- #1
member 587159
My textbook says that if ##X: \Omega \to \mathbb{R}## is discrete stochast (I.e., there are only countably many values that get reached), then it suffices to know the probability function ##p(x) = \mathbb{P}\{X =x\}## in order to know the distribution function ##\mathbb{P}_X: \mathcal{R} \to \mathbb{R}: A \mapsto \mathbb{P}\{X \in A\} = \mathbb{P}(X^{-1}A)##Indeed, if ##S:= \{x : p(x) > 0\}##, then for ##A \in \mathcal{R}##, it follows that $$\mathbb{P}\{X \in A\} = \sum_{x \in S \cap A}p(x)$$
But how do they get this formula?
I tried the following:
$$\mathbb{P}\{X \in A\} = \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in S\cap A}\{a\} \cup\bigcup_{a \in A\setminus S}\{a\}\right)\right) $$
$$= \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in S \cap A}\{a\}\right)\right) + \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in A\setminus S}\{a\}\right)\right) $$
$$=\sum_{a \in S \cap A}p(a) + \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in A\setminus S}\{a\}\right)\right)$$
But how do I show that the probability on the right is zero? I can't use ## \sigma##-additivity on uncountable disjoint unions.
EDIT: ##\mathcal{R}## is the smallest sigma algebra that contains the usual topology on the real numbers, i.e. it are the Borel parts of the reals.
But how do they get this formula?
I tried the following:
$$\mathbb{P}\{X \in A\} = \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in S\cap A}\{a\} \cup\bigcup_{a \in A\setminus S}\{a\}\right)\right) $$
$$= \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in S \cap A}\{a\}\right)\right) + \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in A\setminus S}\{a\}\right)\right) $$
$$=\sum_{a \in S \cap A}p(a) + \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in A\setminus S}\{a\}\right)\right)$$
But how do I show that the probability on the right is zero? I can't use ## \sigma##-additivity on uncountable disjoint unions.
EDIT: ##\mathcal{R}## is the smallest sigma algebra that contains the usual topology on the real numbers, i.e. it are the Borel parts of the reals.
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