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I have a question.

Let Y be a real-valued random variable defined on a probability space ($\Omega$, $F$, $P$) where Y:$\Omega \longrightarrow R$.

Show that the $\sigma$-algebra $\sigma(Y)$ generated by the random variable $Y$ coincides with the $\sigma$-algebra $\sigma(Y^{-1}{(\mathcal{B}}))$ generated by the collection of events $Y^{-1}(\mathcal{B})$= { $ Y^{-1}(B) | B \in \mathcal{B} $ }. The $\mathcal{B}$ is the Borel $\sigma$-algebra.

I don't have much clue about how to approach this question. Could someone comments?

It is just a self study problem that I have.

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1 Answer 1

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Just a game of logic. Use the minimality of the two $\sigma$-algebra's.

Obviously $Y$ is $\sigma(Y^{-1}(\mathcal{B}))$-measurable, so we must have $\sigma(Y) \subset \sigma(Y^{-1}(\mathcal{B}))$ because $\sigma(Y)$ is the smallest $\sigma$-algebra makes $Y$ measurable. For the otehr direction, $\forall A= \{Y\in B\} \in Y^{-1}(\mathcal{B})$, where $B$ is an arbitrary element of $Y^{-1}(\mathcal{B})$, we have $ A \in \sigma(Y)$ by the definition of $\sigma(Y)$. Since $\sigma(Y^{-1}(\mathcal{B}))$ is the smallest $\sigma$-algeba containing $Y^{-1}(\mathcal{B})$, we also have $\sigma(Y^{-1}(\mathcal{B})) \subset \sigma(Y)$.

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  • $\begingroup$ Hello, when you say "obviously Y is $\sigma(Y^{-1} (B))$ measurable", could you explain why? $\endgroup$
    – john_w
    Oct 25, 2019 at 3:15
  • $\begingroup$ $\forall B \in \mathcal{B}$, $\{Y \in B\} \in Y^{-1}(\mathcal{B}) \subset \sigma(Y^{-1}(\mathcal{B}))$, so $Y$ is $\sigma(Y^{-1}(\mathcal{B}))$-measurable. $\endgroup$
    – yuguaw
    Oct 25, 2019 at 3:17
  • $\begingroup$ could you explain why "$\sigma(Y)$" is the smallest sigma algebra that makes $Y$ measurable? $\endgroup$
    – john_w
    Oct 25, 2019 at 3:21
  • $\begingroup$ That's exactly how "$\sigma$-algebra generated by a random variable" defines. See Probability: Theory and Examples Version 5 by Rick Durrett, page 15, Section 1.3 Random Variable $\endgroup$
    – yuguaw
    Oct 25, 2019 at 3:24
  • $\begingroup$ Hello, yes it says $\sigma(Y)$ is the smallest sigma algebra generated by $Y$. They always denote $\sigma(Y)$ as "smallest". But is it possible to know why it is the smallest? how do you know it is the smallest? $\endgroup$
    – john_w
    Oct 25, 2019 at 3:35

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