It is known that if $X_i$ are iid, $E(|X_i|) < \infty$ and $E(X_i) = 0$ then $S_n = \sum_1^n X_i$ is a martingale. Suppose all $X_i$ are defined wrt sample space $\Omega$.

I don't understand why $S_n$ is $\sigma(X_1, .. X_n)$-measurable? Specifically, I am confused about the sample space for $S_n$. To be $\sigma(X_1, .. X_n)$-measurable the sample space of $S_n$ should be $\Omega$. I do not see why. Any help would be appreciated.

  • 3
    $\begingroup$ This is essentially by definition. The filtration is defined with respect to a probability space $(\Omega, \mathcal F, \mathbb P)$. The $X_i$ are (measurable) functions on that space. $S_n$ is then, of course, a function with domain $\Omega$. That it is measurable is simply an application (via induction) of a general theorem stating that the sum of measurable functions is measurable. $\endgroup$ – cardinal May 1 '14 at 22:29

Since all the $X_i$'s are defined on $\Omega$ and are real valued, $S_n\colon \Omega\to\mathbb R$ is well-defined as $S_n(\omega):=\sum_{j=1}^nX_j(\omega)$.

The $\sigma$-algebra $\mathcal A$ generated by $X_j,1\leqslant j\leqslant n$ is the smallest (for inclusion) $\sigma$-algebra making the maps $X_j\colon\Omega\to\mathbb R$ measurable. This means, using projections, that $\omega\mapsto (X_1(\omega),\dots,X_n(\omega))$ is measurable. Here $\mathbb R^n$ is endowed with the Borel $\sigma$-algebra. This implies that for each measurable map $f\colon\mathbb R^n\to\mathbb R$, the map $f(X_1,\dots,X_n)$ is $\mathcal F$-measurable. Using this with $f(x_1,\dots,x_n)=\sum_{j=1}^nx_j$, we are done.


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