Say we have $ x \in R$ and a random variable $ X$.
My lecture notes say that if we look at the following event:
$$ \{\omega \in \Omega \mid X(\omega) = x\} = X^{-1}(x)$$ we can call it the event "$X = x$", and can calculate its probabilty accordingly:
$$ \Pr\lbrack X=x\rbrack = \Pr\lbrack X^{-1}(x)\rbrack = \Pr\lbrack \{\omega \in \Omega \mid X(\omega) = x\} $$
Which does not make sense to be because as I do not undestand what $X^{-1}$ means, can anyone clear this up ?
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Say you throw a pair of three-sided dice, so each outcome is $1,$ $2,$ or $3.$
The possible outcomes are: $$ \begin{array}{ccc} (1,1) & (1,2) & (1,3) \\ (2,1) & (2,2) & (2,3) \\ (3,1) & (3,2) & (3,3) \end{array} $$ Let $X$ be the maximum of the numbers shown by the two dice.
$$ \Pr(X=3) = \Pr(X^{-1}(3)) = \Pr(\{(1,3), (2,3), (3,3),(3,2),(3,1)\}). $$
The set $X^{-1}(3)$ is the set of all outcomes for which $X=3$ , i.e. it is this set of five outcomes: $$ X^{-1}(3) = \{(1,3), (2,3), (3,3),(3,2),(3,1)\}. $$
The lesson of the lecture is that a random variable $X$ is a function on a probability space. The notation $X^{-1}(x)$ means the pre-image of the value $x$, that is, the set of everything in the probability space that maps to the same value $x$.