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Reading this post, I realize that is possible to have another type of PDF (probability density function) in the special case when the sample space is an Euclidean space.

Usually, we have a probability space $(\Omega, \mathcal{F}, P)$, a random variable $X:\Omega\to\mathbb{R}$ and the PDF of $X$, $f_X:\mathbb{R}\to[0, \infty)$. The whole point of the PDF is that we can integrate it over any borel set $A$ (contained in the codomain of $X$) and get the probability $ P[X\in A] = \int_A f_X\ dx$.

What about the case when we have that $\Omega = \mathbb{R}$ and $\mathcal{F}$ is the Borel sigma algebra over $\mathbb{R}$? Looks like it's easier to define the PDF to be a function $f:\Omega\to[0,\infty)$ such that $P[A] = \int_A f\ dx$. So we still can use this PDF to get the probability of events via integration over borel sets. The idea in the book (mentioned on the other post) is that $\Omega$ can be a manifold, so we integrate the PDF over a subset contained in this manifold to get the probability of that subset. If $X:\Omega\to\mathbb{R}$ is a random variable (not necessarily the map $x\to x$), we could talk about $f_X$, the PDF of $X$, such as described before. But in this sample space, we can use the other PDF $f$. So, instead calculating $\int_Af_X\ dx = P[X\in A]$, we can calculate $\int_{\{X\in A\}}f\ dx = P[X\in A]$. Note that the first integral is over the codomain of $X$, but the second is over the sample space $\Omega$.

This article in particular is an example of what I'm saying. Instead defining real random variables $X_{ij}$ as the entries of a random matrix, it's possible to see the sample space as $\mathbb{R}^{n^2}$ and define a probability distribution on $\mathbb{R}^{n^2}$ to work with.

I want to know if there is some technical detail I should be aware of and how common is this. Thank you.

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Your formula $$\int_{\{X\in A\}}f\ dx = P[X\in A]\tag{1}$$ (I guess you wanted to say it should hold for all Borel $A\subseteq\mathbb R$) can be rewritten as $\int_B f\ dx = P(B)$ for all $B$ in $\sigma(X)$, the smallest sigma-algebra with respect to which the random variable (r.v.) $X$ is measurable. So, condition $(1)$ means precisely that $f$ is the density (that is, the Radon--Nikodym derivative) of the restriction $P|_{\sigma(X)}$ of the measure $P$ to $\sigma(X)$ with respect to the restriction $\lambda|_{\sigma(X)}$ of the Lebesgue measure $\lambda$ to $\sigma(X)$: $$f=\frac{dP|_{\sigma(X)}}{d\lambda|_{\sigma(X)}}. \tag{2} $$ Such a density (when it exists) depends on $X$ only through $\sigma(X)$ and thus usually provides little information on the r.v. $X$, and it provides no information on its distribution. To illustrate this point, suppose that $(2)$ holds for a r.v. $X$ and some "alternative pdf" $f$. Let $Y_1:=1-X$, $Y_2:=X^3$, and $Y_3:=\arctan X$. Then $\sigma(Y_1)=\sigma(Y_2)=\sigma(Y_3)=\sigma(X)$ and hence the same "alternative pdf" $f$ will serve all these r.v.'s: $X,Y_1,Y_2,Y_3$.

Somewhat more generally, suppose in addition that the distribution function (d.f.) $F$ of $X$ is continuous and strictly increasing (c.s.i.). Let $G$ be any other c.s.i. d.f. Let $Y:=G^{-1}(F(X))$. Then the d.f. of the r.v. $Y$ is $G$. On the other hand, $Y$ is a c.s.i. transformation of $X$ and hence $\sigma(Y)=\sigma(X)$. So, $Y$ is served by the same "alternative pdf" $f$ as $X$, even though the d.f. $Y$ is an arbitrary c.s.i. d.f. $G$.

Therefore, I don't see why such a "alternative pdf" would be of use.

As for the linked note by Diaconis, it seems to have nothing to do with formula $(1)$, even though he does let $\Omega=\mathbb R^{n^2}$ (nothing wrong or unusual with that). What Diaconis does in that note to answer the question in its title "What is a Random Matrix?" is just provide a particular construction, via Gram--Schmidt orthogonalization, of the Haar measure on the orthogonal group $O_n$, starting with the standard Gaussian measure on $\mathbb R^{n^2}$, which latter is the product of $n^2$ copies of the standard Gaussian measure on $\mathbb R$.

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  • $\begingroup$ Diaconis considered $\mathbb{R}^{n^2}$ as the sample space with the Haar measure $\mu$, if I understand correctly. For instance, if exists a continuous function $f:\mathbb{R}^{n^2}\to[0,\infty)$ such that $\int_{\mathbb{R}^{n^2}}f\ dx = 1$ and $\int_A f\ dx = \mu(A)$ (for all borel sets $A$), could I call that function a PDF? The PDF I'm contemplating goes in this lines. $\endgroup$
    – Integral
    Commented Feb 10, 2016 at 2:26
  • $\begingroup$ No, as I said, Diaconis considered (as a starting point) the standard Gaussian (not Haar) measure on $\mathbb R^{n^2}$. Also, the Haar measure is defined on the sigma-algebra of Borel subsets of the compact group $O_n$, rather than of $\mathbb R^{n^2}$. Further, I suggest that such imprecise phrases as "the sample space with ... a measure" be avoided. Why do you need a "sample space" if you can have a set? $\endgroup$ Commented Feb 10, 2016 at 3:39
  • $\begingroup$ I have added details on why the "alternative pdf" provides no information on the distribution of $X$. $\endgroup$ Commented Feb 10, 2016 at 3:40
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    $\begingroup$ If by $\int_A f\, dx$ you mean the integral with respect to the Lebesgue measure $\lambda$ (say), then the condition that $\int_A f\, dx=\mu(A)$ for all Borel sets $A\subseteq\mathbb R^{n^2}$ means, by definition, that $f$ is the density of the measure $\mu$ with respect to $\lambda$. Of course, then one can construct any number of r.v.'s $X$ with values in $\mathbb R^{n^2}$ such that $f$ is the pdf of $X$, in the conventional sense. One such construction is letting $X$ be the identity mapping of $\mathbb R^{n^2}$ endowed with the Borel measure $\mu$. However, I don't see anything new here. $\endgroup$ Commented Feb 10, 2016 at 4:05
  • $\begingroup$ Yes, that's what I meant. My confusion started reading the post I mentioned first. This one: math.stackexchange.com/questions/1476568/… $\endgroup$
    – Integral
    Commented Feb 10, 2016 at 14:11

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