# What is the definition of a Gaussian random variable?

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?.

• 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 2:27
• 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 2:42
• @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 2:51
• @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 3:03
• 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 3:09

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.