Expectation of Random Variable as Integral with respect to Probability Distribution

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, 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:


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