I am dealing with the following question: given two dependent random variables $X_1,X_2$, I am wondering whether the following equivalence for the generated sigma-algebras holds:
- $$\sigma(X_1)=\sigma(X_1 + c) $$
- $$\sigma(X_1,X_2)=\sigma(X_1 + c,X_2+\mathbb{E}[X_2|X_1])$$
where $c$ is a known constant.
For the first one. I am pretty sure that it holds, as the heuristic is telling us: on LHS "I know" $X_1$ and on the RHS "I know" $X_1$ up to a known constant.
For the second one: the conditional expectation $\mathbb{E}[X_2|X_1]$ is $\sigma(X_1)$-measurable, i.e.
$$ \sigma(\mathbb{E}[X_2|X_1]) \subset \sigma(X_1)$$
and it can be written as $f(X_1)$ a.s., where $f$ is Borel measurable.
So what we need to show is that
$$\sigma(X_1,X_2)=\sigma(X_1 + c,X_2+f(X_1))$$
But as $f$ is deterministic, and given 1., this means that 2. holds, if the following holds
$$\sigma(X_1,X_2)=\sigma(X_1,X_2+X_1)$$
which looks correct to me.
Am I right? Is it enough to say that 2. holds?
I have tried to write it formally with sets but I did not manage to find evidence of 2. At the same time I have not been able to find a counterexample. Any hint or idea would be helpful.
Thank you in advance.
Best Answer
$\sigma(X_1)=\sigma(X_1 + c) $ follows from the fact that $(X_1+c)^{-1}(A)=X_1^{-1}(A-c)$ for any Borel set $A$: $A$ is Borel if and only if $A-c$ is Borel.
It is trivial to check that $\sigma(X_1 + c,X_2+\mathbb{E}[X_2|X_1]) \subseteq \sigma(X_1,X_2)$. For the reverse inclusion we have to show that $X_1$ and $X_2$ are measurable w.r.t. $\sigma(X_1 + c,X_2+\mathbb{E}[X_2|X_1])$. For $X_1$ this is clear. For $X_2$ here is a hint: use the equation $X_2=(X_2+\mathbb{E}[X_2|X_1])- g(X_1+c)$ where $g$ is a measurable function such that $\mathbb{E}[X_2|X_1])=g(X_1+c)$. [Can you see why $g$ exists?].