Expectations as an Fx Measurable Function

Question

In our lecture notes, $X$ and $Y$ are random variable. From what I understand $E[Y|X]$ is an $F_x$ measurable random variable. In our lecture notes the formula for the conditional expectation was written as:

(1) $E[Y|X]=\int_{-\infty}^\infty y \ \ dF_{y|x} \ \ dy=\int_{-\infty}^\infty y \ \ f(y|x) \ \ dy$

What I want to know: Are there any cases were $E[Y|X]$ is not a function of $X$? For example if I write $E[Y|X=x]$, did I change it to become a scalar or a function of $Y$? I would say that is a constant value (assuming finite means) and will never be a function of $Y$.

Is the pdf $f(y|x)$ which should equal $f(x,y)/f(x)$ random and a function of both $(x,y)$? There is a big distinction between upper case $X,Y$ and lower case $x,y$. So should the pdf be $f(X,Y)=f(X,Y)/f(X)$?

Answer 1

$E[Y|X]$ is always a function of $X$. The other issue is a matter of notation/shorthand: $$E[Y|X] \Longleftrightarrow E[Y|X=x].$$

There is a distinction. That's why we use subscripts. The pdf of $Y|X$ is written $$f_{Y|X}(y|x).$$ Sometimes the subscript will be dropped whenever it is clear from the context.

If I have some random variable $X$, and then I have some function of that $X$, then we usually write $g(X)$. For example, if I have $X$ and I square it, then $$g(X) = X^2.$$ This is a function on the random variable $X$ and is not the pdf of $X$. The pdf of $X$ is still $$f_X(x).$$

Answer 2

In the formulation with which I am most familiar, $E[Y \mid X]$ means $E[Y \mid \sigma(X)]$, where $\sigma(X)$ is the minimal $\sigma$-algebra on $\Omega$ such that $X$ is measurable. Here in general if $\mathcal{A}$ is a sub-$\sigma$-algebra of $\mathcal{F}$ then $E[X \mid \mathcal{A}]$ is any random variable which is $\mathcal{A}$-measurable and satisfies the equality $E[E[X \mid \mathcal{A}]1_A]=E[X 1_A]$ whenever $A \in \mathcal{A}$.

Two relevant theorems:

1. $E[X \mid \mathcal{A}]$ exists and is unique up to modification on a set of probability zero.
2. There exists a Borel measurable function $\phi$ such that $E[Y \mid \sigma(X)](\omega)=\phi(X(\omega))$ with probability $1$.