# What does it mean for a random variable to describe an experiment?

I often hear the expression, random variable (or sequence of rd's) to describes an experiment (or sequence of experiments.
But that sound totally unrigorous to me: So we're given a mapping $X:\Omega \rightarrow \mathbb{R}$ and that is supposed to tell us all about our experiment. But it tells us nothing, since there are no probabilities involved.
Let's assume we say "ok, we fix that" and assign probabilities to $X(\Omega)$; but even the this is all unrigorous since there may be events in $\Omega$ whose probability we dont know (take $\Omega=\{1,2,3\}$ with $X:\ \{1,2\}\mapsto 1,\ \{3\} \mapsto 44$ and $P_X (\{1,2\}):=\frac{2}{3}$ and $P_X(\{3\}):=\frac{1}{3}$: The probability of $\{1\}$ is unknown!).
So how can a random variable describe an experiment, if - 1) there is not even, using the mere definition of an r.d. a probability present -2)there can be elements in the sample space whose probability is undefined, if we decide to fix 1) by assigning probabilities to the image of our r.d. ?

To end it by paraphrasing Russell: "My mathematical lecturers never showed me any reason to suppose probability theory was anything but a tissue of fallacies".

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-1 But $\{\emptyset, \{1,2\}, \{3\}, \Omega\}$ is a $\sigma$-algebra of events in which $\{1\}$ is not an event and so has no probability assigned to it by the probability measure. Also, depending on how you choose to interpret $\{1\}$, $$P\{X = 1\} = P\{X \in \{1\}\} = \frac{2}{3}$$ is perfectly well defined. How about trying a different rant? – Dilip Sarwate Jan 7 '13 at 15:30
Where did you get that Russell's quote? A google search "My mathematical lecturers never showed me any reason to suppose probability theory was anything but a tissue of fallacies" leads only to your posts on math.stackexchange.com? – Learner Jan 7 '13 at 15:41
Intuitively, I think perhaps the issue is that in this formalism, we allow for states of the world that can't be distinguished by our experiment $f$- $\sigma(f)$ morally is the information about $\Omega$ accessible from running $f$. – uncookedfalcon Jan 7 '13 at 15:51
@Learner It was an adaptation of the original quote: Replace "lecturers" with "tutors" and probability theory" with "Calculus" – nus Jan 7 '13 at 15:52
@DilipSarwate I didn't knew that making a critique is forbidden in the sciences (but on the brightside downvotes only attract more viewers wanting to see why that question is downvoted :) (I tacitly assumed the $\sigma$-algebra to be the power set) – nus Jan 7 '13 at 15:53

First you define a probability space $(\Omega,\mathcal F,P)$, where $\mathcal F$ is a $\sigma$-algebra on $\Omega$, and $P$ is a probability measure on $\mathcal F$. You've only told us that $\Omega=\{1,2,3\}$, but you haven't said what $\mathcal F$ is. If you're running into contradictions, you need to be explicit about these things. In your question, you only assign probabilities to $\{1,2\}$ and $\{3\}$, but in your comments you say you're assuming $\mathcal F$ to be the power set of $\Omega$.
• If $\mathcal F=\{\emptyset,\{1,2\},\{3\},\Omega\}$, then everything is fine and $P$ is well-defined by specifying $P(\{1,2\})$ and $P(\{3\})$. It makes no sense to ask for $P(\{1\})$, because $P$ is a function from $\mathcal F$ to $[0,1]$, and $\{1\}$ is not an element of $\mathcal F$.
• If $\mathcal F$ is the power set of $\Omega$, then you have not defined $P$ completely. In other words, you have not specified what $P(\{1\})$ is, so obviously $P(\{1\})$ is unknown! This is nobody's fault but yours.
Anyhow, note that $P$ has nothing to do with any random variable $X$, it is a part of the probability space itself, so there is no need to write it as $P_X$.