Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and let $\{A_i\}_{i\in\mathbb{N}}\subseteq\mathcal{F}$ be a countable partition of $\Omega$, where $\mathbb{P}(A_i)=ab^i,\forall i\geq1$. Which conditions must be imposed for $(a,b)$ so as to make $\mathbb{P}$ be a probability measure?

Well, to begin with, I showed the conditions so as to satisfy $\mathbb{P}(\Omega)=1$: $$\mathbb{P}(\Omega)=\mathbb{P}(\cup_{i\geq1}A_i)=\sum_{i\geq1}\mathbb{P}(A_i)=\sum_{i\geq1}ab^i$$ For the sum to converge, $|b|<1$ and the sum goes to $\frac{ab}{1-b}=1$.

Also, all individual probabilities $\mathbb{P}(A_i)\geq0$, so, $a\geq0,b\geq0$.

But, for $\mathbb{P}$ to be a probability measure, we must have that for any disjoint sequence of sets $\{B_i\}_{i\geq1}\subseteq\mathcal{F}$, $\mathbb{P}(\cup_{i\geq1}B_i)=\sum_{i\geq1}\mathbb{P}(B_i)$ holds, not only for the given partition $\{A_i\}_{i\in\mathbb{N}}$.

My question is if this additional constraint can impose any new condition over $(a,b)$.

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    $\begingroup$ Better to choose where you want to ask. Cross-posting is discouraged. $\endgroup$ – StubbornAtom Jun 7 at 8:18
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    $\begingroup$ The assertion beginning "for any disjoint sequence of sets" is part of what it means for $(\Omega,\mathcal{F},\mathbb{P})$ to be a probability space: it need not be proven. One way to see that it imposes no additional conditions on $(a,b)$ is to construct examples of such probability spaces for every possible $(a,b)$ permitted by the criteria you have already derived. $\endgroup$ – whuber Jun 7 at 14:26

Because the answer you received to your cross-posted question is unrevealing, let's get down to the essence of the issue: if for every $(a,b)$ satisfying your criteria we can exhibit a probability space with the given probabilities, we will have shown there can be no (general) additional condition.

We're discussing a geometric distribution, which concerns the chances of observing a sequence of "failures" before the occurrence of the first "success" in a sequence of independent binary experiments. The following construction is motivated by this observation. It is not the simplest possible one, but it is perhaps the most direct and revealing possible model of binary experiments that may extend for an indeterminate number of repetitions.


To this end, let $(a,b)$ be arbitrary numbers for which $0\lt b\lt 1$ and $ab=1-b.$ The constraints $0\lt b\lt 1$ imply $a = (1-b)/b$ is defined and positive, which (at least) shows all the putative probabilities $ab^i$ are positive. You have already established they sum to unity (over all whole numbers $i$), as required of any probability distribution.

Define $\Omega$ to be the set of binary sequences

$$\Omega = \{(\omega_1,\omega_2,\ldots,\omega_n,\ldots\mid \omega_i\in\{0,1\}\ \forall\, i \ge 1\}.$$

Generate the sigma-algebra $\mathcal{F}$ from the basic events given by finite binary sequences. That is, given any finite binary sequence $\eta = (\eta_1 ,\ldots,\eta_n )$ for $n\ge 0,$ define the event $\mathcal{E}(\eta)$ to be all sequences whose initial parts agree with $\eta,$

$$\mathcal{E}(\eta) = \{(\omega_1,\omega_2,\ldots,\omega_n,\ldots\mid \omega_1=\eta_1,\ldots,\omega_n=\eta_n\}.$$

Create a probability measure on $(\Omega,\mathcal F)$ by defining it on the basic events as

$$\mathbb{P}(\mathcal{E}(\eta)) = b^{n-l(\eta)}(1-b)^{l(\eta)}\tag{*}$$

where $l(\eta)$ counts the number of ones appearing in $\eta.$ This is the measure determined by supposing the events $$\mathcal{I}_n = \{\omega\in\Omega\mid \omega_n = 1\},$$ where the $n^\text{th}$ component of the sequence is $1,$ are independent and identically distributed (with common probability $1-b$). This characterization demonstrates $\mathbb P$ is well-defined and is a probability measure: there are no hidden inconsistencies in the formula $(*).$

Application of the construction

With this setup, for any integer $i\ge 1$ set

$$\mathcal{A}_i = \{\omega\in\Omega\mid \omega_1=\omega_2=\cdots=\omega_{i-1}=0;\, \omega_{i}=1\}$$

to be the event that the first $1$ to appear occurs in the $i^\text{th}$ component. This is one of the basic events, $\mathcal{A}_i = \mathcal{E}(0,0,\ldots, 1),$ and so obviously is measurable. Clearly $\{\mathcal{A}_i\mid i\ge 1\}$ is a countable partition of $\Omega.$ Moreover, by virtue of $(*)$ the probabilities in this partition are

$$\mathbb{P}(\mathcal{A}_i) = b^{i-1}(1-b)^1 = \frac{1-b}{b}\,b^i = a\,b^i$$

as intended, QED.

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  • $\begingroup$ Beautifully explained, thanks a lot! $\endgroup$ – Statistical Shiba inu Jun 7 at 20:08
  • $\begingroup$ By the way, I deleted the cross post version of this question, won't do that again. I didn't really know this was discouraged, sorry. $\endgroup$ – Statistical Shiba inu Jun 7 at 20:09

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