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I have trouble understanding the definition of a random variable:

Let $(\Omega, \cal B, P )$ be a probability space. Let $( \mathbb{R}, \cal R)$ be the usual measurable space of reals and its Borel $\sigma$- algebra. A random variable is a function $X : \Omega \rightarrow \mathbb{R}$ such that the preimage of any set $A \in \cal R$ is measurable in $\cal B$: $X^{-1}(A) = \{ w: X(w) \in A \} \in \cal B$. This allows us to define the following (the first P is the new definition, while the 2nd and 3rd Ps are the already-defined probability measure on $ \cal B$):

$ P(X \in A) = P(X^{-1}(A)) = P(\{ w : X(w) \in A \}) $

$ P(X = x ) = P(X^{-1}(x)) = P(\{ w : X(w) = x \}) $

Does it mean that the first P (the leftmost one) is a new probability measure defined on $( \mathbb{R}, \cal R)$? (in contrast with the 2nd and 3rd Ps defined on $(\Omega, \cal B)$)

What is $(\Omega, \cal B, P )$ for a binomial distribution for example?

In which cases do 2 random variables share the same probability space $(\Omega, \cal B, P )$?

usual me
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  • "the first P is the new definition, while the 2nd and 3rd Ps are the already-defined probability measure on B"... Every $P$ (even the first one) is the probability measure on $(\Omega,\mathcal B)$. – Did Oct 06 '13 at 07:41
  • I am confused because from this page it looks like the first $P$ is a pushforward measure on $ ( \mathbb R, {\cal R} ) $. Am I missing something? – usual me Oct 10 '13 at 08:05
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    Note that $P(X\in A)$ is a shorthand for $P(B)$ where $B=[X\in A]$, that is, $B={\omega\in\Omega\mid X(\omega)\in A}$, hence $B$ belongs to $\mathcal B$, not to $\mathcal R$ (except when $(\Omega,\mathcal B)$ is $(\mathbb R,\mathcal R)$, naturally...). – Did Oct 10 '13 at 09:18
  • Got it! I made a confusion between 2 notations: $P(X \in A)$ (which is a shorthand as you explained) and $P(A)$ (which would imply a probability measure over $( \mathbb R, {\cal R }) $ ) – usual me Oct 10 '13 at 09:52

1 Answers1

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The probability of an event $X \in A$ is, by definition, the $P$-measure of the set of "outcomes" $\omega$ for which $X(\omega)$ is in $A$. Strictly speaking, all events are measurable subsets of the sample space, but it's usually simpler to speak of events involving random variables without explicitly mentioning the sample space.

For example, if your random variable $X$ is the number of successes in $n$ independent Bernoulli trials with probability of success $p$ in each one (and thus has the binomial($n$,$p$) distribution), the sample space $\Omega$ could be $\{0,1\}^n$ (where $0$ corresponds to failure and $1$ to success). $\mathcal B$ would be all subsets of $\Omega$, and $P$ gives each outcome with $k$ 1's and $n-k$ 0's probability $p^k (1-p)^{n-k}$.

But then if you want to relate this to some other random variable $Y$ that is not determined by those same Bernoulli trials, you'll need a bigger sample space, where each outcome consists not just of the outcomes of those $n$ trials but also something else that determines $Y$.

EDIT: This is what distinguishes probabilists from real analysts. The analyst is studying real-valued functions on a given space $\Omega$ with a given $\sigma$-algebra $\mathcal B$ and probability measure $P$.
The probabilist will use the rigourous definition of the random variable $X$ in terms of $\Omega$ and $\mathcal B$ if necessary, but really thinks of $X$ in terms of a quantity involved in some (actual or imagined) experiment, and he/she is willing to change $\Omega$ and $\mathcal B$ in midsentence if that becomes convenient.

Robert Israel
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  • Thank you. I think I understand why we need to have $X^{-1}(A) \in \cal B$ in order to be able to write $P(X \in A)$. But why do we need $A \in \cal R$? Why isn't $A \subset \mathbb R$ enough? – usual me Oct 07 '13 at 14:48
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    For a complicated $X$, it may be very difficult to determine exactly which $A \subseteq \mathbb R$ have $X^{-1}(A) \in \mathcal B$. $A$ being Borel is a good sufficient condition for that. Non-Borel sets will hardly ever come up in practice. – Robert Israel Oct 07 '13 at 15:01