I'm taking a course in Random Processes and I have been given in a question that the sample space of the probability space is the set of natural numbers $N$. And the question is to define the entire probability space, i.e., to define a valid sigma algebra. I thought of taking the power set of $N$, $P$($N$). But then I figured, that based on Cantor's theorem, it is going to have cardinality greater than the cardinality of $N$, which is infinity. Hence, is $P$($N$) a valid sigma-algebra for a probability space? If not, then what can be a valid sigma algebra in this case?
Is the power set of natural numbers a valid sigma-algebra to define a probability space?
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probability-theory
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0Yes. It's a commonly used discrete probability space. You define a probability on it by assigning to each natural number $n$ a real number $\mathrm{Pr}[X=n]$ such that $$\sum_{n=0}^\infty \mathrm{Pr}[X=n]=1$$ The induced probability is $$\mathrm{Pr}[X\in A]=\sum_{n\in A}\mathrm{Pr}[X=n]$$ – 2017-02-04
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0Yes, it is a probability space. However, it is notably impossible to assign each $n\in \Bbb N$ the same probability. – 2017-02-04
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0Why would it matter if $P(\mathbb{N})$ has greater cardinality than $\mathbb{N}$? – 2017-02-05
3 Answers
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A probability space is a triple $(X,\mathcal A,\mu)$ where $\mathcal A$ is a $\sigma$-algebra and $\mu$ is a measure with $\mu(X)=1$.
Your example $\mathbb N$ with $\mathcal A=\mathcal P(\mathbb N)$ is indeed a $\sigma$-algebra, so if you find a measure which has $\mu(\mathbb N)=1$ then you're more or less good to go.
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Yes. In fact, if $\mu$ is any probability measure defined on some $\sigma$-algebra of $\mathbb{N}$, then we can extend it to a probability measure on $\mathcal{P}(\mathbb{N})$. There are certain properties such a measure cannot have. For instance, we can't have a non-trivial measure $\mu$ on $\mathbb{N}$ such that $\mu(\{n\}) = 0$ for all $n$ (it'd follow that $\mu(\mathbb{N}) = \sum_{n = 1}^{\infty} \mu (\{n\}) = 0$), but there are certainly plenty of probability measures to define on $\mathcal{P}(\mathbb{N})$.
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If we have an ultrafilter $\mathcal{F}$ on $\mathbb{N}$(use Zorn's lemma to get a free ultrafilter), then we can define a (finitely additive) probability measure on $\mathscr{P}(\mathbb{N})$ by $\mu(A) = 0$ iff $A \notin \mathcal{F}$ and $\mu(A) = 1$ iff $A \in \mathscr{F}$.
To get a $\sigma$-additive one, take any sequence of positive real numbers such that $\sum_n a_n = 1$, and define $\mu(A) = \sum \{a_n: n \in A\}$.
We cannot have a $\sigma$-additive finite measure where all measures $\mu(\{n\})$ are the same : the measure of $\mathbb{N}$ would be either $0$ or $+\infty$.