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I would like to know what should I verifiy in order to show that two probabilities are equal.

Here is the exercice :

Let $F_0$ be an algebra of sets over $\Omega$ and $P$, $P'$ two probabilities over $\sigma(F_0)$ (the $\sigma$-algebra generated by $F_0$).

Show that if $P=P'$ over $F_0$, then $P=P'$.

Should I show that $P(x\in A) = P'(x\in A)$ for all $A\in \Omega$ by double inclusion ? Or that these $A$ have the same mesure in $\Omega$ and $\sigma(F_0)$ ?

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There seems to be some confusion here. You need to show that $P(A)=P'(A)$ for all $A\in\sigma(F_0)$, given that $P(B)=P'(B)$ for all $B\in F_0$. The fact that $P'=P$ over the sigma algebra is related to the Caratheodory Extension Theorem. –  Alex R. Jan 24 '13 at 23:57
    
I have no idea to show that $P(A)=P′(A)$ for all $A∈σ(F_0)$... Any hint ? –  Alan Simonin Jan 25 '13 at 0:06
    
I feel like I mislead you with the Caratheodory extension theorem. See my answer below –  Alex R. Jan 25 '13 at 0:16

3 Answers 3

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In what follows $P,P'$ are assumed to be countably additive. If you want to show $P'=P$ you don't really need anything like Caratheodory extension theorem. In a restatement of A. Blumenthal's answer, consider the set $\mathcal{S}:=\{A: \ P(A)=P'(A)\}$. This set can be shown to be a monotone class. I remind you that a monotone class is closed under increasing unions and decreasing intersections, that is if $A_1\subset A_2\ldots$ then $\cup_i A_i\in \mathcal{S}$ and if $A_1\supset A_2\supset\ldots$ then $\cap_i A_i\in \mathcal{S}$. As well the empty set and $\Omega$ are assumed to be in a monotone class.

The idea of a monotone class is extremely useful from the following: the smallest monotone class generated by a field is the same as the sigma field generated by a field.

If you can show the above statement, then clearly $\mathcal{S}$ must contain the sigma field $\sigma(F_0)$, which in turn implies uniqueness.

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Oh I see... I just proved the theorem which states that the smallest monoton class containing $A$ is exactly the $\sigma$-algebra generated by $A$ (when $A$ is an field of set). I didn't know how to use it to prove my statement. Thanks for the tip, it seems easy now :) –  Alan Simonin Jan 25 '13 at 0:20
    
Okay, I understand and proved the statement. But one thing remain unclear : when you take an increasing sequence $\(A_n\)$ of $S$, then you have $\cup_n P(A_n) = P(\sum_n A_n) = P'(\sum_n A_n) = \cup_n P'(A_n)$ ? And where do we use that $\(A_n\)$ is increasing ? –  Alan Simonin Jan 25 '13 at 0:38
    
So you have to show that $\mathcal{S}$ is a monotone class. To do so you are assuming that you have an increasing sequence of $A_i$ where $P(A_i)=P'(A_i)$ and you need to conclude that $P(\cup_i A_i)=P'(\cup_i A_i)$. You are essentially trying to prove that $P$ commutes with increasing/decreasing limits. To wit, you need to show $\lim_i P(A_i) = P(\lim_i \cup_i A_i)$ whenever $A_i$ are increasing/decreasing. Notice that the left hand side limit must exist since $0<P<1$ and $A_i$ are increasing/decreasing. –  Alex R. Jan 25 '13 at 0:45
    
You are right, I made a mistake : as $n$ is not necessarily finite, I have to use limits instead of sums. (and by the way, I wrote $\bigcup_n$ and $\sum_n$ in the wrong order on my last comment...) –  Alan Simonin Jan 25 '13 at 0:51

This is a special case of a venerable, extremely useful result in measure theory proper: if two finite measures agree on a $\pi$-system (closed under finite intersections) then they agree on the sigma algebra generated by that $\pi$-system.

The proof is a straightforward application of the Dynkin Theorem: take $F_0$ to be a $\pi$-system. Then note that the class of subsets over which $P, P'$ agree is a $\lambda$-system $\Lambda$ (this won't be hard to to). Now the $\pi-\lambda$ theorem states that the smallest $\lambda$-system containing a $\pi$ system $F_0$ also contains $\sigma(F_0)$. As $\Lambda$ is a $\lambda$-system, we have that $\sigma(F_0) \subset \Lambda$ as a consequence.


Edit: I realize after reading @Alex's response that the Caratheodory theorem can be used here. I may have overkilled.

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This result is demonstrated in my book, but how can I use it to show what I want ? –  Alan Simonin Jan 25 '13 at 0:10

You should prove that $P$ and $P'$ agree on all sets which are elements of $\sigma(F_0)$.

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Indeed, but how ? –  Alan Simonin Jan 25 '13 at 0:06
    
Take a set of $\sigma(F_0)$. You should prove that it can be represented by countable union and finite intersection of sets in $F_0$. But $P$ acts good with respect to that operations... –  Emanuele Paolini Jan 25 '13 at 16:29

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