Let $x_n$ be a sequence of random variables on $(\Omega,F,P)$ and $a_n$ be a sequence of real numbers. For any $1\leq i<k<\infty$, I want to evaluate if
$$\sigma(a_mx_m: i\leq m\leq k)\subseteq \sigma(x_m: i\leq m\leq k).$$
My attempt
The sigma-algebra generated by each element of $a_n$ is the trivial sigma-algebra. Then
$$\sigma(a_mx_m: i\leq m\leq k)=\sigma(\cup_{m=i}^k\sigma(a_mx_m))\subseteq \sigma(\cup_{m=i}^k(\sigma(x_m)\cup \{0,\Omega\}))=\sigma(\cup_{m=i}^k \sigma(x_m))$$
if $\sigma(a_mx_m)\subseteq \sigma(a_m,x_m)$ holds (which I'm not sure).
This implies the above result. *The inclusion was justified in the answer of this question (here) (but I am still confused).
Question Do you have any alternative way to show this or to show the mentioned inclusion above?
Best Answer
The general result holds:
Let $Y$ be a random variable with values in space $(E,\mathcal B(E))$, $f:E \to E$ be borel function. Then $\sigma(f(Y)) \subset \sigma(Y)$.
You should know that
$\sigma(Y) = \{ Y^{-1}[A] ; A \in \mathcal B(E) \}$
while
$\sigma(f(Y)) = \{ (f \circ Y)^{-1}[A] ; A \in \mathcal B(E) \} = \{ Y^{-1}[f^{-1}[A]] ; A \in \mathcal B(E)\}$
Since $f^{-1}[A]$ is a borel set, too ($f$ was a borel function), we know that every set $Y^{-1}[f^{-1}[A]]$ is also $Y^{-1}[B]$ for some borel $B$, so every $C \in \sigma(f(Y))$ is also in $\sigma(Y)$, so $\sigma(f(Y)) \subset \sigma(Y)$.
In your case $Y=(X_i,...,X_k)$ and $f(x_i,...,x_k) = (a_ix_i,...,a_kx_k)$ is a borel function (even continuous), so your result holds.
Note that when $f$ is invertible and $f^{-1}$ is borel, the other inclusion holds, too (since $Y= f^{-1}(f(Y))$, so $\sigma(Y) = \sigma(f^{-1}(f(Y))) \subset \sigma(f(Y)) \subset \sigma(Y)$).
By that we get : if every $a_i,...,a_k \neq 0$, then you have equality in those sigma fields (since the inverse then exists)