A random variable $X$ is a measurable function from a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ to the reals:

$X: \Omega \to (-\infty, \infty)$

such that for every Borel set $B$,

$X^{-1}(B)=\{\omega \in \Omega:X(\omega) \in B\} \in \mathcal{F}$

shorthand notation: $\{X \in B\}=\{\omega \in \Omega:X(\omega) \in B\}$

in other words, the fact that value of $X$ belongs to a given Borel has to be an event (has to belong to the set of events $\mathcal{F}$).

Is the probability density function unique for every random variable $X$ in the probabilty space $(\Omega, \mathcal{F}, \mathbb{P})$?

I think the answer is yes, because $\mathbb{P(}{a \le X \le b})$ is defined by the probability function $\mathbb{P}$.

$a \le X \le b$ is an event belonging to $\mathcal{F}$, and $\mathbb{P}$ defines what are probabilities of events in $\mathcal{F}$.



Let it be that $f,g:\mathbb R\to[0,\infty)$ are Borel-measurable functions with: $$\lambda\{x\in\mathbb R\mid f(x)\neq g(x)\}=0$$

where $\lambda$ denotes the Lebesgue measure.

If $f$ serves as PDF for random variable $X$, then so does $g$.

This because for each Borel-measurable set $A$: $$\mathbb P(X\in A)=\int_Afd\lambda=\int_Agd\lambda$$


Firstly, not all random variables have a pdf. But even if they have, the answer is no, the densities are not uniquely determined by the distribution. For example $X \sim U(0,1)$ and $Y\sim U[0,1]$ have the same distribution but different pdf's. It is possible to change the pdf in countable many points and to keep the cdf the same.

However, the cdf uniquely determines the probability law $\mathbb P_X$ of the random variable $X$, that is the probability $$P(X\in B), \quad B \in \mathcal B(\mathbb R)$$ The changes that I mentioned above are in sets of $\mathbb P_X$-measure zero and so do not affect these probabilities.

  • $\begingroup$ The point is that there are many PDFs, but the value of integral $\int_a^b f_x(x) dx$ is equal for all of them? $\endgroup$ – user5539357 Dec 26 '15 at 10:36
  • $\begingroup$ Yes exactly, (assuming a pdf exists). $\endgroup$ – Jimmy R. Dec 26 '15 at 10:36

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