Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Definitions In our course we defined a discrete probability space as a tuple $\left(\Omega,P\right)$, where $P:\mathcal{P}(\Omega)\rightarrow\left[0,1\right]$ and $\Omega$ is at most countable, such that $P\left(\Omega\right)=1$ and for a finite our countable infinite set $I$ we have $P\left(\bigcup_{i\in I}A_{i}\right)=\sum_{i\in I}P\left(A_{i}\right)$ where all $A_{i}$ are pairwise disjoint.

Motivation Now we did several exercise that implicitly involved product spaces. For all these exercises $\Omega$ was finite.

Problem To avoid having to construct for each example separately a product space, I wanted to do it once and for all in the abstract, by proving the following theorem: Let $\left(\Omega_{1},P_{1}\right),\ldots,\left(\Omega_{n},P_{n}\right)$ be discrete probability spaces. Then $\left(\Omega_{1}\times\ldots\times\Omega_{n},P\right)$ where $P:\Omega_{1}\times\ldots\times\Omega_{n}\rightarrow\left[0,1\right],\ P\left(\omega_{1}\ldots\omega_{n}\right)=P_{1}\left(\omega_{1}\right)\cdot\ldots\cdot P_{n}\left(\omega_{n}\right)$ is a discrete probability space as well.

But in trying to prove the second property ($\sigma$-addivity) of discrete probability spaces, I ran into problems: How do I prove, that $$ P\left(\bigcup_{i\in I}A_{i}\right)=P_{1}\left(\text{pr}_{1}\left[\bigcup_{i\in I}A_{i}\right]\right)\cdot\ldots\cdot P_{n}\left(\text{pr}_{n}\left[\bigcup_{i\in I}A_{i}\right]\right)=$$ $$=\sum_{i\in I}P_{1}\left(\text{pr}_{1}\left[A_{i}\right]\right)\cdot\ldots\cdot\sum_{i\in I}P_{n}\left(\text{pr}_{n}\left[A_{i}\right]\right)\quad\left(\star\right) $$

equals $\sum_{i\in I}\left(P_{1}\left(\text{pr}_{1}\left[A_{i}\right]\right)\cdot\ldots\cdot P_{n}\left(\text{pr}_{n}\left[A_{i}\right]\right)\right)=\sum_{i\in I}P\left(A_{i}\right)$ ? ($\text{pr}_{j}\left[\cdot\right]$ denotes the $j$-th projection mapping)

If the $\Omega_i$'s are finite I can prove my theorem, since together with the disjointness of the $A_{j}$'s, this implies that $I$ has to be finite, so everything is fine. But for countable infinite $\Omega_i$'s, the sum from $\left(\star\right)$ can only be evaluated using the Cauchy product for series, and this number wouldn't be the same as $\sum_{i\in I}\left(P_{1}\left(\text{pr}_{1}\left[A_{i}\right]\right)\cdot\ldots\cdot P_{n}\left(\text{pr}_{n}\left[A_{i}\right]\right)\right)$, I think.

Can product spaces maybe be defined in a meaningful way only under the conditions from above ?

share|improve this question
    
Yes, product measures always exist, even for non-discrete spaces. This should be in almost every measure theory textbook, probably in the chapter that discusses Fubini's theorem. –  Nate Eldredge Apr 19 '12 at 21:10
    
Sigma-additivity is not what you write as $(\ast)$. –  Did Apr 19 '12 at 22:34
    
@NateEldredge Yes, but I'm just doing a course in elementary probability theory and know nothing of measure theory, so I don't even know what a measure is - let alone a product measure. It is no use to me, to know that some objects exist, that are probably an abstractization of probability space $(\Omega,P)$, that give something like want I want. All I want is to know, whether there is a meaningful way for countable infinite $\Omega_i$'s, to get a probability space (see my definition), that is defined in the cartesian product of the omegas. –  user26698 Apr 20 '12 at 10:50
    
@Didier I thought it were "$P\left(\bigcup_{i\in I}A_{i}\right)=\sum_{i\in I}P\left(A_{i}\right)$ where all $A_{i}$ are pairwise disjoint sets". If not, what is it then ? –  user26698 Apr 20 '12 at 10:51
1  
@user26698 What you wrote in the comment is correct; but this is not what equation $\star$ says. Suppose $\Omega=\{0,1\}\times\{0,1\}$ and let $A=\{(0,0),(1,1)\}$. You don't get the measure of $A$ using projections. –  Byron Schmuland Apr 20 '12 at 12:18
show 1 more comment

1 Answer 1

up vote 2 down vote accepted

What you did right: You have correctly defined the product space $\Omega=\Omega_1\times\cdots\times \Omega_n$, and assigned probability values to individual points $\omega=(\omega_1,\dots,\omega_n)$ by the product $$P(\omega)=P_1(\omega_1)\cdots P_n(\omega_n).$$ This assigns a non-negative value to every point $\omega\in\Omega$ such that $\sum_{\omega\in\Omega}P(\omega)=1$.

Where you went wrong: Your equation $(\star)$ seems to indicate that you want to assign a probability to an arbitrary subset $A$ of $\Omega$ using projections, as in (2). This is incorrect; you should simply define $$P(A)=\sum_{\omega\in A}P(\omega).\tag1$$ The formula $$P\left( A\right)=P_{1}\left(\text{pr}_{1} A\right)\cdots P_{n}\left(\text{pr}_{n} A\right)\tag 2$$ works only for product sets, but this is a completely separate issue from countable additivity.

What to do next: You want to show that $P$ is countably additive. At this point, you can completely forget about the product structure of $\Omega$. In general, if $\Omega$ is any countable space, and $P$ assigns non-negative values to the points of $\Omega$, where $\sum_{\omega\in \Omega}P(\omega)=1$, then (1) defines a countably additive measure on all subsets of $\Omega$. The proof of this will involve manipulating countably many infinite sums, but I think you can do it.

share|improve this answer
    
Ok, I think I understood where my problem was: First I defined a discrete probability model where $P$ was actually defined on $\mathcal{P}(\Omega)$ and not just $\Omega$ (this was a typo of mine, which I now corrected), and then, when establishing the product space, I switched to defining the space in a different manner, namely via $P$ being defined directly on $\Omega_{1}\times\ldots\times\Omega_{n}$. Wanting to prove the $\sigma$-additivity I mixed the two definitons and wanted to show $\sigma$-additivity (...) –  user26698 Apr 22 '12 at 8:20
    
(...) like in the first probability definition - which, as you said, isn't even necessary, since every mapping $P$ that satisfies $(1)$ gives me a mapping $P'$ that is $\sigma$-additive. But if I had defined my $P$, as it is used in $(\star)$, like this: $$ P:\mathcal{P}(\Omega_{1})\times\ldots\times\mathcal{P}(\Omega_{n})\rightarrow \left[ 0,1 \right] $$ $$P\left(A_{1}\ldots A_{n}\right)=P_{1}\left(\text{pr}_{1}\left[A_{1}\right]\right)\cdot\ldots\cdot P_{n}\left(\text{pr}_{n}\left[A_{n}\right]\right) $$, then I think $(\star)$ would have correctly been the $\sigma$-additivity. –  user26698 Apr 22 '12 at 8:34
    
What do you think ? Is that correct ? And how would I go under this definition (i.e. without using your indicated route to define $P$ like $P\left(\omega\right)=P_{1}\left(\omega_{1}\right)\cdot\ldots\cdot P_{n}\left(\omega_{n}\right)$ and then using a theorem that tells me that whenever a function $P:\Omega \rightarrow \left[0,1\right]$ satisfies $(1)$ I get a function $P'$ on $\mathcal{P}(\Omega_{1})\times\ldots\times\mathcal{P}(\Omega_{n})$ - which obviously works) to prove in the more direct way the sigma additivity (i.e. proving $(\star)$) ? –  user26698 Apr 22 '12 at 8:44
    
There, I think, my concerns about products of infinite sums (as indicated in my question) come in to play...(sorry for the long and numerous comments) –  user26698 Apr 22 '12 at 8:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.