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In probability theory, when having $ E(f(X))=\int_{-\infty}^\infty f(x)\, dg(x) $, an expectation of a measurable function $f$ of a random variable $X$ with respect to its cumulative distribution function $g$,

  1. is it true that it is always a Lebesgue–Stieltjes integral?
  2. Furthermore, is it always a Riemann–Stieltjes integral?

Thanks and regards!

3 Answers 3

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If $f$ is continuous, then $\int_a^b {f(x)dF(x)} = \int_a^b {f(x)d\mu (x)} $, for any $-\infty < a < b < \infty$, where $F$ and $\mu$ are the distribution function and probability distribution of $X$, respectively (they are related by $\mu((s,t])=F(t)-F(s)$, for any $-\infty < s < t < \infty$).

A drawback of the Riemann-Stieltjes integral is illustrated in the following simple example. Suppose that $X=0$ almost surely. Then, $\int {F(x)dF(x)} $ is not defined, whereas $\int {F(x)d\mu (x)} = F(0) = 1$. Of course, ${\rm E}[F(X)] = {\rm E}[F(0)]= 1$.

Another (more significant) drawback is indicated in GWu's answer.

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    @Tim: For arbitrary small \varepsilon > 0, there exist $c_1,c_2 \in [-\varepsilon,\varepsilon]$ such that $F(c_1)[F(\varepsilon)-F(-\varepsilon)] = 0 \cdot 1 = 0$ but $F(c_2)[F(\varepsilon)-F(-\varepsilon)]=1 \cdot 1 = 1$. Hence, $\int_a^b {F(x)dF(x)} $ is not defined if a<0. More generally, consider $\int_a^b {f(x)dF(x)} $ when $f$ and (the cdf) $F$ share a common point of discontinuity $s \in (a,b)$.2011-04-11
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This question is closely related to your other question regarding Riemann-Stieltjes integrals.

In the case that $f$ is continuous, these two types of integral agree provided they are finite.

But there are also other cases. For instance, $f$ is merely Borel measurable, then the Lebesgue-Stieltjes integral $\int f(x) dg(x)$ is defined, but not Riemann-Stieltjes integral because $f$ might be unbounded. This still have probabilistic interpretation because in this case $f(X)$ is still a random variable, and we can still consider the expectation of $f(X)$.

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    I think Riemann integral is only defined for bounded functions.2011-04-10
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It maybe better for you to use the notation $ \mathsf{E}[f(X)] = \int\limits_{\mathbb{R}}f(x)Q(dx) $ where $Q$ is a distribution of an r.v. $X$. Then you do not need to worry if $X$ is a continuous r.v. or a discrete one. Then depending on what you know about $X$ (sumulative distribution function $g$ or a density function $h$) you will rewrite the first equation using $ Q(dx) = dg(x) = h(x)\,dx. $ Usually only Lebesgue (Lebesgue-Stieltjes) integrals are used in the probability theory. On the other hand to calculate them you can use an equivalence of Lebesgue-Stieltjes and Riemann-Stieltjes integrals (provided necessary conditions).

Edited: For the discrete distribution CDF $g$ is a pure jump function with jumps at values $\alpha_i$ of the r.v. $X$ and sizes of jumps $p_i$ such that $p_i = \mathsf{P}(X = \alpha_i)$. Then the integral is again a Lebesgue integral above can be rewritten as a Lebesgue-Stieltjes integral using $g$ or as a sum.

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    Whether or not the integral is R-S depends on what $f$ is. Even if $g$ is continuous, for example, $g(x)=x$, which reduces to the Riemann integral, $\int f(x)dx$ may not make sense if $f$ is not nice enough (namely, Riemann integrable).2011-04-10