Consider two Bayesian updates, where there are two observations. One updates with respect to $x_1$, and then uses the posterior of that as a prior to update with respect to $x_2$. In both cases, $x_1$ and $x_2$ are considered conditionally independent given the parameters (and identically distributed).

The other version updates straight from the start on both examples.

Case a will lead to a posterior $$p(\theta | x_1,x_2) = \frac{p(\theta)p(x_1 | \theta)}{p(x_1)p(x_2)} \times p(x_2 | \theta)$$

Case b will lead to a posterior $$p(\theta | x_1, x_2) = \frac{p(\theta)p(x_1 | \theta)p(x_2 | \theta)}{p(x_1,x_2)}$$

Integrate both sides of both posteriors, and you get 1. Therefore the integrals are equal (to 1). The numerator of the integrals is the same, therefore $$p(x_1,x_2) = p(x_1)p(x_2)!!$$

That seems to me very strange. Should $x_1$ and $x_2$ be independent even when considering their marginal version, and not conditioning on the parameters?


1 Answer 1


This is one of the first results I give in my Bayesian Analysis class. You are confused by notations: using the same symbol $p$ all over is a reason for this confusion and hence let me introduce $\pi(\cdot)$ for the prior, $p_1(x_1|\theta)$ for the density of $X_1$, $p_{2|1}(x_2|\theta,x_1)$ for the conditional density of $X_2$ given $X_1=x_1$, and $p_{12}(x_1,x_2|\theta)$ for the joint density of $(X_1,X_2)$.

  1. In a sequential update of the information on $\theta$,$$\pi(\theta|x_1)= \frac{\pi(\theta)p_1(x_1 | \theta)}{m_1(x_1)}$$and the second update on $\theta$ is\begin{align*}\pi_{x_1}(\theta|x_2) &= \frac{\pi(\theta|x_1)p_{21}(x_2|\theta,x_1)}{m_{2|1}(x_2|x_1)} \\ &=\frac{\pi(\theta)p_1(x_1 | \theta)}{m_1(x_1)}\frac{p_{2|1}(x_2|\theta,x_1)}{m_{2|1}(x_2|x_1)}\\ &= \frac{\pi(\theta)p_1(x_1 | \theta)}{m_1(x_1)}\frac{p_{2|1}(x_2|\theta,x_1)}{\int \frac{\pi(\theta)p_1(x_1|\theta)}{m_1(x_1)}p_{2|1}(x_2|\theta,x_1)\text{d}\theta} \\&= \frac{\pi(\theta)p_1(x_1 | \theta)p_{2|1}(x_2|\theta,x_1)}{\int \pi(\theta)p_1(x_1|\theta) p_{2|1}(x_2|\theta,x_1)\text{d}\theta}\end{align*}
  2. In a joint update, the posterior of $\theta$ is $$\pi(\theta|x_1,x_2)=\frac{\pi(\theta)p_{12}(x_1,x_2|\theta)}{\int \pi(\theta)p_{12}(x_1,x_2|\theta)\text{d}\theta}=\frac{\pi(\theta)p_1(x_1 | \theta)p_{2|1}(x_2|\theta,x_1)}{\int \pi(\theta)p_1(x_1|\theta) p_{2|1}(x_2|\theta,x_1)\text{d}\theta}$$

Therefore both expressions are the same with no assumption on the dependence between both variables. The side result is that $$m_1(x_1)\times m_{2|1}(x_2|x_1)=m_{12}(x_1,x_2)$$

  • 1
    $\begingroup$ thanks. I think you got a typo in the definition of $\pi(\theta | x_1,x_2)$, the second equality, in the integral, you have $x_1$ on both sides of the $|$. $\endgroup$ Jan 25, 2015 at 22:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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