Let $(\Omega,\mathbb{F},P)$ be a probability space and $\epsilon_1,…,\epsilon_n$ be real-valued random variables defined on $\Omega$. Now let $\mathbb{D}_n$ be the sigma-algebra generated by $\epsilon_1,…,\epsilon_n$, i.e
$$
\mathbb{D_n}=\sigma(\epsilon_1,…,\epsilon_n)= \sigma \left( \bigcup_{i=1}^n \{ \epsilon_i^{-1}(B)|B\in \mathbb{B} \} \right)
$$
where $\mathbb{B}=\mathbb{B(\mathbb{R})}$ is the Borel sigma-algebra.
With $X_0=0$ define recursively $$
X_k=\alpha X_{k-1}+\epsilon_k \quad \quad \quad \text{for} \quad k=1,…,n$$
I have shown that each $X$ can be defined as
$$
X_k=\sum_{i=0}^{k-1} \alpha^i\epsilon_{k-i}=\epsilon_k+\alpha \epsilon_{k-1}+ \cdot \cdot \cdot + \alpha^{k-1}\epsilon_1
$$
Now the problem is that i am to show that
$$
\sigma(\epsilon_1,…,\epsilon_k)= \sigma(X_1,…,X_k) \quad \quad \quad \text{for} \quad k=1,…,n$$
So far my strategy for showing the above is to first show that $$\left(\bigcup_{i=1}^k \{\epsilon_i^{-1}(B)|B\in\mathbb{B}\} \right) \subseteq \left(\bigcup_{i=1}^k \{X_i^{-1}(B)|B\in\mathbb{B}\} \right)
$$
and conversely $\supseteq$, by taking an arbitrary element in the first and showing that it is in the other.
Now the great question: How is this done?
Best Answer
Hint: there is a bijective linear map $T$ which maps $(\epsilon_1,\dots,\epsilon_k)$ to $(X_1,\dots,X_k)$.
If $B$ is a Borel subset of $\Bbb R$, this allows us to see that $$X_j^{-1}(B)=(X_1,\dots,X_k)^{-1}(B')$$ where $B'=\Bbb R^{j-1}\times B\times \Bbb R^{k-j}$. Indeed, $X_j(\omega)\in B$ if and only if $(X_1(\omega),\dots,X_k(\omega))\in B'$.