I read in a pdf here: MCMC - Harvard that (pg. 8)

Suppose we want to draw from our posterior distribution $p(\theta|y)$, but we cannot sample independent draws from it. For example, we often do not know the normalizing constant.

I was wondering why not knowing the normalizing constant implies that we can only sample dependent draws. Furthermore, if we knew the normalizing constant, what makes it imply we can sample independent draws?

  • $\begingroup$ Although it is not the same question, this earlier entry on X validated could help. $\endgroup$
    – Xi'an
    Nov 21, 2015 at 14:19

1 Answer 1


This is mostly unrelated: missing normalising constant and dependence have no logical connection. That is to say, you may have a completely defined density and be unable to produce iid samples, and you may have a density with a missing constant and be able to produce iid samples.

If you know a density $f(\cdot)$ up to a normalising constant,$$f(x)\propto p(x)$$there are instances when you can draw independent samples, using for instance accept-reject algorithms: if you manage to find another density $g$ such that

  1. you can simulate from $g$
  2. there exists a known constant $M$ such that$$p(x)\le Mg(x)$$

then the algorithm

  simulate y~g(y)
  simulate u~U(0,1)
until u<p(y)/Mg(y)

produces iid simulations from $f$, even though you only know $p$.

For instance, if you want to generate a Beta B(a+1,b+1) distribution from scratch [with $a,b\ge 1$], the density up to a normalising constant is $$p(x)=x^a(1-x)^b\mathbb{I}_{(0,1)}(x)$$ which is bounded by $1$. Thus, we can use $M=1$ and $g(x)=1$, the density of the uniform distribution in an accept-reject algorithm:


produces a sample [with random size] that is i.i.d. from the Beta B(3.3,4.4) distribution. Histogram of 9781 simulations from a Be(3.3,4.4) distribution with theoretical density

In practice, finding such a $g$ may prove a formidable task and an easier approach is to produce simulations from $f$ by Markov chain algorithms, such as the Metropolis-Hastings algorithm. Given $p$, you chose a conditional distribution with density $q(y|x)$, for instance a Gaussian centred at $x$, and run the following algorithm:

pick an arbitrary initial value x[0]
for t=1,...,T
  simulate y~q(y|x[t-1])
  simulate u~U(0,1)
  compute mh=p(y)*q(x[t-1]|y)/p(x[t-1])q(y|x[t-1])
  if (u<mh) set x[t]=y
  else set x[t]=x[t-1]

It will generate a sequence $x_0,x_t,\ldots$ which has the following properties

  1. It is Markov dependent, that is, the distribution of $x_t$ given the past $x_i$'s depends on the realisation of $x_{t-1}$
  2. It is ergodic wrt to $f$, that is, the distribution of $x_t$ converges to the distribution with density $f$.

To get back to the Beta example, if I use for $q(\cdot|x)$ the normal density with mean $x$ and variance $1$, the code of the Metropolis-Hastings algorithm is

for (t in 2:N)

(where the constant in dbeta(x,a,b) does not matter). The result gives a perfect fit to the Beta B(3.3,4.4) distribution, while the sample is correlated. MCMC sample of size 10⁶ for a Beta B(3.3,4.4) target

Another possibility is to use the uniform U(0,1) as transition, independently of the current value:

for (t in 2:N)

with a similar great and even greater fit. Once again, the sample is made of dependent random variables.

MCMC sample of size 10⁶ for a Beta B(3.3,4.4) target

  • $\begingroup$ Thanks! I was wondering, if I used a function in R to sample from say a beta distribution, using rbeta(), why are these independent? What exactly causes dependence? $\endgroup$
    – user123276
    Nov 19, 2015 at 7:38
  • 2
    $\begingroup$ The common generators are (a) pseudo-random generators (b) that produce independent outputs, in the sense that any statistical test of independence will validate the independence of a sample produced by those generators. $\endgroup$
    – Xi'an
    Nov 19, 2015 at 8:02
  • $\begingroup$ Does this extend to higher dimensions trivially? $\endgroup$ Aug 12, 2019 at 17:52
  • $\begingroup$ @bringingdownthegauss: most of the above applies to any dimension, even though the examples are unidimensional. $\endgroup$
    – Xi'an
    Aug 13, 2019 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.