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I am reading A Second Course in Probability by Ross and Peköz. I came across the following question:

1.10.4. Show that if $X$ and $Y$ are real-valued random variables measurable with respect to some given sigma field, then so is $XY$ with respect to the same sigma field.

My attempt was to first show that:

(i) $X+Y$ is measurable

(ii) $cX$ is measurable for any $c\in\mathbb R$

(iii) $X^2$ is measurable.

Then, since $XY = \frac14[(X+Y)^2-(X-Y)^2]$, we just use the above properties.

Property (i) was proven in Chapter 1. However, I am having a bit of trouble showing property (ii) in the case that $c < 0$. From the problem setup, we know that for any $x\in\mathbb R$, $\{X\leq x\}\equiv\{\omega\in\Omega:X(\omega)\leq x\}\in \mathcal F$.

So $\{-X\leq x\}=\{X\geq -x\} = \{X<-x\}^c$, but I don't know if $\{X<-x\}\in\mathcal F$. If I expand a little more, it seems I essentially need to show that $\{X=x\}\in\mathcal F$, but I don't see how that is necessarily true.

I have seen the question Product of two random variables, though in the answers, they assume knowledge of the fact that $cX$ is measurable, and they also seem to be working with the Borel $\sigma$-algebra.

I have also seen these notes, but they omit the case where $c < 0$.

My Question: How do we show that $\{-X\leq x\}\in\mathcal F$ if we already know that $\{X\leq x\}\in\mathcal F$? Do we need to assume something about the sigma field in order to show this for real-valued random variables (e.g. Borel sigma field)?

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  • $\begingroup$ Do you mean that you want to prove this for an arbitrary sigma field on $\mathbb{R}$? Rather than simply the Borel sigma field? $\endgroup$
    – Dasherman
    Commented Jun 7, 2020 at 21:55
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    $\begingroup$ Hint: write $\lbrace X <-x \rbrace = \bigcup_{n} \lbrace X \leq -x-1/n\rbrace$ $\endgroup$
    – Michh
    Commented Jun 7, 2020 at 21:56
  • $\begingroup$ @Dasherman Maybe that is what I am failing to recognize. The question never states anything about the Borel sigma field, so I assume that I should prove it for an arbitrary sigma field. $\endgroup$ Commented Jun 7, 2020 at 22:00

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Note that $\{X<x\}=\cup_{n=1}^\infty \{X\leq x-\frac{1}{n}\}$. Since a Sigma field is closed under countable unions this implies $\{X<x\}\in\mathcal{F}$ for every $x\in\mathbb{R}$.

A general note: If $X:\Omega\to\mathbb{R}$ is a random variable then we have $X^{-1}(B)\in\mathcal{F}$ for every Borel set $B\subseteq\mathbb{R}$. In many books this is actually the definition of a random variable. The definition you are using, that is $\{X\leq x\}\in\mathcal{F}$ for every $x\in\mathbb{R}$, is an equivalent definition. It is good to know both definitions. To prove the equivalence note that the collection of sets $\{B\subseteq\mathbb{R}: X^{-1}(B)\in\mathcal{F}\}$ is a Sigma field on $\mathbb{R}$, and from your definition it is not too hard to show it contains all open subsets of $\mathbb{R}$, hence must contain the whole Borel sigma field.

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