Probability measures and convex combination

In summary: Thanks again!In summary, \mathcal B(\Omega) is the Borel algebra of \Omega, and a "convex combination" of probability measures is defined as \bigg(\sum_{k=1}^n w_k\mu_k\bigg)(E)=\sum_{k=1}^n w_k\mu_k(E). Every convex combination of probability measures is a probability measure, including the probability measures defined as \mu_s(E)=\chi_E(s)=\begin{cases}1 & \text{ if }s\in E\\ 0 & \text{ if }s\notin E\end{cases} for each s\in\Omega. The set of all probability
  • #1
Fredrik
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Let [itex]\mathcal B(\Omega)[/itex] be the Borel algebra of [itex]\Omega[/itex] (the σ-algebra of Borel sets in [itex]\Omega[/itex]). I understand that if we define a "convex combination" of probability measures by [tex]\bigg(\sum_{k=1}^n w_k\mu_k\bigg)(E)=\sum_{k=1}^n w_k\mu_k(E),[/tex] then every convex combination of probability measures is a probability measure. I'm particularly interested in the probability measures defined in the following way: For each [itex]s\in\Omega[/itex], we define [tex]\mu_s(E)=\chi_E(s)=\begin{cases}1 & \text{ if }s\in E\\ 0 & \text{ if }s\notin E\end{cases}[/tex]
Let [itex]S_0[/itex] be the set of all probability measures defined this way, and let [itex]S[/itex] be the set of all convex combinations of members of [itex]S_0[/itex]. I have found that S is closed under convex combinations.

My question is: Are there any probability measures on [itex]\mathcal B(\Omega)[/itex] that aren't members of [itex]S[/itex]?
 
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  • #2
Not sure if I understand your question. The [itex]\mu_s[/itex] that you define are the Dirac measures, and you want to know if there are probability measures not in the convex hull of the Dirac measures, right??

Well, to give you a counterexample: let [itex]\Omega=[0,1][/itex] equipped with the Borel-sigma-algebra. Then the Lebesgue measure is an example of measure that is not in your convex hull.
 
  • #3
Your post raises interesting questions though, so I will go into some more depth:

Let [itex]\Omega[/itex] be a locally compact Hausdorff space. Then there is an interesting result known as the Riesz representation theorem. This says that the set of all regular Borel measures on [itex]\Omega[/itex] is bijective with the positive bounded linear functionals on [itex]\mathcal{C}_0(\Omega)[/itex].

That is, given [itex]\mu[/itex] a regular Borel measure, then

[tex]\mathcal{C}_0(\Omega)\rightarrow \mathbb{R}:f\rightarrow \int fd\mu[/tex]

is a positive bounded linear functional. (positive just means that a positive function f is sent to a positive number in [itex]\mathbb{R}[/itex]).

So to study all the (regular) measures, it suffices to study the positive functionals on [itex]\mathcal{C}_0(\Omega)[/itex]. And to study the probability measures, it suffices to study the positive functionals with norm 1. These functionals are called states.


Now, your Dirac measures correspond to the functionals

[tex]ev_s:\mathcal{C}_(\Omega)\rightarrow \mathbb{R}:f\rightarrow f(s)[/tex]

These are not only linear functionals, but also multiplicative! That is, they satisfy [itex]ev_s(fg)=ev_s(f)ev_s(g)[/itex]. Such a functionals are called pure states.

Note that we can put a topology on [itex](\mathcal{C}_0(\Omega))^\prime[/itex]: the weak*-topology!

Now, one can prove (use the Krein-Milman theorem), that the set of all states is the weak*-closure of the the convex hull of all the pure states.

That is: gives your Dirac probability measures and take the convex hull. Then you have almost all the probability measures. You only need to take the weak*-closure of them.
 
  • #4
Edit: I wrote this before I saw your second post. I will read the second post now, and reply later (maybe tomorrow).

micromass said:
Not sure if I understand your question. The [itex]\mu_s[/itex] that you define are the Dirac measures, and you want to know if there are probability measures not in the convex hull of the Dirac measures, right??

Well, to give you a counterexample: let [itex]\Omega=[0,1][/itex] equipped with the Borel-sigma-algebra. Then the Lebesgue measure is an example of measure that is not in your convex hull.
Thanks. Yes, that's what I meant. I guess it was much simpler than I thought. (I did expect the answer to be "yes". I just wanted to make sure).
 
  • #5
Thanks again. There's some really good stuff in post #3. Looks like exactly what I need. I started this thread because I'm trying to understand how classical mechanics fits into the idea that a theory of physics assigns probabilities to verifiable statements. So my Ω is the phase space of a classical theory of physics. Its members are often called "states", but "pure states" is more accurate. I found that Ω could be bijectively mapped onto the set I called S0. So I decided to call the members of S0 "pure states", and I was thinking about how to define the set of all states. I was torn between making it the set of all probability measures, and just making it the convex hull of S0. So you told me exactly what I need, and you were even using the terminology I had in mind.

I haven't studied this Riesz representation theorem yet (I just knew it said something about a correspondence between measures and functionals), or the Krein-Milman theorem, but I have studied a few theorems involving the weak* topology. So your reply was perfect for someone with my level of knowledge.
 

1. What is a probability measure?

A probability measure is a function that assigns a numerical value to a set of outcomes in a probability space. It represents the likelihood of an event occurring and must satisfy certain mathematical properties, such as being non-negative and summing to 1.

2. How is a probability measure used in statistics?

In statistics, a probability measure is used to quantify the uncertainty associated with a particular event or outcome. It is used to calculate probabilities, which are then used to make predictions and draw conclusions from data.

3. What is a convex combination?

A convex combination is a way of combining two or more probability measures to create a new probability measure. In a convex combination, the weights assigned to each measure must be non-negative and sum to 1.

4. How is a convex combination different from a weighted average?

A convex combination and a weighted average are similar in that they both involve combining multiple measures. However, a convex combination specifically refers to combining probability measures, while a weighted average can be used to combine any type of numerical values.

5. What is the significance of convex combinations in probability theory?

Convex combinations are important in probability theory because they allow for the creation of new probability measures that can accurately represent complex situations. They also have applications in decision making and optimization problems.

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