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Let $(S, \mathcal{F}, P)$ be a probability space and $X:S \to \{x_1, ..., x_n\}$ a discrete random variable. We say that $\sigma(X) = \{A = X^{-1}(B),\quad B \quad \hbox{is borelian set}\}$ is the $\sigma$-álgebra generated by the random variable $X$. I would like to understand this $\sigma$-álgebra as a vector space. One attempt is the following:

Let $\delta_{x_i}: S \to \{0,1\}$ be the indicator random variable with $\delta_{x_i}(X) = 1$ if the event $\{X = x_i\}$ occurs and $0$ otherwise. We can show that $\{\delta_{x_1}, \delta_{x_2},..., \delta_{x_n} \}$ is linearly independente. Futhermore: $$\sigma(X) = \hbox{Span}\{\delta_{x_1}, \delta_{x_2},..., \delta_{x_n} \}$$

But I can not understand this because $\sigma(X)$ is a set of sets and $\hbox{Span}\{\delta_{x_1}, \delta_{x_2},..., \delta_{x_n} \}$ is a set of random variables.

Some help?

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$\sigma(X)$ is not a vector space. $\sigma(X)$ is a set of sets. I think of it as the set of all legal queries you can obtain probabilities for. Vector spaces are sets of objects that satisfy axioms that don't apply to Borel sets. A $\sigma$-algebra is not defined over a field; there's nothing requiring that $S$ even consist of numbers! (And all sets in $\sigma(X)$ are subsets of $S$.)

$X$ may be a mapping from $S$ to a vector space but that hardly makes $\sigma(X)$ anything resembling a vector space.

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  • $\begingroup$ What about the pages 13 and 14 of this ee.nthu.edu.tw/cschang/Talk01142008.pdf ? $\endgroup$ – orrillo Oct 14 '17 at 5:56
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    $\begingroup$ I think the author of those slides is mistaken. For example, on slide 14, he says that $\sigma(X)$ is a set of measurable functions of $X$, but then says that $\sigma(X)$ is a $\sigma$-algebra. The two cannot both be true. $\sigma$-algebras are collections of sets that include $S$ and are closed under complementation and countable unions. They don't contain functions. Roughly speaking, $Y$ is a measurable function of $X$ if for every $B$ that is measurable, $Y^{-1}(B) \in \sigma(X)$. Notice that $Y^{-1}(B)$ is a set (the pre-image of $B$). $\endgroup$ – cgmil Oct 14 '17 at 6:16
  • $\begingroup$ I think the same thing. $\endgroup$ – orrillo Oct 14 '17 at 13:13

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