Sigma-algebras generated by maps

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?

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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 $(X_1(\omega),\dots,X_k(\omega))\in B'$.

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Okay im not sure how to use that information, anyway i have found the lower triangular matrix mapping $(\epsilon_1,...,\epsilon_k)$ to $(X_1,...,X_k)$. –  Martin Dec 7 '12 at 20:26
Okay, im having trouble seeing how $(X_1,...,X_k)^{-1}(B')=X_j^{-1}(B)$ and not equal to $(\Omega,...,\Omega,X_j^{-1}(B),\Omega,...,\Omega)$. But that aside i dont know how the argument of the inclusion is supposed to continue from there. Anyway thanks for taking the time answering the question. –  Martin Dec 7 '12 at 21:13
Note that $X_l^{—1}(\Bbb R)=\Omega$. You can try to show that the LHS is contained in the RHS, and the converse. Do you see how it helps? –  Davide Giraudo Dec 7 '12 at 21:59
I cant see how $$(X_1,...,X_k)^{-1}(B')=( \stackrel{j-1 times}{\overbrace{ \Omega,..., \Omega}},X_j^{-1}(B), \stackrel{k-j times}{\overbrace{\Omega,...,\Omega}})$$ Could help showing equality between the pre-image sigma-aglebras –  Martin Dec 7 '12 at 22:33
Okay, i will give it another try constructing the argument, based on your hints. Anyway thanks for taking the time and helping me, i'm very grateful for that. –  Martin Dec 8 '12 at 16:22

$\sigma( X_1,...,X_k)$ is the smallest $\sigma$-algebra that makes $X_1,...,X_k$ $\sigma(X_1,...,X_k)/\mathbb{B}(\mathbb{R})$-measurable Note that $\epsilon_1,...,\epsilon_k$ can be written as $$\epsilon_i=-\alpha X_{i-1}+X_i \quad \quad \quad \text{ for } \quad i=1,...,k$$ We know that the constant function $h:\Omega\to \mathbb{R}$ given by $h(\omega)=\alpha$ is $\sigma(X_1,...,X_k)$-measurable because $$h^{-1}(B)= \begin{cases} \Omega &\mbox{if } \alpha \in B \\ \emptyset & \mbox{if } \alpha \notin B \end{cases} \quad \quad \quad \text{for} \quad B\in \mathbb{B(\mathbb{R})}$$ Using that any $\sigma$-algebra on $\Omega$ contains both $\Omega$ and $\Omega^c=\emptyset$.
Furthermore we got that the product of two $\sigma(X_1,...,X_k)$-measurable functions is a $\sigma(X_1,...,X_k)$-measurable function. Now with the property that the sum of two $\sigma(X_1,...,X_k)$-measurable functions is a $\sigma(X_1,...,X_k)$-measurable function, we get that every $\epsilon_1,...,\epsilon_k$ is $\sigma(X_1,...,X_k)$-measurable.
By using that $\sigma(X_1,...,X_k)$ makes every $\epsilon_1,...,\epsilon_k$ a measurable function, and that $\sigma(\epsilon_1,...,\epsilon_k)$ is the smallest $\sigma$-algebra making $\epsilon_1,...,\epsilon_k$ measurable, it must be true that $$\sigma(\epsilon_1,...,\epsilon_k) \subseteq \sigma(X_1,...,X_k)$$ Conversely we know that $\sigma(\epsilon_1,...,\epsilon_k)$ makes every $\epsilon_1,...,\epsilon_k$ measurable, and since $$X_i=\sum_{n=0}^{i-1} \alpha^n \epsilon_{i-n}\quad \quad \quad \text{ for } i=1,...,k$$ Which by the same arguments as before makes $X_1,...,X_k$ the sum of $\sigma(\epsilon_1,...,\epsilon_k)$-measurable functions and thus becoming $\sigma(\epsilon_1,...,\epsilon_k)$-measurable. But since $\sigma(X_1,...,X_k)$ is the smallest $\sigma$-algebra making $X_1,...,X_k$ measurable, it must be true that $$\sigma(X_1,...,X_k) \subseteq \sigma(\epsilon_1,...,\epsilon_k)$$ which implies that $$\sigma(X_1,...,X_k) =\sigma(\epsilon_1,...,\epsilon_k)$$ which was what was wanted.