# Probability and random variable [closed]

Given a probability measure on $\mathbb{R}$, can I assume that it is induced by a random variable $X$ ?

A real random variable $X$ is a measurable function from a measure space $(\Omega , F)$ to $(\mathbb{R}, B)$ where $F$ is the $\sigma$-algebra on $\Omega$ and B is the borel sigma algebra.

The probability measure $P:F\rightarrow[0,1]$ with the proprieties that $P(\emptyset)= 0$ and is $\sigma$-additive. The probability induced on $(\mathbb{R}, B)$ is the function $P_X$ defined by $P_X(A)= P(X^{-1}(A))$, $A$ Borel set.

## closed as off-topic by Shailesh, Sahiba Arora, Xander Henderson, José Carlos Santos, NamasteJun 20 '18 at 22:05

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• ok and moreover, if I have two probability measures on $\mathbb{R}$, can i assume that they are induced by two random variables $X$, $Y$ respectively and and that these two random variables are indipendent? – tommycautero Jun 19 '18 at 11:15
• ok thanks. And would you be able to explain me briefly why? Thank you :) – tommycautero Jun 19 '18 at 11:17
• Yeeeeesssss. DO IIITTTT @AlexFrancisco :) :) – Tony Hellmuth Jun 19 '18 at 11:44
• ahahah ok:) have a nice day! – tommycautero Jun 19 '18 at 11:45
• Why two close votes? I think it's an interesting question. I learned something from drhab's answer. – littleO Jun 19 '18 at 12:02

Starting with a probability space $(\Omega,\mathcal A,\mathsf P)$ a random variable $X:\Omega\to\mathbb R$ is a function that satisfies $X^{-1}(B)\in\mathcal A$ for every Borel set $B\subseteq\mathbb R$.
It induces probability measure $\mathsf P_X$ on measurable space $(\mathbb R,\mathcal B)$ where $\mathcal B$ denotes the $\sigma$-algebra of Borel subsets of $\mathbb R$ and is prescribed by $B\mapsto\mathsf P(X\in B)$. Where $\mathsf P(X\in B)$ is an abbreviation of $\mathsf P(\{\omega\in\Omega\mid X(\omega)\in B\})$
Now if $Q$ is some probability measure on $(\mathbb R,\mathcal B)$ then we can apply the following trick to make it a probability measure induced by a random variable.
Let $(\Omega,\mathcal A,\mathsf P):=(\mathbb R,\mathcal B,Q)$ and prescribe $X:\Omega\to\mathbb R$ by $\omega\mapsto\omega$.
Then $X$ is evidently a random variable and this with: $$\mathsf P_X(B)=\mathsf P(\{\omega\in\Omega\mid X(\omega)\in B\})=Q(B)$$Or shortly:$$Q=\mathsf P_X$$