What is $F_\mu(\mu)$?

Question

Given a measure $\mu$ and the corresponding distribution function $F_\mu$. What happens if one looks at $$F_\mu(\mu)~? $$ One might as well assume that $\mu <<\lambda$.

Thanks in advance!

Answer 1

Distribution functions are really only defined for probability measures on the real line. That is, $$ F_\mu(x)= \mu((-\infty,x]), $$ in which case the distribution function is a map $F_\mu:\mathbb{R} \to [0,1]$. Here we note that the domain is the real numbers and as such is does not make any sense to evaluate the function in a probability measure $\mu$.

If your are instead suggesting we look at the push-forward measure of $F_\mu$ on the Borel probability measure $\mu$, then its another scenario and it makes perfect sense. In this case what you are looking for is probably something like the following: Proposition 3.1 in this pdf. If we insist on avoiding talking about random variables, we can replicate the arguments from the proposition in the following manner:

In your notation and using proposition from the above link, it would become something like this: Note that for any set $(-\infty,a]$ we have that \begin{align*} F_\mu(\mu)((-\infty,a]) :&= \mu(F_\mu^{-1}((-\infty,a]))) \\ &= \mu( \{x\in \mathbb{R}: F_\mu(x)\in (-\infty,a]\}) \\ &=\mu( \{x\in \mathbb{R}: F_\mu(x)\leq a\}) \\ &= \left\{ \begin{array}{ll} \mu(\emptyset)=0, & \text{if } a<0 \\ \mu( \{x\in \mathbb{R}: F_\mu(x)\leq a\}), & \text{if } 0 \leq a \leq 1\\ \mu(\mathbb{R})=1, & \text{if } a>1\end{array} \right., \end{align*} Let $F_\mu^{\leftarrow}$ denote generalized inverse of $F_\mu$ given by $$ F_\mu^{\leftarrow}(y) =\inf\{ x\in\mathbb{R}: F_\mu(x)\geq y \}. $$ Now assume that $F_\mu$ is continuous and let $0\leq a \leq 1$. First realize that the generalized inverse is strictly increasing on $\text{range}(F_\mu)=[0,1]$ (c.f. prop. 2.3 (7)) implying that \begin{align*} \mu( \{ x \in \mathbb{R}: F_\mu(x)\leq a\}) &=\mu( \{ x \in \mathbb{R}: F_\mu^{\leftarrow}(F_\mu(x))\leq F_\mu^{\leftarrow}(a\}) \\& =\mu( \{ x \in \text{supp}(\mu): F_\mu^{\leftarrow}(F_\mu(x))\leq F_\mu^{\leftarrow}(a\}) \end{align*} where we in the last equality used that the support of $\mu$ has measure 1 (All Borel probability measures on a separable metric space has support of full measure). The support is formally defined by $ \text{supp}(\mu) = \{x\in\mathbb{R}| \forall N\in \mathcal{N}_x: \mu(N)>0\}, $ where $\mathcal{N}_x$ is all open neighbourhoods of $x$.

Finally we have to realize that $F_\mu$ is strictly increasing on $\text{supp}(\mu)$ such that prop. 2.3. (3) yields that \begin{align*} \mu( \{ x \in \text{supp}(\mu): F_\mu^{\leftarrow}(F_\mu(x))\leq F_\mu^{\leftarrow}(a\}) &= \mu( \{ x \in \text{supp}(\mu): x\leq F_\mu^{\leftarrow}(a\}) \\&= \mu( \{ x \in \mathbb{R}: x\leq F_\mu^{\leftarrow}(a\}) \\&= F_\mu(F_\mu^{\leftarrow}(a)) \\&= a \end{align*} where we in the last equality used prop 2.3 (4).

Since sets of the form $(-\infty,a]$ is an intersection stable generator of $\mathcal{B}(\mathbb{R})$ and our measure $F_\mu(\mu)$ coincides with the uniform measure on $[0,1]$ on these sets, they must be equal.