Expectation of Random Variable as Integral with respect to Probability Distribution
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Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an integrable real-valued random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $P_X$ be the probability distribution of $X$.
Then:
- $\ds \expect X = \int_\R x \map {\rd P_X} x$
where $\expect X$ is the expected value of $X$.
Proof
From the definition of expectation:
- $\ds \expect X = \int_\Omega X \rd \Pr$
We can write:
- $\ds \int_\Omega X \rd \Pr = \int_\Omega I_\R \circ X \rd \Pr$
where $I_\R$ is the identity map for $\R$.
From the definition of probability distribution, we have:
- $P_X = X_* \Pr$
where $X_* \Pr$ is the pushforward of $\Pr$ on $\tuple {\R, \map \BB \R}$, where $\map \BB \R$ denotes the Borel $\sigma$-algebra on $\R$.
So, from Integral with respect to Pushforward Measure: Corollary, we have:
- $I_\R$ is $P_X$-integrable
and:
- $\ds \int_\Omega I_\R \circ X \rd \Pr = \int_\R I_\R \rd P_X$
That is:
- $\ds \int_\Omega X \rd \Pr = \int_\R x \map {\rd P_X} x$
$\blacksquare$
Sources
- 2013: Donald L. Cohn: Measure Theory (2nd ed.) ... (previous) ... (next): $10$: Probability: $10.1$: Basics