Probability of a Union Event
Suppose that a probability space \(\left( \Omega,\mathcal{F},P\right) \)is given. That is, the family of events \(\mathcal{F}\subset2^{\Omega}\)satisfies the axioms of a measurable space \begin{align*}
& \left( M_{1}\right) \ \mathcal{F}\not =\phi\\
& \left( M_{2}\right) \ \forall A\in\mathcal{F}:A^{c}\in\mathcal{F}\\
& \left( M_{3}\right) \ \forall\left\{ A_{n}\right\} _{n\in\mathbb{N} }\subset\mathcal{F}:\bigcup_{n\in\mathbb{N} }A_{n}\in\mathcal{F},
\end{align*}and the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) satisfies the axioms of probability \begin{align*}
& \left( P_{1}\right) \ \forall A\in\mathcal{F}:P\left( A\right) \geq0\\
& \left( P_{2}\right) \ P\left( \Omega\right) =1\\
& \left( P_{3}\right) \ \forall\text{mutually disjoint }\left\{
A_{n}\right\} _{n\in\mathbb{N} }\subset\mathcal{F}:P\left( \bigcup_{n\in\mathbb{N} }A_{n}\right) =\sum_{n\in\mathbb{N} }P\left( A_{n}\right) .
\end{align*}
Axiom \(\left( P_{3}\right) \) concerns a countable collection of mutually disjoint events. For a sequence \(\left\{ A_{n}\right\} _{n\in\mathbb{N} }\) of countably many events that are not necessarily mutually disjoint, the \(\sigma\)-subadditivity of the probability measure \(P\) gives \[
P\left( \bigcup_{n\in\mathbb{N} }A_{n}\right) \leq\sum_{n=1}^{+\infty}P\left( A_{n}\right) ,
\]that is,
\[
P\left( \bigcup_{n\in\mathbb{N} }A_{n}\right) \leq\lim_{N\rightarrow\infty}\sum_{n=1}^{N}P\left(
A_{n}\right) .
\]
When a countable collection of events satisfies certain conditions, more can be said about the probabilities of their union and intersection. In what follows, we write the sequence of events \(\left\{ A_{n}\right\} _{n\in\mathbb{N} }\) simply as \(\left\{ A_{n}\right\} \).
Monotone Sequences of Events
If a sequence of events \(\left\{ A_{n}\right\} \subset\mathcal{F}\) satisfies
\[
A_{1}\subset A_{2}\subset\cdots,
\]that is,
\[
\forall n\in\mathbb{N} :A_{n}\subset A_{n+1},
\]then \(\left\{ A_{n}\right\} \) is called a monotonically increasing sequence.
If a sequence of events \(\left\{ A_{n}\right\} \subset\mathcal{F}\) satisfies
\[
A_{1}\supset A_{2}\supset\cdots,
\]that is,
\[
\forall n\in\mathbb{N} :A_{n}\supset A_{n+1},
\]then \(\left\{ A_{n}\right\} \) is called a monotonically decreasing sequence.
Monotonically increasing sequences and monotonically decreasing sequences are collectively called monotone sequences. Thus, a sequence of events \(\left\{ A_{n}\right\} \subset\mathcal{F}\) is monotone if it is monotonically increasing or monotonically decreasing, or both.
As the following examples suggest, monotone sequences of events frequently arise in experiments involving repeated trials.
\left( 1,0,0,1,\cdots\right) .
\]If the outcome of the \(n\in\mathbb{N} \)th coin toss is denoted by \[
\omega_{n}\in\{1,0\},
\]then each sample point can be written as
\[
\omega=\left( \omega_{1},\omega_{2},\cdots\right) .
\]The sample space is
\[
\Omega=\{1,0\}^{\mathbb{N} },
\]which is uncountable. Define the general term of the sequence of events \(\{A_{n}\}\) by \[
A_{n}=\text{at least one head occurs in the first }n\text{ tosses}.
\]If a head has occurred by the first \(n\) tosses, then a head has also occurred by the first \(n+1\) tosses. Thus, for an arbitrary index \(n\in\mathbb{N} \) and every sample point \(\omega\in\Omega\), \[
\omega\in A_{n}\Rightarrow\omega\in A_{n+1},
\]so
\[
A_{n}\subset A_{n+1}.
\]Therefore, the sequence of events \(\{A_{n}\}\) is monotonically increasing.
\left( 1,0,0,1,\cdots\right) .
\]If the outcome of the \(n\in\mathbb{N} \)th coin toss is denoted by \[
\omega_{n}\in\{1,0\},
\]then each sample point can be written as
\[
\omega=\left( \omega_{1},\omega_{2},\cdots\right) .
\]The sample space is
\[
\Omega=\{1,0\}^{\mathbb{N} },
\]which is uncountable. Define the general term of the sequence of events
\(\{A_{n}\}\) by \[
A_{n}=\text{all of the first }n\text{ tosses are tails}.
\]If all of the first \(n+1\) tosses are tails, then all of the first \(n\) tosses are also tails. Thus, for an arbitrary index \(n\in\mathbb{N} \) and every sample point \(\omega\in\Omega\), \[
\omega\in A_{n+1}\Rightarrow\omega\in A_{n},
\]so
\[
A_{n+1}\subset A_{n}.
\]Therefore, the sequence of events \(\{A_{n}\}\) is monotonically decreasing.
B_{n}=\bigcup_{i=1}^{n}A_{i}.
\]Then, for every index \(n\in\mathbb{N} \), \begin{align*}
B_{n} & =\bigcup_{i=1}^{n}A_{i}\quad\because\text{definition of }\left\{
B_{n}\right\} \\
& \subset\bigcup_{i=1}^{n+1}A_{i}\quad\because\text{definition of union}\\
& =B_{n+1}\quad\because\text{definition of }\left\{ B_{n}\right\} ,
\end{align*}so \(\{B_{n}\}\) is monotonically increasing. Thus, a monotonically increasing sequence can be constructed from any sequence of events \(\{A_{n}\}\).
B_{n}=\bigcap_{i=1}^{n}A_{i}.
\]Then, for every index \(n\in\mathbb{N} \), \begin{align*}
B_{n} & =\bigcap_{i=1}^{n}A_{i}\quad\because\text{definition of }\left\{
B_{n}\right\} \\
& \supset\bigcap_{i=1}^{n+1}A_{i}\quad\because\text{definition of intersection}\\
& =B_{n+1}\quad\because\text{definition of }\left\{ B_{n}\right\} ,
\end{align*}so \(\{B_{n}\}\) is monotonically decreasing. Thus, a monotonically decreasing sequence can be constructed from any sequence of events \(\{A_{n}\}\).
Continuity from Below of a Probability Measure
Suppose that a sequence of events \(\{A_{n}\}\subset\mathcal{F}\) is monotonically increasing. That is, \[
A_{1}\subset A_{2}\subset\cdots.
\]Since the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) is monotone, the sequence of probabilities \(\left\{ P\left( A_{n}\right) \right\} \) satisfies \[
P\left( A_{1}\right) \leq P\left( A_{2}\right) \leq\cdots.
\]Thus, \(\left\{ P\left( A_{n}\right) \right\} \) is a monotonically increasing sequence of real numbers. Since the event space \(\mathcal{F}\) is closed under countable unions, \[
\bigcup_{n=1}^{+\infty}A_{n}\in\mathcal{F},
\]and therefore the probability \(P\left( \bigcup_{n=1}^{+\infty}A_{n}\right) \) of the union event is guaranteed to be a real number under the probability measure \(P\). In this case, the sequence of real numbers \(\left\{ P\left( A_{n}\right) \right\} \) converges to a finite real number, and \[
P\left( \bigcup_{n=1}^{+\infty}A_{n}\right) =\lim_{n\rightarrow+\infty
}P\left( A_{n}\right) .
\]This property is called the continuity from below of the probability measure \(P\).
P\left( \bigcup_{n=1}^{+\infty}A_{n}\right) =\lim_{n\rightarrow+\infty
}P\left( A_{n}\right) .
\]
\left( 1,0,0,1,\cdots\right) .
\]Suppose that each coin toss is fair and independent. If the outcome of the
\(n\in\mathbb{N} \)th coin toss is denoted by \[
\omega_{n}\in\{1,0\},
\]then each sample point can be written as
\[
\omega=\left( \omega_{1},\omega_{2},\cdots\right) .
\]The sample space is
\[
\Omega=\{1,0\}^{\mathbb{N} },
\]which is uncountable. Define the general term of the sequence of events
\(\{A_{n}\}\) by \[
A_{n}=\text{at least one head occurs in the first }n\text{ tosses}.
\]As shown above, \(\left\{ A_{n}\right\} \) is monotonically increasing. The complement of \(A_{n}\) is \[
A_{n}^{c}=\text{all of the first }n\text{ tosses are tails},
\]so its probability is
\[
P\left( A_{n}^{c}\right) =\left( \frac{1}{2}\right) ^{n}.
\]Therefore,
\begin{align*}
P\left( A_{n}\right) & =1-P\left( A_{n}^{c}\right) \\
& =1-\left( \frac{1}{2}\right) ^{n}.
\end{align*}Consider the union event
\[
\bigcup_{n=1}^{+\infty}A_{n}.
\]For an arbitrary \(\omega\in\Omega\), \begin{align*}
\omega\in\bigcup_{n=1}^{+\infty}A_{n} & \Leftrightarrow\exists n\in\mathbb{N} :\omega\in A_{n}\\
& \Leftrightarrow\exists n\in\mathbb{N} :\text{at least one head occurs in the first }n\text{ tosses}\\
& \Leftrightarrow\text{at least one head occurs}.
\end{align*}By the preceding proposition,
\begin{align*}
P\left( \bigcup_{n=1}^{+\infty}A_{n}\right) & =\lim_{n\rightarrow+\infty
}P\left( A_{n}\right) \\
& =\lim_{n\rightarrow+\infty}\left[ 1-\left( \frac{1}{2}\right) ^{n}\right] \\
& =1-\lim_{n\rightarrow+\infty}\left( \frac{1}{2}\right) ^{n}\\
& =1-0\\
& =1.
\end{align*}Therefore, the probability that at least one head occurs is \(1\).
Continuity from Above of a Probability Measure
An analogous result holds for monotonically decreasing sequences of events. Specifically, it is as follows.
Suppose that a sequence of events \(\{A_{n}\}\subset\mathcal{F}\) is monotonically decreasing. That is, \[
A_{1}\supset A_{2}\supset\cdots.
\]Since the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) is monotone, the sequence of probabilities \(\left\{ P\left( A_{n}\right) \right\} \) satisfies \[
P\left( A_{1}\right) \geq P\left( A_{2}\right) \geq\cdots.
\]Thus, \(\left\{ P\left( A_{n}\right) \right\} \) is a monotonically decreasing sequence of real numbers. Since the event space \(\mathcal{F}\) is closed under countable intersections, \[
\bigcap_{n=1}^{+\infty}A_{n}\in\mathcal{F},
\]and therefore the probability \(P\left( \bigcap_{n=1}^{+\infty}A_{n}\right) \) of the intersection event is guaranteed to be a real number under the probability measure \(P\). In this case, the sequence of real numbers \(\left\{ P\left( A_{n}\right) \right\} \) converges to a finite real number, and \[
P\left( \bigcap_{n=1}^{+\infty}A_{n}\right) =\lim_{n\rightarrow+\infty
}P\left( A_{n}\right) .
\]This property is called the continuity from above of the probability measure \(P\). The proof is analogous to that of the preceding proposition.
P\left( \bigcap_{n=1}^{+\infty}A_{n}\right) =\lim_{n\rightarrow+\infty
}P\left( A_{n}\right) .
\]
\left( 1,0,0,1,\cdots\right) .
\]Suppose that each coin toss is fair and independent. If the outcome of the
\(n\in\mathbb{N} \)th coin toss is denoted by \[
\omega_{n}\in\{1,0\},
\]then each sample point can be written as
\[
\omega=\left( \omega_{1},\omega_{2},\cdots\right) .
\]The sample space is
\[
\Omega=\{1,0\}^{\mathbb{N} },
\]which is uncountable. Define the general term of the sequence of events
\(\{A_{n}\}\) by \[
A_{n}=\text{all of the first }n\text{ tosses are tails}.
\]As shown above, \(\left\{ A_{n}\right\} \) is monotonically decreasing. The probability of \(A_{n}\) is \[
P\left( A_{n}\right) =\left( \frac{1}{2}\right) ^{n}.
\]Consider the intersection event
\[
\bigcap_{n=1}^{+\infty}A_{n}.
\]For an arbitrary \(\omega\in\Omega\), \begin{align*}
\omega\in\bigcap_{n=1}^{+\infty}A_{n} & \Leftrightarrow\forall n\in\mathbb{N} :\omega\in A_{n}\\
& \Leftrightarrow\forall n\in\mathbb{N} :\text{all of the first }n\text{ tosses are tails}\\
& \Leftrightarrow\text{tails continue forever}.
\end{align*}By the preceding proposition,
\begin{align*}
P\left( \bigcap_{n=1}^{+\infty}A_{n}\right) & =\lim_{n\rightarrow+\infty
}P\left( A_{n}\right) \\
& =\lim_{n\rightarrow+\infty}\left( \frac{1}{2}\right) ^{n}\\
& =0.
\end{align*}Therefore, the probability that tails continue forever is \(0\).
Exercises
B_{n} & =\bigcup_{i=1}^{n}A_{i}\\
C_{n} & =\bigcap_{i=1}^{n}A_{i}.
\end{align*}Show that \(\{B_{n}\}\) is monotonically increasing and that \(\{C_{n}\}\) is monotonically decreasing. Also show that \begin{align*}
\bigcup_{n=1}^{\infty}B_{n} & =\bigcup_{n=1}^{\infty}A_{n}\\
\bigcap_{n=1}^{\infty}C_{n} & =\bigcap_{n=1}^{\infty}A_{n}.
\end{align*}
\[
P\left( A_{n}\right) =1-\frac{1}{n+1}.
\]Find
\[
P\left( \bigcup_{n=1}^{\infty}A_{n}\right) .
\]
\[
P\left( A_{n}\right) =\frac{2}{n+2}.
\]Find
\[
P\left( \bigcap_{n=1}^{\infty}A_{n}\right) .
\]
0<p<1.
\]Define the event \(A_{n}\) by \[
A_{n}=\text{at least one success occurs within the first }n\text{ trials}.
\]Show that \(\{A_{n}\}\) is monotonically increasing and find the probability that at least one success eventually occurs.
P\left( \bigcap_{n=1}^{\infty}A_{n}\right) =\lim_{n\rightarrow\infty}P\left(
A_{n}\right) .
\]
\forall n\in\mathbb{N}:P\left( A_{n}\right) =1.
\]The sequence \(\{A_{n}\}\) need not be monotone. Show, using the continuity of the probability measure, that \[
P\left( \bigcap_{n=1}^{\infty}A_{n}\right) =1.
\]
B_{n}=\bigcup_{i=1}^{n}A_{i}.
\]Using finite additivity and continuity from below, show that
\[
P\left( \bigcup_{n=1}^{\infty}A_{n}\right) = \sum_{n=1}^{\infty}P\left(
A_{n}\right) .
\]
\[
\left( 3,1,5,6,3,\cdots\right) .
\]If the outcome of the \(n\in\mathbb{N} \)th roll is denoted by \(\omega_{n}\in\{1,2,3,4,5,6\}\), then each sample point can be written as \[
\omega=\left( \omega_{1},\omega_{2},\cdots\right) .
\]The sample space of this experiment is
\[
\Omega=\{1,2,3,4,5,6\}^{\mathbb{N} },
\]which is uncountable. Show that the probability that \(1\) never appears is \(0\). Assume that each die roll is fair and independent.
