I have conjectured the following:

Let $f:\mathbb{R}\supseteq A \rightarrow B \subseteq \mathbb{R}$ be an injective function. Let $X$ be a random variable with support $A$ and $Y$ be some random variable that is not independent from $X$. Then, $$E[Y | X=x]=E[Y|f(X)=f(x)].$$

Is that correct? If it is correct, is there any "weaker" assumption (weaker than $f$ being injective) that would make this true?

Thanks.

if $Z=f(X)$ and $f$ is injective function so $$\sigma(Z)=\sigma(X)$$

since $Z=f(X)$ so $$\sigma(Z) \subset \sigma(X)$$

and because $f$ is injective so $X=f^{-1}(Z)=g(Z)$ so $$\sigma(X) \subset \sigma(Z)$$

so $\sigma(Z)=\sigma(X)$ or $\sigma(f(X))=\sigma(X)$

now $$E(Y|X)=E(Y|\sigma(X))=E(Y|\sigma(f(X)))=E(Y|f(X))$$

February 19, 2020 20:38 PM

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