Compact support functions and law of a random variable

  • #1
psie
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TL;DR Summary
I don't understand the characterization of the law of a random variable through continuous compact support functions. In particular, I don't understand why we need to consider the whole set of all these functions rather than a subset.
I'm reading in my probability book about characterizations of the law of a random variable, that is, the probability measure ##\mathbb P_X(A)=\mathbb P(X\in A)##. I read the following passage (I'm paraphrasing slightly):

Since the indicator function of an open rectangle (i.e. ##(a_1,b_1)\times (a_2,b_2)\times\cdots\times (a_n,b_n)##, for any choice of the reals ##a_1<b_1,\ldots,a_n<b_n##) is the increasing limit of a sequence of continuous functions with compact support, the probability measure ##\mu## is also determined by the values of ##\int \varphi\,\mu(\mathrm{d}x)## when ##\varphi## varies in the space ##C_c(\mathbb R^n)## of all continuous functions with compact support from ##\mathbb R^n## into ##\mathbb R##.

This extract is basically saying that if $$\mathbb E[\varphi(X)]:=\int \varphi\,\mathbb P_X(\mathrm{d}x)=\int \varphi\,\mathbb P_Y(\mathrm{d}x)=:\mathbb E[\varphi(Y)],\quad \varphi\in C_c(\mathbb R^n),$$then ##\mathbb P_X=\mathbb P_Y##. Indeed, if ##U## denotes an open rectangle and we define $$f_n(x)=\min\{n\cdot d(x,U^c),1\},\quad n\geq 1,$$then ##f_n## are continuous and have compact support (also, they are nonnegative), since the closure of ##U## is a closed rectangle, which is compact in ##\mathbb R^n##. And ##f_n\nearrow \mathbf1_U## as ##n\to\infty##, so by monotone convergence, $$\mathbb P_X(U)=\int \mathbf1_U\,\mathrm d\mathbb P_X=\lim_{n\to\infty}\int f_n\,\mathrm{d}\mathbb P_X=\lim_{n\to\infty}\int f_n\,\mathrm d \mathbb P_Y=\mathbb P_Y(U).$$ So ##\mathbb P_X=\mathbb P_Y## for all open rectangles, and by an argument one can make via the ##\pi-\lambda## theorem, they are equal.

Now, what troubles me is that, why does the author of the text stipulate that ##\varphi## needs to vary over all functions in ##C_c(\mathbb R^d)##? Isn't it enough, as I've shown, to just take the nonnegative continuous functions with compact support?
 
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  • #2
Maybe I'm overthinking things. If we have the information ##\mathbb E[\varphi(X)], \varphi\in C_c(\mathbb R^n)##, then we also have ##\mathbb E[\phi(X)]## for ##\phi## being a nonnegative, continuous function with compact support on ##\mathbb R^n##.

I guess what I was after was making the statement slightly weaker by demanding less; the nonnegative continuous functions with compact support on ##\mathbb R^n## are clearly a subset to ##C_c(\mathbb R^n)##.
 
  • #3
This isn't Durrett anymore, isn't it? It would help to see the entire context and not just your personal view which might already be contaminated by your "overthinking".
 
  • #4
No, it is Measure Theory, Probability and Stochastic Processes by Le Gall. I quote the whole relevant passage, as it is stated in the book. This passage appears in a section on expectation, in particular after a proposition that states that expectation of a function of random variable taking values in ##E## is simply ##\mathbb E[f(X)]=\int_E f(x)\,\mathbb P_X(\mathrm{d}x)##, i.e. the law of the unconscious statistician.

Characterization of the Law of a Random Variable It will be useful to characterize the law of a random variable in different ways. It follows from Corollory 1.19 that a probability measure ##\mu## on ##\mathbb R^d## is characterized by its values on open rectangles, that is, by the quantities ##\mu((a_1,b_1)\times (a_2,b_2)\times\cdots\times (a_d,b_d))##, for any choice of the reals ##a_1<b_1,\ldots,a_n<b_n## (one could restrict to rational values of ##a_i,b_i##, and/or replace open intervals by closed intervals). Since the indicator function of an open rectangle is the increasing limit of a sequence of continuous functions with compact support, the probability measure ##\mu## is also determined by the values of ##\int \varphi\,\mu(\mathrm{d}x)## when ##\varphi## varies in the space ##C_c(\mathbb R^d)## of all continuous functions with compact support from ##\mathbb R^d## into ##\mathbb R##.

Corollary 1.19 reads (only part (i) is relevant, as that part applies to finite measures):

Corollary 1.19 Let ##\mu## and ##\nu## be two measures on ##(E,\mathcal{A})##. Suppose that there exists a class ##\mathcal{C}\subset\mathcal{A}##, which is closed under finite intersections, such that ##\sigma(\mathcal{C})=\mathcal{A}## and ##\mu(A)=\nu(A)## for every ##A\in\mathcal{C}##.

(i) If ##\mu(E)=\nu(E)<\infty##, then we have ##\mu=\nu##.
(ii) If there exists an increasing sequence ##(E_n)_{n\in\mathbb N}## of elements of ##\mathcal{C}## such that ##E=\bigcup_{n\in\mathbb N}E_n## and ##\mu(E_n)=\nu(E_n)<\infty## for every ##n\in\mathbb N##, then ##\mu=\nu##.
 
  • #5
Well, that's - other than Durrett - copyright protected so I cannot look it up and I'm not an expert in the field. Isn't the whole issue about exhausting an open set by a sequence of compact (and therefore measurable) subsets to get a reasonable definition for the probability for open, bounded sets? I'm not sure I get your point.
 
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  • #6
Well, I'm not sure I understand my worries either, but basically I'm reacting to the statement that a probability measure ##\mu## is determined by the values of ##\int \varphi\,\mu(\mathrm{d}x)## when ##\varphi## varies in ##C_c(\mathbb R^d)##. That integral could be negative, but a probability measure outputs only positive values, so I don't see why we'd be interested in varying ##\varphi## in all of ##C_c(\mathbb R^d)##. I think there are functions in ##C_c(\mathbb R^d)## that are not part of any sequence that converge to an indicator function of an open rectangle.

Regarding exhausting an open set by a sequence of compact subsets, I was looking at these exercises here earlier, in particular problem 1(a). I think that is kind of what you meant, but they use the Lebesgue measure.
 
  • #7
psie said:
Well, I'm not sure I understand my worries either, but basically I'm reacting to the statement that a probability measure ##\mu## is determined by the values of ##\int \varphi\,\mu(\mathrm{d}x)## when ##\varphi## varies in ##C_c(\mathbb R^d)##. That integral could be negative, but a probability measure outputs only positive values, so I don't see why we'd be interested in varying ##\varphi## in all of ##C_c(\mathbb R^d)##. I think there are functions in ##C_c(\mathbb R^d)## that are not part of any sequence that converge to an indicator function of an open rectangle.

Regarding exhausting an open set by a sequence of compact subsets, I was looking at these exercises here earlier, in particular problem 1(a). I think that is kind of what you meant, but they use the Lebesgue measure.
Well, that would need a closer look at how these notations are defined. An integral is an oriented volume, so you can always get negative values, simply by the fact ##\int_a^b +\int_b^a =0.## That means you have to solve this system immanent problem anyway if you identify a probability with an integral, i.e. a volume.
 
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  • #8
There is an exercise on p. 122 Rosenthal has an exercise to prove the equivalence of 6 different conditions for measures being equivalent including the ones you mentioned.

I think the importance is that different characterizations will make proving different theorems easier.
 
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  • #9
Hmm, a technical point I may have missed/misunderstood: Given the indicator is ##1## within the open rectangles and ##0## elsewhere, it seems its support is the union of open rectangles and thus not compact. Am I missing something?
 
  • #12
psie said:
Well, I'm not sure I understand my worries either, but basically I'm reacting to the statement that a probability measure ##\mu## is determined by the values of ##\int \varphi\,\mu(\mathrm{d}x)## when ##\varphi## varies in ##C_c(\mathbb R^d)##. That integral could be negative, but a probability measure outputs only positive values, so I don't see why we'd be interested in varying ##\varphi## in all of ##C_c(\mathbb R^d)##. I think there are functions in ##C_c(\mathbb R^d)## that are not part of any sequence that converge to an indicator function of an open rectangle.

Regarding exhausting an open set by a sequence of compact subsets, I was looking at these exercises here earlier, in particular problem 1(a). I think that is kind of what you meant, but they use the Lebesgue measure.
There is such thing as signed measures.
https://en.m.wikipedia.org/wiki/Signed_measure
 
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