conditional expectation given 2 random variables

Question

Let's say there are random variables $A$ and $B$ being independent. And random variable $X$.

Are there any properties to simplify $\mathbb{E}(X \mid (A,B))$ : expectation of $X$ given $A$ and $B$ ?

In particular, do we have a "simplification", like $\mathbb{E}(X \mid (A,B)) = \mathbb{E}( \mathbb{E}(X \mid A) \mid B)$ ?

I think it's incorrect (take X = A*B) but it seems strange as intuitively this would seem to be correct, so I feel like there must be some formula linking $\mathbb{E}(X \mid (A,B))$ and the conditional expectations given a single variable

thanks !

Answer 1

Notice that the random variable $\mathbb{E}(X|A)$ is $\sigma(A)$-measurable, so, being $\sigma(A)$ and $\sigma(B)$ independent: $$\mathbb{E}\left({\mathbb{E}}\left(X|A\right)|B\right)=\mathbb{E}\left({\mathbb{E}}\left(X|A\right)\right)=\mathbb{E}(X),$$ so it is $\mathbb{P}$-a.e. constant.

On the other hand, for example, if $X$ is $\sigma(A,B)$-measurable and it is not $\mathbb{P}$-a.e. constant, then: $$\mathbb{E}\left(X|\sigma(A,B)\right)=X\neq\mathbb{E}(X).$$

So, where did your intuition fall? When you get $\mathbb{E}(X|A)$, you get the best prediction knowing $A$ of $X.$ Now, when you get $\mathbb{E}(\mathbb{E}(X|A)|B)$, you get the best prediction knowing $B$ of [the best prediction knowing $A$ of $X$] and not the best prediction knowing $A$ and $B$ of $X$. While the first prediction has to be constant (because it is a prevision made on information independent on the quantity you are estimating, and so, knowing useless information, the best prediction you can make is to take the best prediction you can make when you know nothing at all, i.e. the expectation of the quantity) the second could be very well non-constant.

Answer 2

What is certain is that if we call:
$$P_{X/(\mathcal{Y},\mathcal{Z})}(A)=P_X[A\mid (\mathcal{Y},\mathcal{Z})],\quad P_{X/\mathcal{Z}}(A)=P_X[A\mid\mathcal{Z}],\quad P_{X/\mathcal{Y}/\mathcal{Z}}(A)=P_{X/\mathcal{Z}}[A\mid\mathcal{Y}]$$ then: $\quad P_{X/(\mathcal{Y},\mathcal{Z})}=P_{X/\mathcal{Y}/\mathcal{Z}}$

Answer 3

$A$ and $B$ being independent is a common, but actually strong assumption. It implies one special property for the joint probability density function:

$$ f_{AB}(a,b)=f_{A}(a)f_{B}(b) $$

If $X$ is a random variable of its own, the joint probability of $X$, $A$ and $B$ is :

$$ f_{XAB}(x,a,b) $$

But the conditional probability of $X$ given $A$ and $B$ has the following property: $$ f_{X|AB}(x|a,b) = \frac{f_{XAB}(x,a,b)}{f_{AB}(a,b)} $$

And therefore: $$ f_{X|AB}(x|a,b)f_{AB}(a,b) = f_{XAB}(x,a,b) $$

And if $A$ and $B$ are independent: $$ f_{X|AB}(x|a,b)f_{A}(a)f_{B}(b) = f_{XAB}(x,a,b) $$

And from that it's possible to derive some of the identities you may need: $$ E(X|(A,B))= \int_{\Omega_X} x f_{X|AB}(x|a,b) dx = \int_{\Omega_X} x \frac{f_{XAB}(x,a,b)}{f_{AB}(a,b)} dx $$

Now, in a general case I believe $E[X|(A,B)] \neq E[E[X|A]|B]$ Because when computing E[X|A] the result is a function $g(a)$, in the sense that is may algebraically depend on the assumed value of $A$, but the result would be no longer a function of the random variable $B$ (nor of an algebraic value $b$), thus: $$ E[E[X|A]|B] = E[g(a)|B] = g(a) = \int_{\omega_{X}} f_{X|A}(x|a)\, dx $$

The expectation given both $A$ and $B$ is a function $h$ of both algebraic values $a$ and $b$: $$ E[X|(A,B)] = \int_{\Omega_{X|(A,B)}} f_{X|AB}(x|a,b) dx = h(a,b) $$

If however, $X$ was assumed independent of both $A$ and $B$, then $E[X] = E[X|(A,B)] = E[E[X|A]|B]$ because the values of $A$ and $B$ wouldn't matter.

If X is not a random variable with its distribution, but a function of $A$ and $B$, then, the principle above holds: $$ X= w(A,B) \Rightarrow $$ $$ E[X|(A,B)] = \int_{\Omega_{A,B}}w(a,b) f_{AB|AB}(a,b)\, da\, db = w(a,b) $$ $$ E[X|A] = \int_{\Omega_{A,B}}w(a,b) f_{B|A}(b|a)\, db = g(a) $$