation

What we have done here is simulate a Markov chain $$(X_1,Y_1), (X_2,Y_2), (X_3,Y_3), \dots$$ , whose stationary distribution is $\Pr(X=x,Y=y)=p_{X,Y}(x,y)$.

To see that the pairs $(X,Y)$ form a Markov chain, note that the simulation of $X_i$ is done using only the previous value $Y_{i-1}$, and the simulation of $Y_i$ is done using only $X_i$. So simulation of $(X_i,Y_i)$ depends on the previous states only through the immediate previous state $(X_{i-1},Y_{i-1})$, which means it is a Markov chain. (And in fact it only depends on $Y_{i-1}$, but that is not so important.)

To see why it has stationary distribution $p_{X,Y}(x,y)$ imagine simulating $X_1,Y_1$ from this distribution, so $\Pr(X_1=x,Y_1=y)= p_{X,Y}(x,y)$, and in particular $\Pr(Y_1=y)= \sum_x p_{X,Y}(x,y) = p_Y(y)$.

Now, what is $\Pr(X_2=x,Y_1=y)$? Well we know $$\Pr(X_2=x,Y_1=y) = \Pr(Y_1=y) \Pr(X_2=x| Y_1=y).$$ And we know from above that $$\Pr(Y_1=y) = p_{Y}(y).$$ And we know that given $Y_1-y$, $X_2$ was simulated from the conditional distribution $p_{X|Y}(x|y)$, so $$\Pr(X_2=x | Y_1=y) = p_{X|Y}(x|y).$$ Putting these together we have $$\Pr(X_2=x,Y_1=y) = p_{Y}(y)p_{X|Y}(x|y) = p_{X,Y}(x,y).$$

Now, essentially the same argument shows that $\Pr(X_2=x, Y_2=y) = p_{X,Y}(x,y)$. [Exercise]

Thus, we have shown that if $\Pr(X_1=x, Y_1=y) = p_{X,Y}(x,y)$ then also $\Pr(X_2=x, Y_2=y) = p_{X,Y}(x,y)$. That is exactly what it means for $p_{X,Y}(x,y)$ to be the stationary distribution: if we start the chain by simulating from that distribution then it remains in that distribution after one step, and so it remains in that distribution for ever.

Of course, we did not start the chain at that distribution. But the above argument shows that this is indeed the stationary distribution. There is a general result that discrete Markov chains converge to their stationary distribution, provided that they are what is called “irreducible and aperiodic” (which this Markov chain is). That is, for large enough $n$, we should see $\Pr(X_n=x, Y_n=y) \approx p_{X,Y}(x,y)$ no matter where we start. Furthermore, in the long-run, the proportion of iterations spent in each state will also converge to this distribution.

This explains the simulation result.

sessionInfo()

R version 3.3.2 (2016-10-31)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Ubuntu 14.04.5 LTS

locale:
 [1] LC_CTYPE=en_US.UTF-8       LC_NUMERIC=C              
 [3] LC_TIME=en_US.UTF-8        LC_COLLATE=en_US.UTF-8    
 [5] LC_MONETARY=en_US.UTF-8    LC_MESSAGES=en_US.UTF-8   
 [7] LC_PAPER=en_US.UTF-8       LC_NAME=C                 
 [9] LC_ADDRESS=C               LC_TELEPHONE=C            
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C       

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

other attached packages:
[1] workflowr_0.4.0    rmarkdown_1.3.9004

loaded via a namespace (and not attached):
 [1] backports_1.0.5 magrittr_1.5    rprojroot_1.2   htmltools_0.3.5
 [5] tools_3.3.2     yaml_2.1.14     Rcpp_0.12.9     stringi_1.1.2  
 [9] knitr_1.15.1    git2r_0.18.0    stringr_1.2.0   digest_0.6.12  
[13] evaluate_0.10