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:

$\ds \int_\Omega X \rd \Pr = \int_\R x \map {\rd P_X} x$

$\blacksquare$


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