Prove a result on expectation with 2 random variables

Question

Part 1

I wish to show that:

$$E(X)=E[(X|Y=y1)*Pr(Y=y1)+(X|Y=y2)*Pr(Y=y2)]$$

where the random variable Y can take 2 possible values.

As many comments and answers have suggested, this may be technically confusing and ill posed, so as an extension I am posting the motivation for writing something like this:

Suppose I want to calculate $E(S(T))$ Where $S(T)$ represents the stock price at time T. However, it so happens that $T$ is also a random variable, so I want to be able to write this as:

$$E[S(T1)*1(T=T1)+S(T2)*1(T=T2)]$$

Note that $S(T1)$ and $S(T2)$ are still random and are stock prices in the future.

Answer 1

I think I disagree with @gunes answer: $X|Y=y$ can be notation used to denote the random variable $X$ conditional on $Y=y$, and although it may not necessarily exist, it does in certain scenarios and has accordingly been used as notation (see notation in the example of the wiki page for conditional continuous distributions).

To answer your question, I will assume that $Y$ is a discrete RV, so that $\mathbb{1}[Y=y]$ makes sense (else it has measure zero and $X|Y=y$ may not be defined or may not make sense). Then $W=\mathbb{1}[Y=y]$ is the rv that is $1$ with probability $p = P(Y=y)$, and $0$ with probability $1-p$.

Independence is defined as $f_{X,Y} = f_Xf_Y$ assuming densities exist (let's assume so). In our case, let $Z = (X|Y=y)$. Then $$f_{Z,W}(z,w) = P(W=w|Z=z)f_{Z}(z) = f_{W}f_{Z} \iff P(W=w|Z=z) = f_W(w)$$

Recalling our notation, we have

$$P(W=w|Z=z) = P(\mathbb{1}[Y=y] = w | (X|Y=y))$$

If $w=1$, we have $P(Y=y | X|Y=y)$. Then clearly if $X=Y$, we have that $$P(Y=y|Y|Y=y) = P(Y=y|Y=y) = 1 \neq P(Y=y) = P(W=1) = p$$

so no, they are not independent, and you can come up with other examples where this also fails.

EDIT

Given your massive edits, I believe you are asking about the property that for an event $B$, $E[Y|B] = E[Y\mathbb{1}[B]]/P(B)$. Suppose that $T$ is either year 1 or year 2 respectively. Then

$$E[S] = E[S*\mathbb{1}[T = 1]] + E[S*\mathbb{1}[T = 2]]$$

It's unclear what $S(T)$ means, hence why we typically don't use that notation with random variables, and precisely introduce random variables defined as conditional on others, ie S|T. If you want to assume that $T$ only affects $S$ through some function $S(t)$, then you can do what you wrote in your question.

Answer 2

X|Y=y is the random variable X when conditioned on the realization of Y=y.

$X|Y=y$ is a notation used inside expressions such as probability, expected value, variance etc. It is not a random variable.

My intuition very strongly agrees, whether Y is y or not should have nothing to do with what I draw from X when I condition on Y=y. However I am not able to formalize it.

When you condition on $Y=y$, you sample from $p_{X|Y}(x|y)$, so you already acknowledged that $Y=y$.