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I found a previous question with a very nice answer, but still there is something that is not completely clear to me.

We start from a space $(X, \Sigma)$, endowed with a $\sigma$-algebra, and we let $\Delta (X)$ denote the set of all probability measures over $X$.

Can we say that a specific probability density function is a way to graphically represent an element $p \in \Delta (X)$?

Thank you for your time.

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Kolmin
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  • Typically, there is no such density for a lot of elements of $\Delta(X)$, at least not in the classical sense. But when there is, yes, it can be thought of as a graphical representation of the measure if you like. – Ian May 02 '15 at 14:57
  • Thanks for the comment. You know, it is just to have in mind some representation to have a vague idea of what is going on in more abstract settings. Basically I will try to use this intuition to hopefully get what the topology of weak convergence really is in probabilistic settings (see my previous: http://math.stackexchange.com/questions/1259341/random-variables-and-the-topology-of-weak-convergence). – Kolmin May 02 '15 at 15:06

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"To measure density one must have a way to measure volume as density is the proportion of mass per unit of volume."

Therefore we need a special $\nu\in\Delta(X)$ which is going to be the "volume" measure.

I will do the explanation with the most common example of $X=\mathbb{R}$ (or $\mathbb{R}^n$). On $\mathbb{R}$ (or $\mathbb{R}^n$) we have the Lebesgue measure $\nu$ which is the standard measure of length.

For an element $p\in\Delta(\mathbb{R})$ to have a density with respect to $\nu$ we need that $p$ is absolutely continuous with respect to $\nu$. This means that if $E$ is a measurable set and $\nu(E)=0$ then $p(E)=0$ too. It happens that not all $p\in\Delta(\mathbb{R})$ have a density.

For those $p\in\Delta(\mathbb{R})$ that are absolutely continuous with respect to $\nu$ there is a function $f$ (by Radon-Nikodym's theorem) such that $p(E)=\int_E f(x)d\nu(x)$. That function is what is called the density of $p$ (with respect to $\nu$).

The connection with visualization is that in $\mathbb{R}^n$ you can graph the function $f$. But there is another part of the story.

Suppose that $X$ is not necessarily $\mathbb{R}^n$. But we have a random variable $Y$ (a measurable function) on $X$ that takes values on $\mathbb{R}^n$, $Y:X\to\mathbb{R}^n$. Then one can use $Y$ to export measures on $X$ to measures on $\mathbb{R}^n$, and use those measures to study $Y$. If $E\subset\mathbb{R}^n$ is measurable we can put $P(E):=p(Y^{-1}(E))$. This gives us a measure $P$ on $\mathbb{R}^n$. Since on $\mathbb{R}^n$ we have the special measure, the Lebesgue measure $\nu$ then we can consider (if it exists) the density of $P$ with respect to $\nu$.

Alamos
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  • Thanks. As I wrote in a comment, I think I needed this intuition to hopefully get what the topology of weak convergence really is in probabilistic settings (see my previous: http://math.stackexchange.com/questions/1259341/random-variables-and-the-topology-of-weak-convergence). Now, If you have the feeling that I am partially trying to advertise another question of mine... I have to admit you would not be that far from the truth. :) . Btw, again thanks a lot, in particular for the reference to the Radon-Nikodym theorem, that I am aware of, but I do not really know. – Kolmin May 02 '15 at 15:10