Radon-Nikodym and Conditional Expectation Construction

Welcome back. In our previous lesson, we established the formal, measure-theoretic definition of conditional expectation, $E[Y|\mathcal{G}]$. We saw that it is a random variable, not a number, and that it rigorously resolves the paradoxes that arise when trying to condition on events of zero probability. We concluded by noting that the existence of this object is guaranteed by the Radon-Nikodym theorem.

Today, we will unpack that guarantee. Our focus is to understand precisely how the Radon-Nikodym theorem provides the concrete mechanism for constructing $E[Y|\mathcal{G}]$. This lesson will forge the crucial link between the abstract definition of conditional expectation and the powerful machinery of densities and derivatives we developed in the first module. By the end, you'll see how conditional expectation is, in essence, a specific kind of Radon-Nikodym derivative.

The Construction: From a Random Variable to a Measure

Let's begin with the setup from last time. We have a probability space $(\Omega ,\mathcal{F},P)$, an integrable random variable $Y$, and a sub-$\sigma$-algebra $\mathcal{G}\subset \mathcal{F}$ representing our available information. We are searching for a $\mathcal{G}$-measurable random variable, which we'll call $Z$, that satisfies the defining property:
${\int }_{A}ZdP={\int }_{A}YdP\text{for all }A\in \mathcal{G}$
The constructive proof of existence for $Z$ hinges on a clever idea: we use the random variable $Y$ to define a new measure on our information space $(\Omega ,\mathcal{G})$.

Let's define a function $\nu$ that maps sets in $\mathcal{G}$ to real numbers. For any set $A\in \mathcal{G}$, we define:
$\nu (A):={\int }_{A}YdP$
For now, let's assume $Y$ is non-negative, so $\nu (A)$ will also be non-negative. This function $\nu$ is, in fact, a measure on the measurable space $(\Omega ,\mathcal{G})$. It captures the expected value of $Y$ when restricted to various events $A$ that are "knowable" under the information structure $\mathcal{G}$.

Checking the Prerequisite: Absolute Continuity

We now have two measures defined on the same space $(\Omega ,\mathcal{G})$: our new measure $\nu$ and the original probability measure $P$ (or, more formally, $P$ restricted to the sub-$\sigma$-algebra $\mathcal{G}$).

For the Radon-Nikodym theorem to apply, we must verify that $\nu$ is absolutely continuous with respect to $P$, denoted $\nu \ll P$. This means we need to show that for any set $A\in \mathcal{G}$, if $P(A)=0$, then it must follow that $\nu (A)=0$.

Let's check this condition. By definition, $\nu (A)={\int }_{A}YdP$. A fundamental property of the Lebesgue integral is that the integral of any function over a set of measure zero is zero. Therefore, if $P(A)=0$, then ${\int }_{A}YdP=0$, which means $\nu (A)=0$. The condition holds.

The following video provides a concise and clear explanation of this exact construction.

Applying the Radon-Nikodym Theorem

Since we have established that $\nu \ll P$ on the space $(\Omega ,\mathcal{G})$, the Radon-Nikodym theorem applies. It guarantees the existence of a $\mathcal{G}$-measurable function—let's call it $Z$—which acts as the "density" of $\nu$ with respect to $P$. This function $Z$ is the Radon-Nikodym derivative, $Z=\frac{d\nu }{dP}$.

The theorem states that for any set $A\in \mathcal{G}$, the measure $\nu (A)$ can be recovered by integrating this derivative $Z$ over the set $A$ with respect to the measure $P$:
$\nu (A)={\int }_{A}ZdP$
Now, let's combine this with our original definition of $\nu (A)$. We have two ways of expressing $\nu (A)$:

1. $\nu (A)={\int }_{A}YdP$ (our definition)
2. $\nu (A)={\int }_{A}ZdP$ (from the Radon-Nikodym theorem)

Equating these gives us:
${\int }_{A}ZdP={\int }_{A}YdP$
This identity holds for all $A\in \mathcal{G}$. Notice that this is exactly the defining property of the conditional expectation of $Y$ given $\mathcal{G}$. The function $Z$ we found is $\mathcal{G}$-measurable (a property guaranteed by the theorem) and it satisfies the integral identity.

Therefore, this Radon-Nikodym derivative is precisely the conditional expectation we've been seeking:
$E[Y|\mathcal{G}]=Z=\frac{d\nu }{dP}$
This is the "concrete mechanism" that connects the abstract definition to a constructive process rooted in the theory of measures and densities.

The following resource provides a formal write-up of this exact line of reasoning.

To reinforce this point, another excellent text summarizes the application in its appendix.

Conclusion

In this lesson, we have demystified the existence of conditional expectation by showing how it is constructed. The Radon-Nikodym theorem is not just an abstract guarantee; it provides a specific recipe.

Key Takeaways:

• The conditional expectation $E[Y|\mathcal{G}]$ is constructed by first defining a new measure, $\nu (A)={\int }_{A}YdP$, on the conditioning $\sigma$-algebra $\mathcal{G}$.
• This new measure $\nu$ is absolutely continuous with respect to the original probability measure $P$.
• The Radon-Nikodym theorem guarantees that $\nu$ has a density (a derivative) with respect to $P$. This derivative is a $\mathcal{G}$-measurable random variable.
• This very Radon-Nikodym derivative, $d\nu /dP$, satisfies the defining properties of conditional expectation and therefore is $E[Y|\mathcal{G}]$.

We now have the complete theoretical picture: the paradox of conditioning on zero-probability events forces us to define conditional expectation as a $\mathcal{G}$-measurable random variable, and the Radon-Nikodym theorem provides the engine to construct it.

In our next lesson, we will bring this full circle. We will apply this machinery to a concrete financial example involving a bivariate normal distribution. You will see how this abstract process of defining measures and taking derivatives rigorously reproduces the familiar formula $f(y|x)=f(x,y)/f_X(x)$ from your econometrics work, solidifying the final piece of the argument for your essay.