I always make confusion when a measure has to be changed in some other measure. This time I'm stuck on a change of measure in the definition of conditional expectation of a random variable.

If $Z$ is a r.v. with values in a countable set, we define the conditional law of $X$ given $Z$ as:

$n(z,A)=P(X\in A\ | \ Z = z)=\frac{P(X\in A\ , \ Z=z)}{P(Z=z)}$

Then we define the conditional expectation of $X$ given $Z=z$ as

$E(X \ | \ Z =z )=\int\limits_{\mathbb{R}} x \ n(z,dx)$

So I would like to change this formula $\int\limits_{\mathbb{R}} x \ n(z,dx)$ into this: $\frac{1}{P(Z=z)}\int\limits_{\{Z=z\}} X(\omega) \ P(d\omega)$

Intuitively is quite clear, but formally I don't get the right passage. My try:

$\int\limits_{\mathbb{R}} x \ n(z,dx) = \int\limits_{\mathbb{R}} x \ \frac{P(X\in dx\ , \ Z=z)}{P(Z=z)}=\frac{1}{P(Z=z)}\int\limits_{\mathbb{R}} x \ P(X\in dx, Z=z)$ and now?

How do I switch from the integration in $\mathbb{R}$ to the integration in $\Omega$ (the probability space)? Why I can pull out $Z=z$ from $P(X\in dx, Z=z)$ trasforming it into $\mathbb{1}_{\{Z=z\}}(\omega)$?


This follows from a standard argument which is often used in measure theory: To show that $$ \int_{\mathbb{R}} f(x)\, n(z,\mathrm dx)=\frac{1}{P(Z=z)}\int_{\Omega} f(X)\mathbf{1}_{Z=z}\,\mathrm dP \tag{1} $$ holds for all measurable functions $f$, we first show that it holds for all indicator functions and then we extend to the more general case.

We see that $(1)$ holds for all $f=\mathbf{1}_A$ for some $A\in\mathcal{B}(\mathbb{R})$ since $$ \begin{align} \int_{\mathbb{R}}\mathbf{1}_A(x)\, n(z,\mathrm dx)&=n(z,A)=\frac{1}{P(Z=z)}P(X\in A,Z=z)\\ &=\frac{1}{P(Z=z)}\int_\Omega \mathbf{1}_{A}(X)\mathbf{1}_{Z=z}\,\mathrm dP. \end{align} $$ Now, note that if $(1)$ holds for two measurable functions $f$ and $g$, then $(1)$ also for $\alpha f+\beta g$ for all $\alpha,\beta\in\mathbb{R}$. That is, the functions satisfying $(1)$ constitutes a vector space. Lastly, if $(f_n)_{n\geq 1}$ is a non-decreasing sequence of measurable functions satisfying $(1)$ such that $\sup_n f_n(x)<\infty$ for all $x$, then also $\sup_n f_n$ satisfies $(1)$. The last is a consequence of the monotone convergence theorem.


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.