How are the Probability Measure and Cumulative Distribution Function linked when calculating the Expectation of a RV X?

Question

Given a probability space $(\Omega, \Sigma, P)$ and a measure space $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$, let $X: \Omega\rightarrow\mathbb{R}$ be a RV which is $(\Sigma, \mathcal{B}(\mathbb{R}))$-measurable and let $F:\mathbb{R}\rightarrow[0,1]$ be its CDF.

I have trouble understanding the following, due to my lack of understanding of measure theory I suppose. Let us consider the expectation of $X$,

$\mathbb{E}_P(X) = \int^{}_{\Omega}X(\omega)P(d\omega) = \int^{}_{\mathbb{R}}tdF(t)$.

I understand both integrals independently (I think) but how are they linked? At first I thought one might use the Radon–Nikodym theorem, but than $F$ and $P$ would have to be defined on the same measurable space which they are not. Futhermore the RHS can be understood as a Riemann-Stieltjes ingegral whereas the LHS is a Lebesgue integral.

Since this is a rather measure theory heavy question I would like to additionally ask about the following notation I came across. Does $dP(\omega)$ and $P(d\omega)$ mean the same? Coming from measure theory I would expect the former.

Answer 1

To answer your last question: They are used interchangably. Authors don't agree on what the best notation for integration is.

As for how they're linked: The two integrals output the same number. There's not really anything else to it. The link is really between the coupling of the map $X$, the measure $P$ (on $\Omega$ as you say) and the measure $dF$ (at least you can view it as the measure given by $dF((a,b])=F(b)-F(a)$).

This is also a case of the abstract change of variables formula, since $dF$ is the push-forward measure of $P$ under $X$.

Answer 2

To see how the two expressions are linked, I think it is useful to look at the right hand side instead as a Lebesgue-Stieltjes integral. (Given some continuity conditions on the integrand, the Lebesgue-Stieltjes and the Riemann-Stieltjes integrals are the same; see discussion by Billingsley at the end of Section 17). By definition, then, your right hand side is a Lebesgue integral: $$ \int_a^b h(t) \, dF(t) := \int_{[a,b]} h \, d\mu_F $$ where, in your case, the measureable map $h:\mathbb{R} \rightarrow \mathbb{R}$ is the identity map, ${\rm id}_{\mathbb{R}}$, and $\mu_F$ is the uniquely-defined "Lebesgue-Stieltjes measure" whose value on any interval is $$ \mu_F\left(\,(s,t] \,\right) = F(t) - F(s)\,. $$ As an intermediate step to make the connection to your left-hand side, we can show that the above-defined measure is in fact the so-called probability function, $P^{(X)}$, associated to the random variable $X$. This function is defined as the pushforward of the probability measure on $\left( \Omega, \Sigma, P\right)$ under $X:\Omega \rightarrow \mathbb{R}$, i.e., $$ P^{(X)}:\mathbb{R} \rightarrow \mathbb{\bar{R}^+_0},\quad P^{(X)}(u) = P\left({\rm preim}_X(u)\right). $$ This "probability function" makes the measureable target space of the random variable into the probability space $\left(\mathbb{R},\mathcal{B}\left(\mathbb{R}\right), P^{(X)}\right)$, and the distribution function of the random variable, $F_X$, is defined by it (see, e.g., Kolmogorov), as $$ F_X: \mathbb{R} \rightarrow [0,1], \quad F_X(a) = P^{(X)}\left( (-\infty, a) \right)\,. $$ From this we can see that $$ P^{(X)}\left((s,t]\right) = F_X(t) - F_X(s). $$ Thus, $P^{(X)}$ is the Lebesgue-Stieltjes measure $\mu_F$.

Summarizing, we can write $$\begin{eqnarray} \int_{-\infty}^{+\infty} t \, dF_X(t) &=& \int_{-\infty}^{+\infty} {\rm id}_{\mathbb{R}}(t) \, dF_X(t) \\ &=& \int_{\mathbb{R}} {\rm id}_{\mathbb{R}} \, d\mu_F = \int_{\mathbb{R}} {\rm id}_{\mathbb{R}} \, dP^{(X)} = \int_{\Omega} \left({\rm id}_{\mathbb{R}} \circ X\right) \, dP = \int_{\Omega} X \, dP \,, \end{eqnarray}$$ where:

• The integrals in the first line (your right-hand side) are Lebesgue-Stieltjes integrals (or Riemann-Stiltjes, I think, since ${\rm id}_{\mathbb{R}}$ is continuous).
• The integrals in the second line are Lebesgue integrals, where the first two differ only in the name given to the probability measure.
• The second-to-last equality — taking an integral over the random variable's target probability space $\left(\mathbb{R},\mathcal{B}\left(\mathbb{R}\right), P^{(X)}\right)$ into an integral over its domain $\left(\Omega,\Sigma, P\right)$ — follows from the change of variable theorem for Lebesgue integrals (see, e.g, Billingsley, Section 16): $$ \int_{\Omega} \left( f \circ T \right) \, d\mu = \int_{\Omega^{\prime}} f \, d\mu_T^{\prime} $$ where $T:\Omega \rightarrow \Omega^{\prime}$ is taken here to be $X$ with $\Omega^{\prime} = \mathbb{R}$, $f:\Omega^{\prime}\rightarrow \mathbb{R}$ is ${\rm id}_{\mathbb{R}}$, and the measures are related by $\mu_T^{\prime}\left(A^{\prime}\right) := \mu\left( {\rm preim}_T \left(A^{\prime}\right)\right)$, making it clear that $\mu_T^{\prime}$ is $P^{(X)}$, defined exactly as stated above via $\mu = P$.
• The last expression is your left-hand side integral (without expression of the dummy variable).