# Is either $P(A)=\sum_{x \in A} p(1-p)^x$ or $P(A)=1$ if $A$ has a finite number of elements a probability on (S, B)?

My GA proposed the following question in class. I am not too sure which textbook this question came from, so if you can identify which textbook she is sampling, I would enjoy doing some additional readings.

In a random sample, let $$S$$ be the set of all nonnegative integers and the sample space, while $$B$$ is the $$\sigma$$-algebra of $$S$$. Check whether $$P$$ is a probability on $$(S, B)$$:

(i) For $$A \in B$$, $$P(A) = \sum_{x \in A} p(1-p)^x$$, where $$0

(ii) For $$A \in B$$, $$P(A) = 1$$ if $$A$$ has a finite number of elements. Otherwise, $$P(A)=0$$.

My work:

(i) Since $$1>p>0$$, $$1-p > 0$$, so the first Kolmogorov Axiom is satisfied. To show the second axiom, that is $$P(S)=1$$, I would assume that you would need to manipulate the series to show that $$P(A)$$ sums to 1. However, I am not too sure how to manipulate this series. The third axiom also does not follow easily for me, in which we assume that $$A_i$$ are pairwise disjoint.

(ii) Since $$P(A)$$ only has two options (namely, 0 and 1), then the first axiom is satisfied ($$P(A) \ge 0$$). Clearly, $$P(A) = 1$$ if $$A$$ has a finite number of elements, but can we assume that A does? As in (i), I am not too sure how to prove the third Kolmogorov Axiom for this item.

For (i): Write down the sum explicitly (show it to us in the comments) and then you should read about the geometric series... Unfortunately, the wikipedia article is really shitty: there is a lot of blahblah but the most important formula is not present: The geometric series essentially states that if $$q$$ is a number such that $$0 < q < 1$$ then $$\sum_{n=0}^\infty q^n = \frac{1}{1-q}$$

For (ii): Consider $$A_n = \{n\}$$. The $$A_n$$ are pairwise disjoint. What is $$\sum_{n \in \mathbb{N}_0} P(A_n)$$ and what is $$P(\mathbb{N}_0)$$?

EDIT: More hints: Since the $$A_n$$ are pairwise disjoint, we must have $$P(\bigcup_{n \in \mathbb{N}_0} A_n) = \sum_{n \in \mathbb{N}_0} P(A_n)$$. So, in order to see whether or not $$P$$ is a probability measure in the case (ii) we should simply compare both sides. Each set $$A_n$$ is finite, hence $$P(A_n)=...$$. The set $$\mathbb{N}_0$$ is not finite, hence $$P(\mathbb{N}_0)=...$$ and therefore the equation ... (holds/does not hold). You have to fill in the blanks :-)

• I see. Since $\Sigma_{x \in A} p(1-p)^x = p\Sigma_{x \in A} (1-p)^x$, and $\Sigma_{x \in A} (1-p)^x$ is a geometric series that converges to $\frac{1-p}{1-(1-p)}$, this becomes $(1-p) \ne 1$, since $0 < p <1$. Commented Sep 3, 2019 at 23:59
• For (ii), would it be sufficient to say that since $S=\{1,2,3,...\}$, $S$ is not finite. Therefore, $P(A)=0 \forall A$? Or do some subsets $A_i$ have a finite number of elements? Can you please expand a bit on your explanation for (ii)? I am a bit confused on your $P(N_0)$ work. Commented Sep 4, 2019 at 0:01
• Thank you for the additional hints. How is each set $A_n$ finite? I am under the impression that since $A_n \in B$, where $B$ is the $\sigma$-algebra of an infinite set, then some $A_n$ will be infinite while others will be finite. Commented Sep 4, 2019 at 20:27
• $A_n=\{n\}$ consists of just one element and is therefore finite... Commented Sep 5, 2019 at 6:03
• I understand now. Thank you for your help! Commented Sep 5, 2019 at 15:24