Suppose we have $n$ experiments performed independently. Every $j$-th experiment has a probability space $(\Omega_j, F_j, P_j)$ associated to it. The outcome of $n$ experiments is a sequence $(w_1, \ldots, w_n$) - $w_j$ is the outcome of $j$-th experiment.

Then our sample space (we conduct $n$ experiments) is $\Omega = \Omega_1 \times \Omega_2 \times \cdots \times \Omega_n$. $(1) F=F_1\otimes \cdots \otimes F_n$.

Then probability $P$ should satisfy the property that $P(A_j) = P_j(A_j)$ where $A_j \in F_j$, i.e.

\begin{align} & (2) P(A_1 \times A_2 \times \cdots \times A_n) = P(A_1 \times \Omega_2 \times \cdots \times \Omega_n) \times \cdots \times P(\Omega_1 \times \cdots \times A_n) \\ = {} & P_1(A_1) \cdot P_2(A_2)\cdot \cdots \cdot P_n(A_n)) \end{align}

There exists only one $P$, namely $(3) P=P_1\otimes P_2\otimes P_3 \otimes \cdots \otimes P_n$.

Could you explain what does it all mean? It makes sense to me that if we have several experiments with respective sample spaces, then conducting them all gives a sample space that is thier cartesian product:

1) What does (1) mean and why is it true?

2) I understand the first term - the probability of cartesian product of events. Why is it equal to the right side and how we obtain the second and third term in this equality (by term I mean the part after the 1st and 2nd '=' sign). For example, after 1st equality sign there is a cartesian product of probabilities. How can we have cartesian product of numbers? Isn't it a mistake? Anyway, this equality follows from the above statement 'Then probability $P$ should satisfy the property that $P(A) = P_j(A_j)'$, but I'm not sure what's the connection.

3) Again, what does it mean and why is that?

  • $\begingroup$ "Suppose we have $n$ independent experiments. Every $j$-th experiment has a probability space $(\Omega_j, F_j, P_j)$ associated to it." No, absolutely no. To say that events $A_j$ are independent is to assume that $A_j\in\mathcal F$ for some given probability space $(\Omega,\mathcal F,P)$ (the same for every $j$) and that $P(A_1\cap\cdots\cap A_n)=P(A_1)\cdots P(A_n)$ (note the same probability $P$ everywhere). Is the text of your question coming from some specific source? $\endgroup$ – Did Apr 15 '15 at 19:07

First, this construction of the common probability space is valid only if the experiments are independent. If they are, then by definition and (from the theoretical point of view) only by definition, we have

$$P(A_1\cap A_2 \cap \cdots \cap A_n)=P_1(A_1)P_2(A_2)\cdots P_n(A_n) \ \ \ \ (*)$$

if the occurrence of $A_j$, can be decided by observing only the $j^{th}$ experiment; that is when $A_j\in F_j$ (for all$j$).

However, when we consider $\Omega_1 \times \Omega_2 \times \cdots \times \Omega_n$ we may have to compute the probability of events that cannot be written in the form of $(*). $ That is, when the probability cannot be calculated as a product of probabilities.

The best is to consider the simplest possible example: flipping a coin $n$ times. Then the event that during the $n$ experiments there occurred exactly one tails is not an event whose occurrence could be decided by observers who can see only one of the experiments. Let this event be denoted by $T$. That is, let

$$T=\{\text{only one tails occurred during the experiments.}\}$$

So, when we consider $F_1 \times F_1 \times \cdots \times F_n$ we talk about events like the one mentioned above.

If, for instance $n=3$ and $$\Omega_1=\{h_1,t_1\},\Omega_2=\{h_2,t_2\},\Omega_3=\{h_3,t_3\}$$ and $$F_1=\{\emptyset, \{h_1\},\{t_1\},\{h_1,t_1\}\},$$ $$F_2=\{\emptyset, \{h_2\},\{t_2\},\{h_2,t_2\}\},$$ $$F_3=\{\emptyset, \{h_3\},\{t_3\},\{h_3,t_3\}\}$$ then

$$T\not \in F_j, \ j=1,2,3$$.

because $$T=\{(t_1,h_2,h_3),(h_1,t_2,h_3),(h_1,h_2,t_3)\}\in F_1\times F_2 \times F_3.$$

So, $F_1\times F_2 \times F_3$ consists of those events that can be defined if all the experiments are taken into account.

There are many questions that arise when one wants to create these sets and when one wants to extend the probabilities given in the form of $(*)$ to the events that get generated when it comes to the events in $F_1 \times F_1 \times \cdots \times F_n$.

To understand all the details you will have to study measure theory, the theoretical basis of the calculations related to probability. Go google or wiki!

EDITED (Equasion 2)

Let's stick to the example of flipping a coin three times. Say, $A_1=\{t_1\}, A_2=\{t_2\},A_3=\{t_3\}$. (That is, all the three experiments resulted in tails. Let $P_1(A_1)=P_2(A_2)=P_3(A_3)=\frac {1}{2}$.) We have a specific event (set) now: $$\{(t_1,t_2,t_3)\}\subset\Omega_1 \times \Omega_2 \times \Omega_3.$$ A possible notation for this event is $$\{t_1,t_2,t_3\}=A_1\times A_2 \times A_3.$$

The fact that the first result was tails ($A_1$) can be expressed in $\Omega_1 \times \Omega_2 \times \Omega_3$ as $$"A_1"=\{(t_1,t_2,t_3),(t_1,t_2,h_3),(t_1,h_2,t_3),(t_1,h_2,h_3)\}.$$ Obviously, the event that the first result was a tails has two representations now: The first one was $A_1=\{t_1\}$ (in $\Omega_1$ ) and the second one was $"A_1"$ (in $\Omega_1 \times \Omega_2 \times \Omega_3 $). I used quotation marks because theoretically $A_1$ and $"A_1"$ are two very different things: two different representations of the same event. So, saying "the same event" is quite dangerous if one does not understand the theory behind.

Said all that, we have $$P_1(A_1)=P("A_1")=P(\{(t_1,t_2,t_3),(t_1,t_2,h_3),(t_1,h_2,t_3),(t_1,h_2,h_3)\})=\frac{1}{2}.$$ You can write the same, if you like, the following way: $$\{(t_1,t_2,t_3),(t_1,t_2,h_3),(t_1,h_2,t_3),(t_1,h_2,h_3)\}=A_1\times\Omega_2 \times \Omega_3$$ or $$P(\{(t_1,t_2,t_3),(t_1,t_2,h_3),(t_1,h_2,t_3),(t_1,h_2,h_3)\})=P(A_1\times\Omega_2 \times \Omega_3)=\frac{1}{2}.$$

Apply the notation described above for $A_2$ and $A_3$ and use the assumption that the consecutive events are independent then you will have $$P(A_1\times A_2 \times A_3)=P(\{(t_1,t_2,t_3)\})=P_1(A_1)P_2(A_2)P_3(A_3)=\frac{1}{2^3}.$$


I've just realized that it may be a problem to see why $$A_1 \times A_2 \times A_3=$$ $$=(A_1 \times \Omega_2 \times \Omega_3) \cap (\Omega_1 \times A_2 \times \Omega_3) \cap (\Omega_1 \times \Omega_2 \times A_3).$$

If you think in terms of triplets like $(t_1,h_2,t_3)$ then you will understand why $$P("A_1"\cap "A_2" \cap "A_3")=P(A_1 \times A_2 \times A_3).$$


How to justify the product rule (in the case of classical probabilities.)

In the case of a classical probability field, the number of elements in $\Omega$ is a finite number, say, $\Omega=\{1,2,\cdots,N\}$. The possible events are modeled by the class of all sets in $\Omega$ and the events of one single elements are equally likely: $P(\{i\})=\frac{1}{N}, \ i=1,2, \cdots N.$

When we model two experiments of this kind then we consider $\Omega \times \Omega=\{(i,j): i,j=1,2,\cdots, N\}$, a set of $N^2$ elements, (and we consider all the possible subsets of $\Omega \times \Omega$).

This is not yet a classical probability space because we have not yet MADE A DECISION about the probabilities of the simplest events of the kind of $\{(i,j)\}$. If we DECLARE that the new probability space is still a classical one then, by definition, we say that $$P'(\{(i,j)\})=P'(\{(k,l)\})=\frac{1}{N^2}=\frac{1}{N}\times \frac{1}{N}=P(\{i\})P(\{j\}).$$

So, from the point of view of the philosophy behind this act of creation, it is not the product rule that we use, but it is the theoretically still non-justify-able hypothesis that the product space is a classical one. Experiments can falsify our hypothesis but it can never be verified apodictically.

I have spoken!

  • $\begingroup$ Thanks, but maybe you could answer my specific questions? Does the part I've rewritten from my book make sense? I can't understand equation 2 for example... $\endgroup$ – user4205580 Apr 15 '15 at 18:07
  • $\begingroup$ Sounds like no to me, but anyway, I can start a bounty and give you some extra points. If it was so easy for me to just check a few wiki pages to figure things out, I probably wouldn't even ask here. But I've spent a lot of time on it. $\endgroup$ – user4205580 Apr 15 '15 at 18:29
  • $\begingroup$ I appreciate your efforts and did my best to further explain. Please, take a look again. OK $\not =NO$ $\endgroup$ – zoli Apr 15 '15 at 19:02
  • $\begingroup$ Nice, that makes sense. However, back to my equation 2, shouldn't $P(A_1 \times \Omega_2 \times \cdots \times \Omega_n) \times \cdots \times P(\Omega_1 \times \cdots \times A_n)$ be $P(A_1 \times \Omega_2 \times ... \times \Omega_n) \cdot \cdots \cdot P(\Omega_1 \times \cdots \times A_n)$ ? There should be symbol of multiplication instead of cartesian product. $\endgroup$ – user4205580 Apr 15 '15 at 19:23
  • $\begingroup$ Yes, but $\times$ may have two meanings: (1) Descartes product for sets and (2) normal (number) product for probabilities. $\endgroup$ – zoli Apr 15 '15 at 19:27

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