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 $0<p<1$

(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.


1 Answer 1


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 :-)

  • $\begingroup$ 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 $0 < p <1$. $\endgroup$
    – Ron Snow
    Commented Sep 3, 2019 at 23:59
  • $\begingroup$ 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. $\endgroup$
    – Ron Snow
    Commented Sep 4, 2019 at 0:01
  • $\begingroup$ 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. $\endgroup$
    – Ron Snow
    Commented Sep 4, 2019 at 20:27
  • 1
    $\begingroup$ $A_n=\{n\}$ consists of just one element and is therefore finite... $\endgroup$ Commented Sep 5, 2019 at 6:03
  • $\begingroup$ I understand now. Thank you for your help! $\endgroup$
    – Ron Snow
    Commented Sep 5, 2019 at 15:24

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