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Some people define a Gaussian random variable as a random variable that has a Gaussian p.d.f., which is defined (for the univariate case) as

$$ {\displaystyle f(x)={\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}}} $$

Now, this is fine, but $f$ above is not the Gaussian random variable, or is it? A random variable must take values from the sample space $\Omega$ to measurable space, but isn't the Gaussian p.d.f. defined from $\mathbb{R}$ to $\mathbb{R}$? So, what is the formal definition of a Gaussian random variable (i.e. do not tell me that it's a random variable with p.d.f. $f$). I want to know how it is formally defined. For example, a Bernoulli r.v. is defined as

$$ {\displaystyle Y(\omega )={\begin{cases}1,&{\text{if }}\omega ={\text{heads}},\\[6pt]0,&{\text{if }}\omega ={\text{tails}}.\end{cases}}} $$

What is the equivalent definition of a Gaussian r.v.?

I am asking this question after having asked these ones: Can we really compose random variables and probability density functions? and Why is the exact relationship between a Gaussian p.d.f. and its associated probability measure and random variable?.

StubbornAtom
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  • $f$ is density function for Gaussian random variable i.e. probability for random variable to belong to some $B$ set is calculated using $P(B)=\int\limits_{B}f(x)dx$. – zkutch Jul 26 '20 at 02:21
  • @zkutch Right, I know that, but how would you define the r.v. itself? –  Jul 26 '20 at 02:22
  • What is wrong with the definition using p.d.f.? It is completely rigorous (modulo basic knowledge about Lebesgue measure and absolute continuity of measures). – Sangchul Lee Jul 26 '20 at 02:22
  • @SangchulLee Because that definition doesn't explicitly tell me what that r.v. accepts (i.e. domain) and produces (i.e. codomain) and how it transforms inputs to outputs. A r.v. is a function, so it maps inputs to outputs in some way. I want to know this for the Gaussian r.v. –  Jul 26 '20 at 02:23
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    I guess that you are more concerned about how to construct (or realize) a gaussian random variable, rather than how to verify whether a given random variable is gaussian or not. There is a standard method that allows to realize any probability measure on $\mathbb{R}$ as the distribution of a random variable. – Sangchul Lee Jul 26 '20 at 02:27
  • If I correctly understand you would like to know concrete examples of processes/functions which have Gaussian density function? If yes, then take look at https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator or https://en.wikipedia.org/wiki/Diffusion. In pure mathematics central limit theorem gives a lot of possibilities. If I am mistaken, explain, please. – zkutch Jul 26 '20 at 02:29
  • @zkutch Actually, no, I want to know the actual definition of the Gaussian r.v., as I say above in the post. –  Jul 26 '20 at 02:30
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    Gaussian random variable is not defined in a constructive way, so the definition alone does not even entail the existence of any such random variable. This is pretty much how a definition is developed in modern mathematics, i.e., separating the core properties from everything else. For instance, the axiomatic approach to completely ordered field allows us to formalize the field of real numbers $\mathbb{R}$, although it never tells us whether any such object exists. (Of course, we can construct a model of real numbers, which is why we can safely use them without worry.) – Sangchul Lee Jul 26 '20 at 02:42
  • @SangchulLee What's the domain and codomain of a Gaussian r.v.? I am not asking you to define all possible mappings, if that's not possible, but to define it in the way that is possible without saying that it's a r.v. that has a Gaussian pdf, which means zero (or epsilon) to me. The only thing that helps me is that I will then be multiplying Gaussian pdfs rather than other pdfs when e.g. computing joints, but I would rather live without r.v.s at this point, if you can even define them, i.e. you introduce something and you don't even know what you're talking about. –  Jul 26 '20 at 02:45
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    @nbro, Codmain is affirmatively $\Bbb{R}$, but the domain can by any probability space rich enough to host a continuous r.v. However, there is a fairly simple method of constructing a Gaussian r.v. We choose $Ω=(0,1)$, $\mathcal{F}=\mathcal{B}(Ω)$, and $P=[\text{Lebesgue measure restricted to }Ω]$. Let $$F(x)=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x'-\mu)^2}{2\sigma^2}} , \mathrm{d}x'.$$ Since $F$ is strictly increasing with $F(\Bbb{R})=(0,1)=Ω$, it has an inverse function $X:Ω\to\Bbb{R}$. Now we can check that $X\sim\mathcal{N}(\mu,\sigma^2)$. – Sangchul Lee Jul 26 '20 at 02:51
  • @SangchulLee Are you saying that the inverse of a Gaussian c.d.f. is the Gaussian r.v.? –  Jul 26 '20 at 02:54
  • I am saying that, precisely under the above construction of the probability space $(\Omega,\mathcal{F},P)$ and the random variable $X$, this $X$ becomes a Gaussan random variable. Indeed, for any $x\in\mathbb{R}$, $$P(X\leq x)=\text{Leb}({ω\in Ω:X(ω)\leq x})=\text{Leb}({ω\in Ω:ω\leq F(x)})=\text{Leb}((0,F(x)])=F(x),$$ proving that $F$ is the c.d.f. of $X$. Changing any of $X$, $P$ or $\Omega$ will very much likely render this conclusion invalid. – Sangchul Lee Jul 26 '20 at 02:56
  • Lets take cumulative distribution function with Gaussian density as random variable - do you accept such example? – zkutch Jul 26 '20 at 02:57
  • @SangchulLee Why would it be invalidated if we changed any of those? –  Jul 26 '20 at 02:58
  • @SangchulLee I believe your upper limit of integration in the definition of $F(x)$ ought to be $x$ rather than $+\infty$. – Nap D. Lover Jul 26 '20 at 02:59
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    @NapD.Lover, Oh, you are right. I can't edit that comment anymore, though. :s / OP, You are now asking the very heart of the concept of probability space and random variable. Since I do not want to spend my time explaining stuffs that should be already covered in many other textbook in much greater details and/or examples, I would like to direct you to whatever resource that you are studying now. – Sangchul Lee Jul 26 '20 at 03:03
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    Only briefly explaining, changing the function $X$ (which includes changing the set $\Omega$) means that you are considering a new random variable, which of course makes anything that we concluded from the previous r.v. irrelevant. Moreover, $P$ contains all the information about randomness on $\Omega$, so changing $P$ also renders any past conclusions irrelevant at best. If this sounds too abstract, note that conditional probabilities are also probability measures and then recall how the distribution of a r.v. might change through conditioning. – Sangchul Lee Jul 26 '20 at 03:09

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As my comment, possibly, lost behind other ones, let me write it here. Let's take $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ with Lebegue measurable sets and define random variable $X:\mathbb{R} \to\mathbb{R} $ which for each $x$ gives "measured" length for interval $(-\infty, x)$ by formula $X(x)= \frac {1}{{\sqrt {2\pi }}}\int_{-\infty}^{x}e^{-t^2/2}dt$. Such random variable $X$ is Gaussian random variable.

zkutch
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