# A Measure Theoretic Formulation of Bayes' Theorem

I am trying to find a measure theoretic formulation of Bayes' theorem, when used in statistical inference, Bayes' theorem is usually defined as:

$$p\left(\theta|x\right) = \frac{p\left(x|\theta\right) \cdot p\left(\theta\right)}{p\left(x\right)}$$

where:

• $$p\left(\theta|x\right)$$: the posterior density of the parameter.
• $$p\left(x|\theta\right)$$: the statistical model (or likelihood).
• $$p\left(\theta\right)$$: the prior density of the parameter.
• $$p\left(x\right)$$: the evidence.

Now how would we define Bayes' theorem in a measure theoretic way?
So, I started by defining a probability space:

$$\left(\Theta, \mathcal{F}_\Theta, \mathbb{P}_\Theta\right)$$

such that $$\theta \in \Theta$$.
I then defined another probability space:

$$\left(X, \mathcal{F}_X, \mathbb{P}_X\right)$$

such that $$x \in X$$.
From here now on I don't know what to do, the joint probability space would be:

$$\left(\Theta \times X, \mathcal{F}_\Theta \otimes \mathcal{F}_X, ?\right)$$

but I don't know what the measure should be.
Bayes' theorem should be written as follow:

$$? = \frac{? \cdot \mathbb{P}_\Theta}{\mathbb{P}_X}$$

where:

$$\mathbb{P}_X = \int_{\theta \in \Theta} ? \space \mathrm{d}\mathbb{P}_\Theta$$

but as you can see I don't know the other measures and in which probability space they reside.
I stumbled upon this thread but it was of little help and I don't know how was the following measure-theoretic generalization of Bayes' rule reached:

$${P_{\Theta |y}}(A) = \int\limits_{x \in A} {\frac{{\mathrm d{P_{\Omega |x}}}}{{\mathrm d{P_\Omega }}}(y)\mathrm d{P_\Theta }}$$

I'm self-studying measure theoretic probability and lack guidance so excuse my ignorance.

• Bayes' Theorem is not about "prior", "posterior", "likelihood", "evidence". Bayes Theorem is about marginal and conditional probabilities. Later research mapped this theorem to the concepts you mention. – Alecos Papadopoulos Jan 10 at 11:09

One precise formulation of Bayes' Theorem is the following, taken verbatim from Schervish's Theory of Statistics (1995).

The conditional distribution of $$\Theta$$ given $$X=x$$ is called the posterior distribution of $$\Theta$$. The next theorem shows us how to calculate the posterior distribution of a parameter in the case in which there is a measure $$\nu$$ such that each $$P_\theta \ll \nu$$.

Theorem 1.31 (Bayes' theorem). Suppose that $$X$$ has a parametric family $$\mathcal{P}_0$$ of distributions with parameter space $$\Omega$$. Suppose that $$P_\theta \ll \nu$$ for all $$\theta \in \Omega$$, and let $$f_{X\mid\Theta}(x\mid\theta)$$ be the conditional density (with respect to $$\nu$$) of $$X$$ given $$\Theta = \theta$$. Let $$\mu_\Theta$$ be the prior distribution of $$\Theta$$. Let $$\mu_{\Theta\mid X}(\cdot \mid x)$$ denote the conditional distribution of $$\Theta$$ given $$X = x$$. Then $$\mu_{\Theta\mid X} \ll \mu_\Theta$$, a.s. with respect to the marginal of $$X$$, and the Radon-Nikodym derivative is $$\tag{1} \label{1} \frac{d\mu_{\Theta\mid X}}{d\mu_\Theta}(\theta \mid x) = \frac{f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, d\mu_\Theta(t)}$$ for those $$x$$ such that the denominator is neither $$0$$ nor infinite. The prior predictive probability of the set of $$x$$ values such that the denominator is $$0$$ or infinite is $$0$$, hence the posterior can be defined arbitrarily for such $$x$$ values.

Edit 1. The setup for this theorem is as follows:

1. There is some underlying probability space $$(S, \mathcal{S}, \Pr)$$ with respect to which all probabilities are computed.
2. There is a standard Borel space $$(\mathcal{X}, \mathcal{B})$$ (the sample space) and a measurable map $$X : S \to \mathcal{X}$$ (the sample or data).
3. There is a standard Borel space $$(\Omega, \tau)$$ (the parameter space) and a measurable map $$\Theta : S \to \Omega$$ (the parameter).
4. The distribution of $$\Theta$$ is $$\mu_\Theta$$ (the prior distribution); this is the probability measure on $$(\Omega, \tau)$$ given by $$\mu_\Theta(A) = \Pr(\Theta \in A)$$ for all $$A \in \tau$$.
5. The distribution of $$X$$ is $$\mu_X$$ (the marginal distribution mentioned in the theorem); this is the probability measure on $$(\mathcal{X}, \mathcal{B})$$ given by $$\mu_X(B) = \Pr(X \in B)$$ for all $$B \in \mathcal{B}$$.
6. There is a probability kernel $$P : \Omega \times \mathcal{B} \to [0, 1]$$, denoted $$(\theta, B) \mapsto P_\theta(B)$$ which represents the conditional distribution of $$X$$ given $$\Theta$$. This means that

• for each $$B \in \mathcal{B}$$, the map $$\theta \mapsto P_\theta(B)$$ from $$\Omega$$ into $$[0, 1]$$ is measurable,
• $$P_\theta$$ is a probability measure on $$(\mathcal{X}, \mathcal{B})$$ for each $$\theta \in \Omega$$, and
• for all $$A \in \tau$$ and $$B \in \mathcal{B}$$, $$\Pr(\Theta \in A, X \in B) = \int_A P_\theta(B) \, d\mu_\Theta(\theta).$$

This is the parametric family of distributions of $$X$$ given $$\Theta$$.

7. We assume that there exists a measure $$\nu$$ on $$(\mathcal{X}, \mathcal{B})$$ such that $$P_\theta \ll \nu$$ for all $$\theta \in \Omega$$, and we choose a version $$f_{X\mid\Theta}(\cdot\mid\theta)$$ of the Radon-Nikodym derivative $$d P_\theta / d \nu$$ (strictly speaking, the guaranteed existence of this Radon-Nikodym derivative might require $$\nu$$ to be $$\sigma$$-finite). This means that $$P_\theta(B) = \int_B f_{X\mid\Theta}(x \mid \theta) \, d\nu(x)$$ for all $$B \in \mathcal{B}$$. It follows that $$\Pr(\Theta \in A, X \in B) = \int_A \int_B f_{X \mid \Theta}(x \mid \theta) \, d\nu(x) \, d\mu_\Theta(\theta)$$ for all $$A \in \tau$$ and $$B \in \mathcal{B}$$. We may assume without loss of generality (e.g., see exercise 9 in Chapter 1 of Schervish's book) that the map $$(x, \theta) \mapsto f_{X\mid \Theta}(x\mid\theta)$$ of $$\mathcal{X}\times\Omega$$ into $$[0, \infty]$$ is measurable. Then by Tonelli's theorem we can change the order of integration: $$\Pr(\Theta \in A, X \in B) = \int_B \int_A f_{X \mid \Theta}(x \mid \theta) \, d\mu_\Theta(\theta) \, d\nu(x)$$ for all $$A \in \tau$$ and $$B \in \mathcal{B}$$. In particular, the marginal probability of a set $$B \in \mathcal{B}$$ is $$\mu_X(B) = \Pr(X \in B) = \int_B \int_\Omega f_{X \mid \Theta}(x \mid \theta) \, d\mu_\Theta(\theta) \, d\nu(x),$$ which shows that $$\mu_X \ll \nu$$, with Radon-Nikodym derivative $$\frac{d\mu_X}{d\nu} = \int_\Omega f_{X \mid \Theta}(x \mid \theta) \, d\mu_\Theta(\theta).$$
8. There exists a probability kernel $$\mu_{\Theta \mid X} : \mathcal{X} \times \tau \to [0, 1]$$, denoted $$(x, A) \mapsto \mu_{\Theta \mid X}(A \mid x)$$, which represents the conditional distribution of $$\Theta$$ given $$X$$ (i.e., the posterior distribution). This means that
• for each $$A \in \tau$$, the map $$x \mapsto \mu_{\Theta \mid X}(A \mid x)$$ from $$\mathcal{X}$$ into $$[0, 1]$$ is measurable,
• $$\mu_{\Theta \mid X}(\cdot \mid x)$$ is a probability measure on $$(\Omega, \tau)$$ for each $$x \in \mathcal{X}$$, and
• for all $$A \in \tau$$ and $$B \in \mathcal{B}$$, $$\Pr(\Theta \in A, X \in B) = \int_B \mu_{\Theta \mid X}(A \mid x) \, d\mu_X(x)$$

Edit 2. Given the setup above, the proof of Bayes' theorem is relatively straightforward.

Proof. Following Schervish, let $$C_0 = \left\{x \in \mathcal{X} : \int_\Omega f_{X \mid \Theta}(x \mid t) \, d\mu_\Theta(t) = 0\right\}$$ and $$C_\infty = \left\{x \in \mathcal{X} : \int_\Omega f_{X \mid \Theta}(x \mid t) \, d\mu_\Theta(t) = \infty\right\}$$ (these are the sets of potentially problematic $$x$$ values for the denominator of the right-hand-side of \eqref{1}). We have $$\mu_X(C_0) = \Pr(X \in C_0) = \int_{C_0} \int_\Omega f_{X \mid \Theta}(x \mid t) \, d\mu_\Theta(t) \, d\nu(x) = 0,$$ and $$\mu_X(C_\infty) = \int_{C_\infty} \int_\Omega f_{X \mid \Theta}(x \mid t) \, d\mu_\Theta(t) \, d\nu(x) = \begin{cases} \infty, & \text{if \nu(C_\infty) > 0,} \\ 0, & \text{if \nu(C_\infty) = 0.} \end{cases}$$ Since $$\mu_X(C_\infty) = \infty$$ is impossible ($$\mu_X$$ is a probability measure), it follows that $$\nu(C_\infty) = 0$$, whence $$\mu_X(C_\infty) = 0$$ as well. Thus, $$\mu_X(C_0 \cup C_\infty) = 0$$, so the set of all $$x \in \mathcal{X}$$ such that the denominator of the right-hand-side of \eqref{1} is zero or infinite has zero marginal probability.

Next, consider that, if $$A \in \tau$$ and $$B \in \mathcal{B}$$, then $$\Pr(\Theta \in A, X \in B) = \int_B \int_A f_{X \mid \Theta}(x \mid \theta) \, d\mu_\Theta(\theta) \, d\nu(x)$$ and simultaneously \begin{aligned} \Pr(\Theta \in A, X \in B) &= \int_B \mu_{\Theta \mid X}(A \mid x) \, d\mu_X(x) \\ &= \int_B \left( \mu_{\Theta \mid X}(A \mid x) \int_\Omega f_{X \mid \Theta}(x \mid t) \, d\mu_\Theta(t) \right) \, d\nu(x). \end{aligned} It follows that $$\mu_{\Theta \mid X}(A \mid x) \int_\Omega f_{X \mid \Theta}(x \mid t) \, d\mu_\Theta(t) = \int_A f_{X \mid \Theta}(x \mid \theta) \, d\mu_\Theta(\theta)$$ for all $$A \in \tau$$ and $$\nu$$-a.e. $$x \in \mathcal{X}$$, and hence $$\mu_{\Theta \mid X}(A \mid x) = \int_A \frac{f_{X \mid \Theta}(x \mid \theta)}{\int_\Omega f_{X \mid \Theta}(x \mid t) \, d\mu_\Theta(t)} \, d\mu_\Theta(\theta)$$ for all $$A \in \tau$$ and $$\mu_X$$-a.e. $$x \in \mathcal{X}$$. Thus, for $$\mu_X$$-a.e. $$x \in \mathcal{X}$$, $$\mu_{\Theta\mid X}(\cdot \mid x) \ll \mu_\Theta$$, and the Radon-Nikodym derivative is $$\frac{d\mu_{\Theta \mid X}}{d \mu_\Theta}(\theta \mid x) = \frac{f_{X \mid \Theta}(x \mid \theta)}{\int_\Omega f_{X \mid \Theta}(x \mid t) \, d\mu_\Theta(t)},$$ as claimed, completing the proof.

Lastly, how do we reconcile the colloquial version of Bayes' theorem found so commonly in statistics/machine learning literature, namely, $$\tag{2} \label{2} p(\theta \mid x) = \frac{p(\theta) p(x \mid \theta)}{p(x)},$$ with \eqref{1}?

On the one hand, the left-hand-side of \eqref{2} is supposed to represent a density of the conditional distribution of $$\Theta$$ given $$X$$ with respect to some unspecified dominating measure on the parameter space. In fact, none of the dominating measures for the four different densities in \eqref{2} (all named $$p$$) are explicitly mentioned.

On the other hand, the left-hand-side of \eqref{1} is the density of the conditional distribution of $$\Theta$$ given $$X$$ with respect to the prior distribution.

If, in addition, the prior distribution $$\mu_\Theta$$ has a density $$f_\Theta$$ with respect to some (let's say $$\sigma$$-finite) measure $$\lambda$$ on the parameter space $$\Omega$$, then $$\mu_{\Theta \mid X}(\cdot\mid x)$$ is also absolutely continuous with respect to $$\lambda$$ for $$\mu_X$$-a.e. $$x \in \mathcal{X}$$, and if $$f_{\Theta \mid X}$$ represents a version of the Radon-Nikodym derivative $$d\mu_{\Theta\mid X}/d\lambda$$, then \eqref{1} yields \begin{aligned} f_{\Theta \mid X}(\theta \mid x) &= \frac{d \mu_{\Theta \mid X}}{d\lambda}(\theta \mid x) \\ &= \frac{d \mu_{\Theta \mid X}}{d\mu_\Theta}(\theta \mid x) \frac{d \mu_{\Theta}}{d\lambda}(\theta) \\ &= \frac{d \mu_{\Theta \mid X}}{d\mu_\Theta}(\theta \mid x) f_\Theta(\theta) \\ &= \frac{f_\Theta(\theta) f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, d\mu_\Theta(t)} \\ &= \frac{f_\Theta(\theta) f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_\Theta(t) f_{X\mid\Theta}(x\mid t) \, d\lambda(t)}. \end{aligned} The translation between this new form and \eqref{2} is \begin{aligned} p(\theta \mid x) &= f_{\Theta \mid X}(\theta \mid x) = \frac{d \mu_{\Theta \mid X}}{d\lambda}(\theta \mid x), &&\text{(posterior)}\\ p(\theta) &= f_\Theta(\theta) = \frac{d \mu_\Theta}{d\lambda}(\theta), &&\text{(prior)} \\ p(x \mid \theta) &= f_{X\mid\Theta}(x\mid\theta) = \frac{d P_\theta}{d\nu}(x), &&\text{(likelihood)} \\ p(x) &= \int_\Omega f_\Theta(t) f_{X\mid\Theta}(x\mid t) \, d\lambda(t). &&\text{(evidence)} \end{aligned}

• Why should $\Omega$ be a Borel space instead of some other measure space? – Dave Jan 9 at 22:53
• @Dave Borel spaces are easier to work with for technical reasons, while also being fairly general. For example, conditional distributions of random variables taking values in a Borel space always exist, whereas they might not exist for random variables taking values in a non-Borel space. Fortunately, most spaces in practice are Borel spaces. For example, every Borel subset of a complete, separable metric space is a Borel space. – Artem Mavrin Jan 9 at 23:11
• I just checked the edit, your answer is extremely clear and detailed and helped me a lot, thank you very much for the time and effort @ArtemMavrin. – Blg Khalil Jan 10 at 0:14
• @BlgKhalil yes, you could call it that if you want, and it would be more consistent with the rest of the notation – Artem Mavrin Jan 11 at 21:16
• @BlgKhalil glad to help :) – Artem Mavrin Jan 11 at 21:19