2
$\begingroup$

I have a conceptual question regarding the law of a random variable in probability theory.

Suppose we begin with a probability space $(\Omega,F,P)$. We also endow a measurable space of reals with its Borel $\sigma$-algebra, $(\mathbb{R},B(\mathbb{R}))$.

Given $X$ is a real-valued random variable, we define the distribution measure of law of $X$ as

$$P_X(B)=P(X^{-1}(B))=(P\circ X^{-1})(B)-P(\omega:X(\omega)\in B)=P(X\in B).$$

Then, the law of $X$ is entirely characterized by its cumulative distribution function such that:

$$F_X(x)=P_X((\infty,x])=P(X\leq x).$$

My understanding is:

$P:F\rightarrow [0,1]$.
$P_X:B(\mathbb{R})\rightarrow[0,1].$

Hence, the law of $X$ is a probability density on the measruable space $(\mathbb{R},B(\mathbb{R})).$

(1) Does this mean any two random variables that are from distinct distributions have their own probability distribution on the space $(\mathbb{R},B(\mathbb{R}))$? In other words, is there distinct laws for each distinct random variable?

(2) A random variable is just a measurable function from one measurable space to another, say $(E,A),(F,B)$. Is it possible for the law of $X$ not be a non-negative and $\sigma$-finite measure defined on $(F,B)$. In other words, does the way we define and characterize the law of $X$ in probability have to do with some "nice" properties of reals or the Borel sigma algebra of reals?

$\endgroup$

1 Answer 1

2
$\begingroup$

(1) Yes, by definition $X\stackrel{d}{=} Y$ ($X$ is equal to $Y$ in distribution) iff the laws of $X$ and $Y$ coincide, that is $P_X = P_Y$.

But it is possible that $X \neq Y$ yet $P_X = P_Y$. For example, if $X\sim\mathcal{N}(0,1)$, then $-X \sim \mathcal{N}(0,1)$ as well but $X \neq -X$.

(2) A random variable is by definition a measurable function $X:(\Omega, \mathcal{F}) \to (E, \mathcal{E})$ where $(\Omega, \mathcal{F}, P)$ is a probability space. In this case, the law of $X$ is a measure (this is a special case of image measure)

$$P_X: \mathcal{E} \to [0,1]$$

and in particular, $P_X$ is positive and finite, so also $\sigma$-finite.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .