Condition for Existence of Expectation of Real-Valued Measurable Function composed with Absolutely Continuous Random Variable

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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:

$\ds \expect {\map h X} = \int_\R \map h x \rd \map {P_X} x$

$\blacksquare$


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