A few questions on conditional probability I have a few questions relating to conditional expectation. It is covered very briefly (half a side of A4-size paper) in my lecture notes, but I feel like I should know about it a bit more in depth.
Essentially, the things I know are:


*

*The definition. Given (i) a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, (ii) a sub-$\sigma$-algebra $\mathcal{G} \subseteq \mathcal{F}$ which is generated by a countable set $(G_i\, :\, i \in I)$ with $\bigcup_i G_i = \Omega$, and (iii) a square-integrable random variable $X : \Omega \to \mathbb{R}$, we define the conditional expectation of $X$ given $\mathcal{G}$ by $\boxed{\mathbb{E}(X|\mathcal{G}) = \displaystyle  \sum_{i \in I} \mathbb{E}(X|G_i)1_{G_i}}$, where $\mathbb{E}(X|G_i) = \dfrac{\mathbb{E}(X 1_{G_i})}{\mathbb{P}(G_i)}$.


*The fact that $\mathbb{E}(X|\mathcal{G})$ is the orthogonal projection of $X$ onto the (closed, complete) subspace $L^2(\Omega, \mathcal{G}, \mathbb{P})$ of $L^2(\Omega, \mathcal{F}, \mathbb{P})$.

This is all fine. But a problem has arisen where I need to calculate this explicitly, and it's thrown me a bit. This has made me wonder a few things:


*

*If $Y \in L^2$ is a random variable and $\mathcal{G}=\sigma(Y)$, we have an uncountable generating set for $\mathcal{G}$, namely $\{ Y^{-1}(B)\, :\, B \in \mathcal{B}(\mathbb{R}) \}$. But if $\mathcal{A} \subseteq \mathcal{B}(\mathbb{R})$ generates $\mathcal{B}(\mathbb{R})$, can we therefore take $\sigma(G) = \{ Y^{-1}(A)\, :\, A \in \mathcal{A} \}$? I'd presume we can since $Y^{-1}$ preserves set operations. If so, this becomes compatible with the above definition because we can choose a countable generating set, such as $\{ [a,b)\, :\, a,b \in \mathbb{Q},\ a < b \}$. Is this correct?


*Do we define $\mathbb{E}(X|Y) = \mathbb{E}(X|\sigma(Y))$ for random variables $Y$? If not, how is this defined?


*In Q25J on page 14 here (PDF) we have $(G,X) \sim N\left( (\mu, \nu), \begin{pmatrix} u & v \\ v & w \end{pmatrix} \right)$, $\mathcal{F} = \sigma(G,X)$ and $\mathcal{G} = \sigma(X)$. It is slightly unclear when it says to "find $Y$ explicitly in this case" $-$ I presume that it means I should find $\mathbb{E}( U\, |\, \mathcal{G})$ for a general $\mathcal{F}$-measurable $U \in L^2$, since this is (an) orthogonal projection. But I can't work out what I can say about the distribution of $Y$ in this case.


*If $\mathcal{F} = \sigma(X)$ or $\sigma(X,Y)$ or so on, what can I say about the distribution of an $\mathcal{F}$-measurable function?

I hope I'm not trying to ask too much. Any input at all would be appreciated, so feel free to reply to as small or large a portion of my post as you like.
 A: *

*The projection version is mostly always good,( not always because you need square integrable.  But you are right, unless Y  is discrete it is not going to be of the form in definition 1.

*Yes

*for that exact problem, the joint normal, you can do the  "projection " version by observing that if $X ,Y$ joint normal you can write $Y = a X + Z$ where $Z$ is independent of $X$.  This easy because for joint normal independent = uncorrelated.  Then it is easy to see what $\mathbb E(Y \vert X)$ is .    The projection formulation shows that if $X, Z$ independent, $\mathbb E(Z \vert X) = \mathbb E(Z)$.  You can prove this as follows:  for any function of X, by independence $\mathbb E(Zg(x)) = \mathbb E(Z) \mathbb E(g(x)) $ so $\mathbb E((Z - \mathbb E(Z))g(x))  = 0$, so $Z - \mathbb E(Z)$ is orthogonal to any function of X, and that make $ \mathbb E(Z)$  the projection of $Z$ on the space of $\sigma(X) $ measurable functions.

*Almost nothing.  It's not much of a constraint on the distribution of a random variable.  If X is discrete, it has to be discrete.
