Suppose $X, Y, Z$ are random variables that live on $(\Omega,\mathscr{F},\mathbb{P})$ and $A\in \mathscr{F}$ such that $\mathbb{P}(A) >0$. My first question is formally what does the conditional expectation:
\begin{equation*} \mathbb{E}\left[ X | Z, A\right] \end{equation*}
actually mean?
I understand that $\mathbb{E}\left[ X|A\right] = \frac{\mathbb{E}\left[ X1_A\right]}{\mathbb{P}(A)}$ and I also understand that $\mathbb{E}\left[ X | Z\right] = \mathbb{E}\left[ X | \sigma(Z)\right]$ is formally defined via the kolmogorov definition) to be a $\sigma(Z)$-measurable random variable such that
\begin{equation*} \mathbb{E}\left[ \mathbb{E}\left[ X | Z\right] 1_E\right] = \mathbb{E}\left[ X1_E\right] \qquad \forall E\in \sigma(Z) \end{equation*}
So what does it mean for when you condition on both an event $A$ and a random variable $Z$? Is this well defined in some way?
I am asking this first question in order to understand whether or not I can then use the towering property to see why it is the case that I can write something as follows:
\begin{equation*} \mathbb{E}\left[ X | Y= y\right] = \mathbb{E}\left[ \mathbb{E}\left[ X | Z, Y = y\right]| Y = y\right] \end{equation*}
(At least I think I can write something like that....hopefully it is true!). If this second statement above is true, then could you please show how you prove it rigorously via the definition of the kolmogorov expectation?
Many thanks.
I find it much more useful to rewrite the conditioning on event A as conditioning on the random variable that is the indicator for event A (and then evaluating the resulting conditional on the said event), since the measure-theoretic definition of conditional expectation is well-defined for $\sigma$-algebras, and not individual events.
I argue that $\mathbb{E}[\mathbb{E}[X \mid Z, Y] \mid Y] = \mathbb{E}[\mathbb{E}[X\mid Z, Y=y]]$ on $\{Y=y\}$, since the right hand side is a function of $y$, and that is measurable w.r.t $Y$ by definition of $Y$ ($\{y\}$ is a Borel set). They agree on this particular set since integrating over $ \{Y=y\}$, $\mathbb{E}[\mathbb{E}[X\mid Z, Y]\mathbb{1}_{\{Y=y\}}] = \mathbb{E}[\mathbb{E}[X\mid Z, Y=y]\mathbb{1}_{\{Y=y\}}]$. (To see it, you can write $\mathbb{E}[X\mid Z, Y] = h(Z,Y)$, and now when you integrate over only the event $\{Y=y\}$, you can replace that by $h(Z,y)$, also $\mathbb{E}[\mathbb{E}[X]]$ = $\mathbb{E}[X]$).
I think this argument is right, but I would definitely want someone to double check it.