I am trying to relate the conditional probability (a random variable) $P(A|G)$ to the probability measure $P(A)$ ( a measure). It is weird to me that while the probability $P$ is a measure, the conditional probability is a random variable.

Take a probability space $(\Omega, F, P)$ In my book on probability theory, the conditional probability of an event with respect to a sigma algebra: $P(A|G)$ is defined to be a random variable, such that

  1. $P(A|G)$ is $G$-measurable

  2. $\int_g P(A|G)dP=P(A \cap g), \quad \forall g\in G$

I am wondering, what if we take $G=F$ (i.e. the original sigma algebra of the entire space).

I wondered at first, whether we then would get the following $$P(A|G)(\omega)=P(A) \quad \forall \omega \in \Omega $$

But this is of course not what happens, since we get in fact:

$$P(A|G)(\omega)=\cases {1 \quad \text {if }\omega\in A\\0 \quad \text {otherwise}}$$

So then I wondered, what if we take the minimal sigma algebra $G$ instead: $G=\{\emptyset, \Omega\}$.

Then $$\int_\Omega P(A|G)dP=P(A) $$

But this does not restrict $P(A|G)$ at all, except that for a given $A$ it must have a total integral equal to $P(A)$, there are still an enormous amount of possibilities to achieve that (and they are not all equal not even almost-surely).

So my question is: (Under what conditions) is it possible for $P(A|G)$ to be related more closely to $P(A)$? For example, my intuition says that $P(A|G)$ would be the "probability density function" with respect to the probability measure $P$, since this is what it suggests if you look at equation $2$. (even though $\Omega$ does not have to be $R^n$ so it would be a more general type than the classical probability density function, but it would have to be almost-everywhere equal to the actual probability density function in those cases).

  • $\begingroup$ Why do you think that $P(A|G)$ is a random variable? What book are you using? $\endgroup$
    – jbowman
    Dec 27, 2017 at 15:07
  • $\begingroup$ @jbowman, p. 430 of Billinsley 1995 Probability theory and Measure $\endgroup$
    – user56834
    Dec 27, 2017 at 15:12
  • 2
    $\begingroup$ Ah, I see what he's saying. You've misunderstood it somewhat, but I have to leave for work now so, if no-one else explains it, I'll get back to you later. The short form is that its value is a random variable because the event $G$ is random, but in the sense you're thinking about it it is actually a probability distribution just like $P(A)$. $\endgroup$
    – jbowman
    Dec 27, 2017 at 15:19
  • $\begingroup$ @jbowman (Didn't check the book but) as defined in the question, $G$ is a sigma-algebra, not an event $\endgroup$ Dec 28, 2017 at 8:05

1 Answer 1


I will first answer the general question in your title: Can we define for a probability measure, a random variable to be its “density function”? If there is a reference measure that dominates your probability measure (the probability measure is absolutely continuous to some $\sigma$-finite measure) then yes, you can find a random variable that acts as a density. This is a special case of the Radon-Nikodym theorem.

Now to the more specific question about conditional probabilities (where the conditioning is with respect to a sub-$\sigma$-algebra). You can define them using the conditional expectation with respect to sub-$\sigma$-algebras, and most modern textbooks will define them in this way.

Your definition is also valid, but we need to invoke the Radon-Nikodym theorem to show that a random variable with properties 1. and 2. exists.

To apply the theorem, we note that $\forall A\in F$, $P(A\cap \cdot)$, as a measure defined on $G$, is absolutely continuous with respect to $P|_G$, the restriction of $P$ to the $\sigma$-algebra $G$. Regarding both as measures defined on $G$, by the Radon-Nikodym theorem there is a $G$-measurable random variable $X_A$ such that $\forall g\in G$ $$P(A\cap g)=\int_g X_A dP|_G=\int_gX_AdP.$$

Thus $X_A$ satisfies properties 1. and 2. It follows that $X_A=P(A|G)$ $P$-almost everywhere. Since $X_A$ is the density or Radon-Nikodym derivative of $P(A\cap\cdot)$ with respect to $P|_G$, we also have

$$P(A|G)=\frac{dP(A\cap\cdot)}{dP|_G}, \quad P\text{-almost everywhere.}$$

Thus your intuition was correct; the conditional probability can indeed be interpreted as a density of appropriate probability measures.

While for fixed $A\in F$, $P(A|G)$ is a random variable, you can verify that the map $P(\cdot|G)(\omega): F\to[0,1]$ given by $A\mapsto P(A|G)(\omega)$ is a probability measure for $P$-almost all fixed $\omega\in\Omega$.


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