Is it true that $\sigma(X_1,X_1+X_2)=\sigma(X_1,X_2)$?

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Given a sequence $(X_k)_{k\in\mathbb{N}}$ of real centered r.v.'s, and a sequence $(Y_n)_{n\in\mathbb{N}}$ defined by $$Y_n = X_1 + ..+X_n$$

it is stated, in a chapter on martingales of the book I'm studying, that $$\sigma(Y_1,..,Y_m)=\sigma(X_1,...X_m)$$

without any hint for a proof, as if it were trivial.

All I know, in general, is that $\sigma(X_n+X_m)\subseteq\sigma(X_n,X_m)$, which should imply that $\sigma(X_n,X_n+X_m)\subseteq\sigma(X_n,X_m)$. But how to go from here?

The source is: P. Baldi, Stochastic Calculus - An Introduction Through Theory and Exercises, Springer, Section 5.1, page 109.

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Elaborating on my comment:

For a measurable function $f$, $$\sigma(f(X)) \subseteq \sigma(X),$$ since a set $\{\omega : f(X(\omega)) \in B\}$ can be expressed as $\{\omega : X(\omega) \in f^{-1}(B)\}$. This also holds if $X$ is a random vector $X=(X_1, \ldots, X_m)$.


Since $Y_i = X_1 + \cdots + X_i$, we therefore have $\sigma(Y_i) \subseteq \sigma(X_1, \ldots, X_m)$ for $i \le m$. Consequently, $\sigma(Y_1, \ldots, Y_m) \subseteq \sigma(X_1, \ldots, X_m)$.

Conversely, since $X_1 = Y_1$ and $X_i = Y_i - Y_{i-1}$ for $i \ge 2$, we have $\sigma(X_i) \subseteq \sigma(Y_1, \ldots, Y_m)$ for $i \le m$, and thus $\sigma(X_1, \ldots, X_m) \subseteq \sigma(Y_1, \ldots, Y_m)$.