Is there anything special about a transforming a random variable according to its density/mass function? Lets say that $X\sim p$, where $p:x\mapsto p(x)$ is either a pmf or a pdf. Does the following random variable possess any unique properties:
$$Y:=p(X)$$
It seems like $E[Y]=\int f^2(x)dx$ is similar to the Entropy of $Y$, which is:
$$ H(Y):=E[-\log(Y)]$$
It seems like we can always make this transformation due to the way random variables are defined. Has someone studied or run across literature that discusses the properties of random variables transformed by their own distribution functions?
One interpretation I can see is that $E[Y]$ gives the degree of "concentration" of the random variable $X$. How about the variance?:
$$\int (p(x)-\mu_Y)^2p(x) dx$$
This seems to measure the degree of uniformity of the function (e.g., it will be 0 for a uniform RV)
I did a couple numerical studies of some common distributions. See below:








 A: (I'm editing this because it seems you use $F$ to denote the PDF, not the CDF.  Is that so?)
To expand Rahul's comment somewhat, let $X$ be a real-valued random variable with a continuous (atomless) distribution—that is, its CDF $F_X$ is a continuous function.  Let $a < b$ be any two distinct values that $X$ can take on.  Then
$$
P(a \leq X \leq b) = F_X(b)-F_X(a)
$$
Since $F_X$ is monotonically non-decreasing, if $Y = F_X(X)$, then
$$
P(F_X(a) \leq Y \leq F_X(b)) = P(a \leq X \leq b) = F_X(b)-F_X(a)
$$
and more generally, since $F_X$ is onto $(0, 1)$ (because we assume that it is continuous), for $u, v \in (0, 1)$,
$$
P(u \leq Y \leq v) = v-u
$$
Therefore $Y$ is a uniformly distributed random variable between $0$ and $1$.
If, instead, we let $Y = f_X(X)$ (where $f_X$ is the PDF of $X$, and again $X$ is atomless, to keep $f_X$ real-valued), then $E(Y)$ does indeed measure, in some sense, the concentration of $X$.  This question has been visited at least once before, with a somewhat terse response:
Name/significance of integral of the square of a probability density function
For what it's worth.
