Condition for Existence of Expectation of Real-Valued Measurable Function composed with Absolutely Continuous Random Variable
Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an absolutely continuous random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $P_X$ be the probability distribution of $X$.
Let $\map \BB \R$ be the Borel $\sigma$-algebra of $\R$.
Let $h : \R \to \R$ be a $\map \BB \R$-measurable function.
Let $\lambda$ be the Lebesgue measure on $\struct {\R, \map \BB \R}$.
Then $\map h X$ is integrable if and only if $h$ is $\lambda$-integrable function.
In this case:
- $\ds \expect {\map h X} = \int_\R \map h x \rd \map {P_X} x$
where $\expect {\map h X}$ denotes the expectation of $\map h X$.
Proof
From Composition of Measurable Mappings is Measurable, we have:
- $\map h X$ is $\Sigma$-measurable.
So:
- $\map h X$ is a real-valued random variable.
From the definition of the probability distribution, we have:
- $P_X = X_\ast \Pr$
where $X_\ast \Pr$ denotes the pushforward measure of $\Pr$ under $X$.
Since $X$ is an absolutely continuous random variable, it is $\Sigma$-measurable.
So, from Integral with respect to Pushforward Measure: Corollary:
- $\map h X$ is integrable if and only if $h$ is $\lambda$-integrable function.
If $\map h X$ is integrable, Integral with respect to Pushforward Measure: Corollary gives:
- $\ds \int \map h X \rd \Pr = \int_\R \map h x \rd \map {P_X} x$
that is:
- $\ds \expect {\map h X} = \int_\R \map h x \rd \map {P_X} x$
$\blacksquare$
Sources
- 1991: David Williams: Probability with Martingales ... (previous) ... (next): $6.12$: The 'elementary formula' for expectation