Probability Distribution is Probability Measure
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Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\struct {S, \Sigma'}$ be a measurable space.
Let $X$ be a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S, \Sigma'}$.
Let $P_X$ be the probability distribution of $X$.
Then:
- $P_X$ is a probability measure on $\struct {S, \Sigma'}$.
Proof
From the definition of probability distribution, we have:
- $P_X = X_* \Pr$
where $X_* \Pr$ is the pushforward $X_* \Pr$ of $\Pr$, under $X$, on $\Sigma'$.
From Pushforward Measure is Measure, we have:
- $P_X$ is a measure.
We then have:
\(\ds \map {P_X} S\) | \(=\) | \(\ds \map \Pr {X^{-1} \sqbrk S}\) | Definition of Probability Distribution | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \Pr \Omega\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | Definition of Probability Measure |
so:
- $P_X$ is a probability measure.
$\blacksquare$