Define a probability space $(\Omega,\cal F,\Bbb P)$ and a $\cal F$ measurable random variable $X$, the conditional expectation given a sub $\sigma$-algebra $\cal F_0 \subseteq \cal F$ is a random variable $X_0=\Bbb E(X|\cal F_0)$ satisfying the following two conditions:

  1. $X_0$ is $\cal F_0$ measurable.
  2. For any $E\in \cal F_0$, it holds that $\int_E {X} d\Bbb P =\int_E {X_0} d\Bbb P$.

My confusion is that, since $X_0$ is a random variable, it must be a function defined on a sample space, then what is the sample space for $X_0$? Further the sample space must be equipped with a $\sigma$-algebra to form a measurable space, so what is the $\sigma$-algebra? Since $X_0$ is $\cal F_0$ measurable, so I suppose the $\sigma$-algebra associated with the sample space is $\cal F_0$?

Thank you!

  • $\begingroup$ Yes, assuming that $X_0\colon F_0 \to \mathbf{R}$ is sensible. Stronger assumptions can make the definition vacuous. $\endgroup$ – nomen Aug 28 '15 at 18:31
  • $\begingroup$ Conditional expectation $X_0 \colon \Omega \to \mathbb{R}$ is random variable defined on the same probability space $(\Omega,\cal F,\Bbb P)$ as random variable $X$. $\endgroup$ – Zoran Loncarevic Aug 28 '15 at 19:32
  • $\begingroup$ @ZoranLoncarevic Thank you for the reply! But I am still confused. If the sample space is still $\Omega$, how does $X_0$ map elements of $\Omega$ to $\Bbb{R}$? For example, suppose $\Omega$ is the outcome of a dice, i.e. $\Omega=\{1,2,3,4,5,6\}$, and $X(\omega) = \omega$ for any $\omega \in \Omega$. If $\cal F_0$ $=\{\emptyset,\Omega, A,A^C \}$ where $A=\{1,3,5\}$, then how $X_0$ should look like? $\endgroup$ – Tony Aug 28 '15 at 20:11
  • $\begingroup$ @ZoranLoncarevic $X(\omega)=\omega$ for every $\omega$ in $\Omega$. $\endgroup$ – Tony Aug 28 '15 at 20:22
  • 2
    $\begingroup$ You can check for yourself that this function satisfies the conditions 1 and 2 from your own definition of conditional expectation. $\endgroup$ – Zoran Loncarevic Aug 28 '15 at 21:52

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