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Here is another question from the book of V. Rohatgi and A. Saleh. I would like to ask help again. Here it goes:

Let $\mathcal{A}$ be a class of subsets of $\mathbb{R}$ which generates $\mathcal{B}$. Show that $X$ is an RV on $\Omega\;$ if and only if $X^{-1}(A)$ $\in \mathbb{R}$ for all $A\in \mathcal{A}$.

I actually do not know how/where to start. I hope someone can help. Thanks.

math_stat_enthusiast
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    You want $X^{-1}(A)\in\mathcal{F}$ where $\mathcal{F}$ is your sigma-algebra on $\Omega$. – Stefan Hansen Oct 29 '13 at 08:38
  • "I actually do not know how/where to start"... This is actually amazing, no? – Did Oct 29 '13 at 08:43
  • @Stefan: Actually the exercise gave it as $\mathbb{R}$. Does that make a difference? – math_stat_enthusiast Oct 29 '13 at 08:43
  • @Did: I am sorry. I am just starting with probability theory right now, and I think I lack the adequate measure theory knowledge. That is why I am trying to learn this on a step-by-step approach. By the way is my question in any way related to this? http://math.stackexchange.com/questions/60623/necessary-and-sufficient-conditions-for-random-variables -- except that in the question I referenced in (which you answered), $\mathcal{A}$ is taken to be $\mathcal{A} = { (-\infty, x]: x \in \mathbb{R} }$ (a special case)? – math_stat_enthusiast Oct 29 '13 at 11:39

2 Answers2

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Let $(\Omega,\mathcal{F},P)$ be a probability space and let $X:\Omega\to\mathbb{R}$ be a mapping from $\Omega$ to $\mathbb{R}$. By definition, $X$ is a random variable if $X^{-1}(A)\in\mathcal{F}$ for all $A\in\mathcal{B}$, where $\mathcal{B}$ denotes the Borel sets on $\mathbb{R}$.

Clearly, if $X$ is a random variable, then $X^{-1}(A)\in\mathcal{F}$ for all $A\in\mathcal{A}$ (why?).

To show the other direction, we can check that

  1. $\Sigma:=\{A\subseteq\mathbb{R}\mid X^{-1}(A)\in \mathcal{F}\}$ is a sigma-algebra on $\mathbb{R}$, and

  2. since $\mathcal{A}\subseteq\Sigma$ we can conclude that..

Stefan Hansen
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  • Correct me if I am wrong, so the first one "if $X$ is a random variable, then $X^{-1}(A)\in \mathcal{F}$ for all $A\in \mathcal{A}$ since this is the definition of an RV right? – math_stat_enthusiast Oct 29 '13 at 08:48
  • No, the definition is that $X^{-1}(A)\in\mathcal{F}$ for all $A\in\mathcal{B}$ but since $\mathcal{A}$ generates $\mathcal{B}$ we must have $\mathcal{A}\subseteq\mathcal{B}$, right? – Stefan Hansen Oct 29 '13 at 09:29
  • Okay. I am sorry if I am causing you a hard time. It is just that I find it hard to get a grasp of these concepts since I am just starting to take these up, and I lack the necessary knowledge in measure theory. so now, I would like to present you what I think I have: (1) If $X$ is a random variable, then by definition, $X^{-1}(A)\in\mathcal{F}$ for all $A\in\mathcal{B}$. Since $\mathcal{A}$ generates $\mathcal{B}$, then $\mathcal{A}\subseteq\mathcal{B}$. Hence, for any $A\in\mathcal{A},A\in\mathcal{B}$. Thus, $X^{-1}(A)\in\mathcal{F}$ as long as $A\in\mathcal{A}$. Is this correct? – math_stat_enthusiast Oct 29 '13 at 11:17
  • @math_stat_enthusiast: Yes that is essentially correct, however, I would word it in another way since it's almost a trivial observation. For instance: "... Since $\mathcal{A}\subseteq\mathcal{B}$ then $X^{-1}(A)\in\mathcal{F}$ for all $A\in\mathcal{A}$ obviously." – Stefan Hansen Oct 29 '13 at 11:40
  • Yey! So my problem is simply the other direction. Now, from what you have give so far, "we can conclude that..." $\sigma \mathcal{A}=\Sigma$. Is this what I need? If so, how do I then proceed with saying that $X$ is an RV? – math_stat_enthusiast Oct 29 '13 at 11:58
  • $\Sigma$ is a sigma-algebra containing $\mathcal{A}$, and $\sigma(\mathcal{A})$ is the smallest sigma-algebra containing $\mathcal{A}$, and hence we must necessarily have $\sigma(\mathcal{A})\subseteq\Sigma$. What does this mean? What is $\sigma(\mathcal{A})$? – Stefan Hansen Oct 29 '13 at 12:01
  • So $\sigma(\mathcal{A})=\mathcal{B}$, i.e. Borel space in $\mathbb{R}$? Thus, the definition of an RV holds and $X$ is an RV? – math_stat_enthusiast Oct 29 '13 at 12:04
  • Yes, it's assumed that $\sigma(\mathcal{A})=\mathcal{B}$ from the very beginning. Hence for every $A\in\mathcal{B}\subseteq\Sigma$ we have $X^{-1}(A)\in\mathcal{F}$ from which we conclude that $X$ is a random variable. – Stefan Hansen Oct 29 '13 at 12:07
  • Thank you so much! I guess I still have to study a lot on measure theory, but at least, I can say I got this problem well. :) – math_stat_enthusiast Oct 29 '13 at 12:09
  • You're very welcome :) – Stefan Hansen Oct 29 '13 at 12:14
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Hint: Pre-images preserve arbitrary unions. i.e. for $\{A_\alpha\}$ where $\alpha \in I$ where $I$ is an index set, then $X^{-1}(\bigcup_{\alpha \in I}A_{\alpha}) = \bigcup_{\alpha \in I}X^{-1}(A_{\alpha})$

Let me know if you need more help.

Gautam Shenoy
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  • @Guatam: Ok. So let me clarify on what I have to do: (1) Showing the "only if" part s easy because it follows from the definition of an RV right? (2) Next, to show the "if" part, I have to show that the collection of pre-images $X^{-1}(A)$ forms a $\sigma$-algebra, then it follows that $X$ is an RV, is that correct? – math_stat_enthusiast Oct 29 '13 at 09:06