 # Let's say we conduct a random experiment. The possible outcomes may be too complicated to describe, Peyton Velez 2022-06-15 Answered
Let's say we conduct a random experiment. The possible outcomes may be too complicated to describe, so we instead take some measurement (in terms of real numbers) from it which are of interest to us. Mathematically we have defined a random variable X from the sample space/measurable space $\left(\mathrm{\Omega },\mathcal{A}\right)$ to $\left(\mathbb{R},\mathcal{B}\right)$ where $\mathcal{B}$ is the Borel sigma field. We then choose an appropriate probability for each Borel set, via a distribution function. Then, how does that distribution function determine the probability space $\left(\mathrm{\Omega },\mathcal{A},P\right)$ in theory or proves its existence? What result is implicitly used here?
I understand the probability space on $\left(\mathbb{R},\mathcal{B}\right)$ is well defined, and that is what practically concerns us, but in theory we also have a probability space $\left(\mathrm{\Omega },\mathcal{A},P\right)$ which pushes forward its measure to the Borel sets. Hence given the distribution on $X$ we are pulling back to $\left(\mathrm{\Omega },\mathcal{A}\right)$ via $P\left(X\in B\right)={P}_{X}\left(B\right)$. But what theorem guarantees that such a space exists?
Edit for clarification:For example, let us take weather $\mathrm{\Omega }$ of a city as the original sample space and the recorded temperature $X$ as the random variable. Suppose the distribution of $X$ is decided upon, based on empirical data. That's fine and now we can answer questions like 'what is $P\left(a'. We may not care about the probability space on $\mathrm{\Omega }$ since all the probabilities we wish to know relate to $X$, but how do we know for sure that there exists a probability space on $\mathrm{\Omega }$ in the first place, and further, how do we know that a probability space on $\mathrm{\Omega }$ exists which would push its probability to yield the distribution of $X$? Unless we know that there exists such a probability space in theory, we will not be able to talk about expectation of $X$, so the knowledge that such a space exists is crucial.
*Further edit: Note that we may not have an explicit complete mathematical model of the weather ever, but that does not bother us. We do have the ability however, of taking measurements of different types and thereby getting distributions related to temperature, pressure, wind speed, historical records etc. So practically we can answer questions related to these measurements. This is so what happens in practice I think. What I want is some mathematical theorem which assures me that there exists a model of the weather space, even though it is hard/difficult/impossible to find, which pushes its probability on to these distributions.
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As you mentioned in the example, a "probability space" is a model of an experiment. In some cases it is easy to create such a model. In the rolling a dice example, one can specify explicitly $\mathrm{\Omega }$, $\mathcal{A}$, and $\mathsf{P}$. In your example, it is not clear at all what $\left(\mathrm{\Omega },\mathcal{A}\right)$ is. So one can "model" it by specifying the joint distribution of all the observables. Suppose that you measure temperature $X$ and pressure $Y$ and specify their joint cdf $F$. Then you set $\mathrm{\Omega }=\mathbb{R}×\mathbb{R}$, $\mathcal{A}=\mathcal{B}\left(\mathbb{R}\right)\otimes \mathcal{B}\left(\mathbb{R}\right)$, and $\mathsf{P}$ such that
$\mathsf{P}\left(\left(\mathrm{\infty },a\right]×\left(\mathrm{\infty },b\right]\right)=F\left(a,b\right).$
Put $X\left(\omega \right)={\omega }_{1}$ and $Y\left(\omega \right)={\omega }_{2}$. Then $X$ and $Y$ has the correct joint distribution:
$\mathsf{P}\left(X\le x,Y\le y\right)=\mathsf{P}\left(\left\{\omega :{\omega }_{1}\le x,{\omega }_{2}\le y\right\}\right)=F\left(x,y\right).$