*Def Probability space/Sample space: A measure space $(\Omega,\mathcal{F},p)$ where $\Omega$, the sample set, is the set of all possible outcomes in a given experiment. $\mathcal{F}$ is a sigma algebra defined on the set $\Omega$ and $p$ is a measure on the measurable space $(\Omega,\mathcal{F})$ such that it countably additive and the measure of the whole sample set is $1$.
*Def Random variable: A random variable is a measurable function $X$ defined on $(\Omega,\mathcal{F},p)$ such that $X : \Omega \to \mathbb{R}$, the measure space $\mathbb{R}$ with Lebesgue measure?
*Def Push Forward measure: For a random variable it is defined by $p_{X}(A)=p(X^{-1}(A))$ for any measurable set $A$ of $\mathbb{R}$.

My question is that under these set of definition can I view Random variable with its push forward measure as something that 'Transforms' the given probability space into a probability space defined by the triplet $(\mathbb{R},\mathcal{F^{'}},p_X)$ where $\mathcal{F}'$ is the set of all measurable subsets of $\mathbb{R}$. Notice that $\mathbb{R}$ can be replaced with $Im(\Omega)$ under $X$ with suitable adjustments to the measure.

In short, have I just decided to view my old space under a new lens where I have reduced the amount of unrequired information? And when we talk about the transformation of a random variable, are we ultimately transforming the probability spaces into new ones with their natural push forward measures?


Yes, you could think of the pushforward measure as giving you a particular view of your original probability space. How much information it gives you depends on the random variable $X$ (e.g., $X=\text{const}$ gives you no information whatsoever).

A transformation of a random variable, say $f(X)$, is just a function of your random variable. So it gives you another function $\Omega \to \mathbb{R}$, i.e., another random variable. This random variable comes with its own pushforward measure, which as above can be thought of as giving you certain information about your original probability space.

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  • $\begingroup$ Can it be said if $X$ is one-one on it's image then we are kind of working with the same space? Like some isomorphic spaces? $\endgroup$ – Mann Apr 28 '19 at 14:52
  • $\begingroup$ No, a measure space isomorphism is stricter than that. First, note that when we say a random variable is measurable we mean with respect to the Borel sigma algebra on $\mathbb{R}$, not Lebesgue measure. So for instance suppose $(\Omega, \mathcal{F},P)$ is $[0,1]$ with Lebesgue measurable sets and Lebesgue measure. Take $X:\Omega \to \mathbb{R}$ to be the inclusion map ($X(\omega)=\omega$). Even if we restrict $X$ to be a map $[0,1]\to[0,1]$, the pushforward measure $p_X$ is Borel measure, while the original measure $P$ is Lebesgue measure. $\endgroup$ – kccu Apr 28 '19 at 15:36
  • $\begingroup$ A measure space isomorphism $f:(X, \mathcal{F}, \mu) \to (Y, \mathcal{G}, \nu)$ requires that if $f(A)=B$, then $A \in \mathcal{F} \Leftrightarrow B \in \mathcal{G}$; and in addition $\mu(A)=\nu(B)$. If $f$ is measurable (like a random variable), then you have $B \in \mathcal{G} \Rightarrow A \in \mathcal{F}$, but the forward implication need not hold in general. $\endgroup$ – kccu Apr 28 '19 at 15:37
  • $\begingroup$ I see now, thanks! $\endgroup$ – Mann Apr 28 '19 at 16:58

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