Random variable with 2 distribution functions Just a question here,
Given a random variable $X$ defined in a probability space, is it possible to have more than one distribution function $F$ ?
 A: The CDF $F$ of a random variable $X$ is unique, since it is defined as
$$F(x) = P(X \le x).$$
A: As you've stated the problem, the answer is "no".  You wrote: "Given a random variable $X$ defined in a probability space, is it possible to have more than one distribution function $F$?"  A probability space is a triple $(\Omega,\mathcal F, P)$, where $\Omega$ is the set of outcomes, $\mathcal F$ is the set of events (i.e. measurable subsets of $\Omega$, and $P$ is a probability measure on $\mathcal F$, and a random variable is a measurable function $X:\Omega\to\mathbb R$ (although sometimes one puts something else in place of $\mathbb R$).  Then we have the cumulative probability distribution function
$$
F(x) = P(X\le x) = P\{\omega\in\Omega : X(\omega)\le x \}.
$$
However, in statistics one often considers a(n often parametrized) family of probability measures $P_\theta$ for $\theta$ in some specified parameter space, with a corresponding parametrized family $F_\theta$ of distribution functions.
A: To elaborate on @Dominik's answer, $\mathbb P(X\leqslant x)$ is really shorthand for
$$\mathbb P\left(X^{-1}((-\infty,x]) \right) = \mathbb P\left(\{\omega\in\Omega : X(\omega)\leqslant x\}\right), $$
where we are working in a probability space $(\Omega, \mathcal F,\mathbb P)$. That is, $\Omega$ is a nonempty set called the sample space, $\mathcal F$ is a $\sigma$-algebra of subsets of $\Omega$ whose elements are called events, and $\mathbb P$ is a probability measure on $\mathcal F$, i.e. a function $\mathbb P:\mathcal F \to [0,\infty]$ which satisfies


*

*$\mathbb P(\varnothing)=0$

*$\mathbb P(\Omega)=1$.

*If $\{E_n : n\in\mathbb N\}$ is a disjoint sequence of events, i.e. $E_n\in\mathcal F$ for all $n$ and $i\ne j$ implies $E_i\cap E_j=\varnothing$, then
$$\mathbb P\left(\bigcup_{n=1}^\infty E_n\right) = \sum_{n=1}^\infty \mathbb P(E_n). $$
A random variable is a function $X:\Omega\to\mathbb R$ which is $(\mathcal F, \mathcal B(\mathbb R))$-measurable, i.e. for every Borel set $B$, 
$$X^{-1}(B) = \{\omega\in\Omega : X(\omega)\in B \}. $$
Since $\mathcal B(\mathbb R)$ is generated by the collection of sets of the form $(-\infty, x]$ (a good exercise to prove!), this is equivalent to
$$X^{-1}((-\infty,x])\in\mathcal F $$
for all $x\in\mathbb R$. This means that $F:\mathbb R\to[0,1]$ with
$$F(x) = \mathbb P(X^{-1}((-\infty,x]),$$
called the distribution function of $X$.


Now, if $X$ is a (absolutely) continuous random variable, i.e. it admits a density $f:\mathbb R\to[0,\infty)$ which satisfies
$$F(x) = \int_{-\infty}^x f(t)\ \mathsf dt$$
for each $x\in\mathbb R$, this density need not be unique. For example, consider a random variable $X$ that is uniformly distributed over $[0,1]$. Then
$$F(x) = \mathsf 1_{\{0\leqslant x\leqslant 1\}} x + \mathsf 1_{\{x>1\}} $$
where $\mathsf 1$ denotes the indicator function, i.e. in general
$$\mathsf 1_A(y) = \begin{cases} 1,& y\in A\\ 0,& y\notin A.\end{cases} $$
Since changing the value of a function at finitely many points does not change its Lebesgue integral, both $\mathsf 1_{(0,1)}(x)$ and $\mathsf 1_{[0,1]}(x)$ are densities of $F$. So you can have more than one density function - although necessarily these must be equal almost everywhere.
