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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!

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3 Answers 3

up vote 4 down vote accepted

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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Thanks! (1) Is r.v. $X$ also required to be continuous? (2) Does requiring $f$ and $X$ both to be continuous rule out many other possibilities of expectation? –  Tim Apr 10 '11 at 20:14
    
(1) No, since the distribution function of any random variable $X$ is monotone increasing and right-continuous. –  Shai Covo Apr 10 '11 at 20:17
    
Thanks! I wonder what conclusions stated in Wikipedia or elsewhere you are using? –  Tim Apr 10 '11 at 20:19
    
(2) $X$ is not required to be continuous, so... –  Shai Covo Apr 10 '11 at 20:19
    
As for the last question, the section "Riemann–Stieltjes integration and probability theory" in the first link. –  Shai Covo Apr 10 '11 at 20:22

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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Thanks! When saying "not Riemann-Stieltjes integral because $f$ might be unbounded", for $f$, are boundedness and continuity imply each other? –  Tim Apr 10 '11 at 20:44
    
No, continuity implies boundedness since we are considering functions on bounded intervals first. Then possibly taking the limit $a\to-\infty$ and $b\to\infty$ if it exists. That's why Lebesgue-Stieltjes is more convenient. –  GWu Apr 10 '11 at 20:48
    
Thanks! I still wonder why $\int f(x) dg(x)$ is "not Riemann-Stieltjes integral because $f$ might be unbounded"? –  Tim Apr 10 '11 at 21:49
    
@Tim: Consider $X$ uniform$(0,1)$, with distribution function $F$ and probability distribution $\mu$. Then, $\int {f(x)dF(x)} = \int_0^1 {f(x)dx} $ is an ordinary Riemann integral, whereas $\int {f(x)d\mu (x)} = \int_0^1 {f(x)dx} $ is Lebesgue integral. The former is not defined if $f$ is (for example) unbounded. –  Shai Covo Apr 10 '11 at 22:17
    
I think Riemann integral is only defined for bounded functions. –  GWu Apr 10 '11 at 22:19

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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Thanks! You raised one point I forgot to ask, which is whether or not my question also applies to discrete distribution. For discrete distribution: is it only $Q(dx) = dg(x)$ that makes sense, and is the integral still Riemann-Stieltjes and Lebesgue-Stieltjes? –  Tim Apr 10 '11 at 20:40
    
I've edited my answer. –  Ilya Apr 10 '11 at 20:45
    
Thanks! For a discrete distribution, is the expectation still Riemann-Stieltjes? –  Tim Apr 10 '11 at 20:47
    
since $g$ is discontinuous, I think that there are counterexamples on $f$ in which R-S sums do not converge. I advise in this case to use Lebesgue integral. In fact in probability theory it's better always to use Lebesgue integrals (since Kolmogorov used it) - an use Riemann only if you can prove that they are equal. –  Ilya Apr 10 '11 at 20:52
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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). –  GWu Apr 10 '11 at 20:56

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