I have a probability space $(\Omega, S, P)$, where $\Omega$ is a sample space, $S$ is a sigma-algebra, $P$ is probability measure, $P : S \to [0,1]$. Now I consider some other space $(A, B)$, where $B$ is a sigma-algebra on $A$. This allows me to define a random variable as a measurable function $X : \Omega \to B$ such that for every $\mathcal{B} \in B$ I have $X^{-1} (\mathcal{B}) = \{ \omega \in \Omega : X(\omega) \in \mathcal{B} \} \in S$. Finally I can bring in the probability distribution $\mathbb{P}$ which is another measure, this time defined as $\mathbb{P} : B \to [0,1]$, such that for any $\mathcal{B} \in B$ we have $\mathbb{P} (\mathcal{B}) = P(X^{-1} (\mathcal{B}))$. This is, I believe, a pushforward measure, literally from the definition, correct?
So now I have some question regarding notation, for example in some textbooks I may encounter random variable $X \sim N(0,1)$ and some exercise to calculate $P ( -1 < X <2 )$. My question is what is $P$ in this case? Is it my original $P : S \to [0,1]$, or is it what I've defined as $\mathbb{P}$? Or maybe is it some not-exactly-formal thing to make the notation easier on the eyes?


In $P(-1<X<2)$ $P$ stands for the original probability measure on $\Omega$. In terms of $\mathbb P$ we can write $P(-1<X<2)=\mathbb P ((-1,2))$ since $X^{-1}((-1,2))=\{\omega \in \Omega: X(\omega) \in (-1,2)\}=\{\omega: -1<X(\omega) <2\}$.

  • $\begingroup$ So do I get this right that $P(-1 < X < 2)$ is really $P(\{\omega : -1 < X(\omega) < 2\})$ ? $\endgroup$ – gabe Sep 13 '20 at 23:38
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    $\begingroup$ @gabe Yes, that is a standard notation in Probability Theory. $\endgroup$ – Kavi Rama Murthy Sep 13 '20 at 23:40
  • $\begingroup$ Also, I get that the concept of random variable is useful, because we can work with quantities instead of sometimes not so clear elements of sample space. But why do we even need probability distribution? Not in a common meaning like normal distribution, but in a formal sense, like the $\mathbb{P}$ from my original post. Couldn't we define everything, or almost everything in probability in terms of the original measure $P$? I haven't dug that deep into probability yet so I might be lacking the big picture perspective, hence my question. $\endgroup$ – gabe Sep 13 '20 at 23:45
  • $\begingroup$ do you have any response to that? I would appreciate it $\endgroup$ – gabe Sep 15 '20 at 10:10
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    $\begingroup$ Most of the time we don't know what the random variables or the probability space and we don't even care. We are interested in probabilities , expectations etc. and these things depend only on distributions. Life gets lot more coplicated when we start looking at explicitly defined random variables on a probability space. $\endgroup$ – Kavi Rama Murthy Sep 15 '20 at 10:19

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