I am currently self studying probability theory.
My question how would one compute $E(X|2X)$ only knowing that X is an RV?
First attempt I tried to approach it using CE formula:
$E(X|2X)=\sum x\frac{P(X=x \cap 2X=y)}{P(2X=y)}=\sum x\frac{P(X=\frac{y}{2})}{ P(X=\frac{y}{2})}=\sum x$.
However the answer seem off so I tired using CE definition:
$E(X|2X)=E(X|\sigma(2X))=E(X|\sigma(X))=X$.
But I am not sure if $\sigma(X) = \sigma(2X)$, so I don’t if my answer is correct.
PS: I haven’t study measure theory before probability theory.