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The online course that I am studying has used the following notation several times and it is confusing me:

Let:

  • $A =\left \{ S_n \le t \right \}$, be the event $S_n$ has occured by time $t$; and
  • $B =\left \{ S_{n+1} > t \right \}$, be the event that $S_{n+1}$ has occured after $t$.

It is said that $B^c \subset A$ where $B^c =\left \{ S_{n+1} \le t \right \}$.

How can $B^c$ be a strict subset of $A$ when $A$ is based on the event $S_n$ but $B^c$ the event $S_{n+1}$?

Thanks in advance.

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  • $\begingroup$ The notation "$\subset$" does not imply equality is impossible. See p. 3 of Halmos, e.g. $\endgroup$
    – whuber ♦
    Mar 28, 2019 at 12:56
  • $\begingroup$ In some counting process models, there is an assumption that two occurrences cannot occur simultaneously (e,g. Poisson process models) and so $B^c$, the event that the $(n+1)$-th occurrence was at $t$ or earlier, implies that the $n$-th occurrence must have been striictly before $t$ $\endgroup$
    – Dilip Sarwate
    Mar 28, 2019 at 13:36
  • $\begingroup$ @Dilip I believe the assumption usually is that a simultaneous occurrence has zero probability. $\endgroup$
    – whuber ♦
    Mar 28, 2019 at 15:58

2 Answers 2

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Let's use the word arrivals for what the OP calls occurrences and let $S_n$ denote the time of the $n$-th arrival. Thus, $S_n$ is a real-valued nonnegative random variable. and it must be that $$S_1 \leq S_2 \leq \cdots \leq S_n \leq S_{n+1} \leq \cdots $$ where the event that any of those $\leq$ signs are actually $=$ signs is an event of probability $0$ as @whuber points out in response to my comment on the main question.

Now, in ordinary probability theory where we begin discussions of probability spaces and what is meant by $(\Omega, \mathcal F, P)$, one of the notions drummed into us is that an event is an element of the $\sigma$-algebra $\mathcal F$ and is thus a specially blessed subset of the sample space $\Omega$ to which the probability measure $P$ assigns a real number called the probability of the event (subject to the axioms etc but no matter). Now, when the experiment is performed, one outcome $\omega \in \Omega$ occurs, but we also say that every event that contains $\omega$ has also occurred -- on any trial of the experiment, a single outcome occurs but multiple events occur. I will likely get down-voted for the following but what the heck: on any trial of the experiment, exactly "half" the events in $\mathcal F$ occur and the other "half" don't in the sense that there is a one-to-one correspondence between events that have occurred and the events that haven't -- one of $A$ and $A^c$ in $\mathcal F$ occurs and the other doesn't, with the reminder to ourselves that $\Omega \in \mathcal F$ always occurs and $\emptyset \in \mathcal F$ never occurs. Note that if event $B$ is a subset (proper subset or not, it doesn't matter) of event $A$ -- that is, $B \subset A$ -- and we are told that $B$ has occurred, then we are can be sure that event $A$ also has occurred: $\omega \in B \implies \omega \in A$.

With this in hand, let us consider the events $C = \{S_n \leq t\}$ and $D = \{S_{n+1} \leq t\}$. If we are told that the event $D$ has occurred, that is, the $(n+1)$-th arrival occurred at or before time $t$, then we know for sure that since $S_n \leq S_{n+1}$, it must be that the $n$-th arrival also occurred at or before the time $t$, that is, the event $C$ also occurred. Event $D = \{S_{n+1} \leq t\}$ is indeed a subset of event $C = \{S_n \leq t\}$.

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  • $\begingroup$ Thanks very much for your reply and going out on a limb to risk being voted down! Would I be correct in restating your answer as follows: At n, there are a more possible events that could occur in the future (as less is known) in comparison to n+1 where more is known and therefore the remaining events that can occur is fewer (at n+1 compared to n). As such, the complement of B is a strict subset of A? $\endgroup$
    – Tejay Lovelock
    Mar 29, 2019 at 6:55
  • $\begingroup$ There is no "at $n$" in this model. If the event $B^c = \{S_{n+1} \leq t\}$ has occurred, then we are guaranteed that the event $A = \{S_n \leq t\}$ has also occurred. In fact, with probability $1$, the event $\{S_n < <t\}$ has occurred, but that is a subset of $A$ and so $A$ has definitely occurred when $B^c$ has occurred. Heck, the event $\hat{A}=\{S_n \leq 3.1415926t\}$ has also occurred, and so on. So, every outcome of the experiment that results in $B^c$ having occurred also results in $A$ (and all supersets of $A$) having occurred, that is, $B^c \subset A$. $\endgroup$
    – Dilip Sarwate
    Mar 29, 2019 at 13:08
  • $\begingroup$ Thank you very much for your time and effort! $\endgroup$
    – Tejay Lovelock
    Mar 30, 2019 at 3:17
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This troubled me too in the beginning. There are multiple notational conventions:

either $\subset$ for strict subset, $\subseteq$ for non-strict subset

or $\subsetneq$ for strict subset, $\subset$ for non-strict subset

(or $\subsetneq$ for strict subset, $\subseteq$ for non-strict subset, but I never encountered this combination in the wild).

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