0

I'm wondering what are sufficient conditions I cas ask for a function $f:\mathbb{R} \mapsto \mathbb{R} $ s.t given ab abs.continous R.V $X:\Omega \mapsto \mathbb{R}$, the composition $ f \circ \Omega$ will remain an abs continous R.V .
Is it enough that f will be continous?

Ron Abramovich
  • 618
  • What do you mean by an absolutely cont. random variable? – Jonathan Hole Dec 25 '20 at 17:01
  • there is non negative density function $f_x(t)$ s.t the cummaltive distribution function of X is given by $ F_X(t) = \int_{-\infty}^t f_X(t)$ – Ron Abramovich Dec 25 '20 at 17:05
  • I don't think your suggestion of $f$ continuous is sufficient, as even e.g. the composition of two absolutely continuous functions need not be absolutely continuous: https://math.stackexchange.com/questions/2406796/composition-of-absolutely-continuous-function-on-mathbbr – Prasiortle Dec 25 '20 at 17:21

1 Answers1

1

If $A$ is a Borel set, then $\mathbb{P}\{F(X) \in A\}$ is given by \begin{equation*} \mathbb{P}\{f(X) \in A\} = \int_{F^{-1}(A)} f_{X}(x) \, dx. \end{equation*} Hence the question boils down to: when is it possible to change variables in the right-hand side of the expression above?

There is a somewhat well-known sufficient condition, though I haven't seen it written down formally anywhere. Roughly speaking, if $F^{-1}(\{x\})$ is finite for every $x \in \mathbb{R}$, then $F(X)$ has the following PDF: \begin{equation} \label{E: perron frobenius density} f_{F(X)}(y) = \sum_{x \in F^{-1}(\{y\})} \frac{f_{X}(x)}{|F'(x)|}. \end{equation}

When $f$ is a $C^{1}$ diffeomorphism of $\mathbb{R}$, this follows directly from the change-of-variables formula: \begin{equation*} \int_{F^{-1}(A)} f_{X}(x) \, dx = \int_{A} f_{X}(F^{-1}(y)) |(F^{-1})'(y)| \, dy = \int_{A} \frac{f_{X}(F^{-1}(y))}{|F'(F^{-1}(y))|} \, dy. \end{equation*}

More generally, one can prove the following (e.g. using Sard's Theorem):

If $F : \mathbb{R} \to \mathbb{R}$ is a piecewise $C^{1}$ map (i.e. $C^{1}$ on $\mathbb{R} \setminus D$, where $D$ is a closed set with no accumulation points) with the property that $F^{-1}(N)$ is a Lebesgue null set whenever $N$ is a Lebesgue null set, then, for any random variable $X$ with absolutely continuous law, the random variable $F(X)$ also has absolutely continuous law given by \begin{equation*} f_{F(X)}(y) = \sum_{x \in F^{-1}(\{y\})} \frac{f_{X}(x)}{|F'(x)|}. \end{equation*}

The case where $F$ is piecewise constant shows that the assumption concerning null sets is necessary to get absolute continuity of the law of $F(X)$. (If $F$ is piecewise constant with countable range, then $F(X)$ has a singular law.)

In ergodic theory, the operator $f \mapsto \sum_{x \in F^{-1}(\{y\})} \frac{f_{X}(x)}{|F'(x)|}$ is called the Perron-Frobenius transfer operator; that's where I encountered it first. I've also seen this formula (which is clearly of interest elsewhere) in a probability book for electrical engineers.