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I just want to be sure about this:

If I read the phrase ' a random variable is exponentially distributed'( which is often said in probability theory and then it is never explictely stated what $X$ actually is and it is just looked at the probability itself) this could rigorously mean, that there is a map $X: \Omega \rightarrow \mathbb{R}$ and that there is a probability space $(\Omega = \mathbb{R}_{>0}, X^{-1}(\mathscr{B}) , P)$ (where $X^{-1}(\mathscr{B})$ is the inverse map of the sigma algebra of the Borel sets) such that $P(\{X^{-1}(E)\}):= \int_\mathbb{R} \chi_E(x) \lambda e^{-\lambda x}dx$?

This is one interpretation that came to my mind by thinking about what it could mean. Could anybody here now explain to me where I am wrong and what it actually means? ( So my goal is to give all the details to the phrase ' a random variable $X$ is exponentially distributed)

Does anybody here have an idea? If there is anything unclear about my question, please let me know.

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A real valued random variable $X$ is exponentially distributed with parameter $\lambda$ if and only if its cumulative distribution function (CDF) is $$F(x)=P(X \leq x) = \left\{\begin{array}{lr}1- e^{-\lambda x}&,x\geq 0\\0&,o.w.\end{array}\right. $$

Note that having a particular distribution (in this case, exponential) doesn't imply $X$ lives on any particular probability space. It does, however, specify $P(X \in B)$ where $B$ is a Borel-measurable subset of $\mathbb{R}$.

Given a CDF $F$, you can easily construct an appropriate probability space for which a random variable with that CDF lives on (this is easy to do - get the probability space $(\Omega,\mathcal{F},P)$ with $\Omega = (0,1)$, $\mathcal{F}$ the Borel sigma algebra on $\Omega$ and $P$ be the lebesgue measure, and let $X(\omega)=\sup\{ y : F(y) < \omega \}$ is probably the simplest construction of a probability space and random variable with distribution $F$). However, you can come up with infinitely many such spaces and random variables (with the same distribution), which can look very different.

Having a particular distribution is not a property of any particular underlying probability space - this is why you can have sequences of random variables which live on different probability spaces "converge in distribution" in a meaningful way!

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  • $\begingroup$ Actually, I noticed that although we don't know what $\Omega$ is, $X: \Omega \rightarrow Y$ comes along with some restrictions on $Y$. Therefore, $Y$ should always be the set that belongs to the distribution, so $\{1,...,n\}$ for discrete uniform, $\mathbb{R}_{\ge 0}$ for exponential distribution and so on, right? $\endgroup$
    – user66906
    Mar 1, 2014 at 22:09
  • $\begingroup$ Typically, you just use a measure space rich enough for your purposes for $Y$ - usually, $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ with the Borel measure. This simplifies things greatly, and with the additive structure on $\mathbb{R}$, you can easily handle addition and multiplication and what not for discrete and continuous random variables jointly. Note that all the sets youd be interested in this case are measurable (and many that you aren't interested in just pull back to a set of measure $0$). $\endgroup$
    – Batman
    Mar 2, 2014 at 3:21
  • $\begingroup$ That is, a random variable does not need to take on all the values in $Y$ (in the cases you're interested in, youd use $(\mathbb{R},\mathcal{B}(R))$ and $Y$ would not take all values). In case of discrete uniform r.v. $X$ on $\{1,\ldots,n\}$ , say with $(\Omega=\{1,\ldots,n\}, \mathcal{F}=2^{\Omega}, P)$ with $P$ being the counting measure divided by n, we can let $X^{-1}(B) = \text{# of elements of } \{1,\ldots,n\} \text{ in }B$ where $B$ Is a borel-measurable set in $\mathbb{R}$. You can see that if $B$ doesn't contain an element of $\{1,\ldots,n\}$, then it just pulls back to the empty set. $\endgroup$
    – Batman
    Mar 2, 2014 at 3:28

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