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		<title>Interchangeability of Limits and Probability of Increasing or Decreasing Sequence of Events</title>
		<link>https://yutsumura.com/interchangeability-of-limits-and-probability-of-increasing-or-decreasing-sequence-of-events/</link>
				<comments>https://yutsumura.com/interchangeability-of-limits-and-probability-of-increasing-or-decreasing-sequence-of-events/#respond</comments>
				<pubDate>Mon, 20 Jan 2020 18:54:31 +0000</pubDate>
		<dc:creator><![CDATA[Yu]]></dc:creator>
				<category><![CDATA[Probability]]></category>
		<category><![CDATA[axiom of probability]]></category>
		<category><![CDATA[decreasing sequence of events]]></category>
		<category><![CDATA[increasing sequence of events]]></category>
		<category><![CDATA[interchangeable]]></category>
		<category><![CDATA[limit]]></category>
		<category><![CDATA[probability]]></category>

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				<description><![CDATA[<p>A sequence of events $\{E_n\}_{n \geq 1}$ is said to be increasing if it satisfies the ascending condition \[E_1 \subset E_2 \subset \cdots \subset E_n \subset \cdots.\] Also, a sequence $\{E_n\}_{n \geq 1}$ is&#46;&#46;&#46;</p>
<p>The post <a href="https://yutsumura.com/interchangeability-of-limits-and-probability-of-increasing-or-decreasing-sequence-of-events/" target="_blank">Interchangeability of Limits and Probability of Increasing or Decreasing Sequence of Events</a> first appeared on <a href="https://yutsumura.com/" target="_blank">Problems in Mathematics</a>.</p>]]></description>
								<content:encoded><![CDATA[<h2> Problem 744</h2>
<p>	A sequence of events $\{E_n\}_{n \geq 1}$ is said to be <strong>increasing</strong> if it satisfies the ascending condition<br />
	\[E_1 \subset E_2 \subset \cdots \subset E_n \subset \cdots.\]
	Also, a sequence $\{E_n\}_{n \geq 1}$ is called <strong>decreasing</strong> if it satisfies the descending condition<br />
	\[E_1 \supset E_2 \supset \cdots \supset E_n \supset \cdots.\]
<p>	When $\{E_n\}_{n \geq 1}$ is an increasing sequence, we define a new event denoted by $\lim_{n \to \infty} E_n$ by<br />
	\[\lim_{n \to \infty} E_n := \bigcup_{n=1}^{\infty} E_n.\]
<p>	Also, when $\{E_n\}_{n \geq 1}$ is a decreasing sequence, we define a new event denoted by $\lim_{n \to \infty} E_n$ by<br />
	\[\lim_{n \to \infty} E_n := \bigcap_{n=1}^{\infty} E_n.\]
<p><strong>(1)</strong> Suppose that $\{E_n\}_{n \geq 1}$ is an increasing sequence of events. Then prove the equality of probabilities<br />
	\[\lim_{n \to \infty} P(E_n) = P\left(\lim_{n \to \infty} E_n \right).\]
	Hence, the limit and the probability are interchangeable.</p>
<p><strong>(2)</strong> Suppose that $\{E_n\}_{n \geq 1}$ is a decreasing sequence of events. Then prove the equality of probabilities<br />
	\[\lim_{n \to \infty} P(E_n) = P\left(\lim_{n \to \infty} E_n \right). \]
<p><span id="more-7188"></span></p>
<h2> Proof. </h2>
<h3>Proof of (1)</h3>
<p>Let $\{E_n\}_{n \geq 1}$ be an increasing sequence of events. Then we define new events $F_n$ as follows.<br />
		\begin{align*}<br />
		F_1 &#038;= E_1\\<br />
		F_n &#038; = E_n \setminus E_{n-1} &#038;&#038; \text{ for any } n \geq 2<br />
		\end{align*}<br />
		The event $F_n$ is depicted as a yellow region in the figure below.</p>
<p>		<img src="https://i2.wp.com/yutsumura.com/wp-content/uploads/2020/01/Event_f_definition.jpg?resize=600%2C600&#038;ssl=1" alt="definition of the set F_n" width="600" height="600" class="alignnone size-full wp-image-7189" srcset="https://i2.wp.com/yutsumura.com/wp-content/uploads/2020/01/Event_f_definition.jpg?w=600&amp;ssl=1 600w, https://i2.wp.com/yutsumura.com/wp-content/uploads/2020/01/Event_f_definition.jpg?resize=300%2C300&amp;ssl=1 300w, https://i2.wp.com/yutsumura.com/wp-content/uploads/2020/01/Event_f_definition.jpg?resize=150%2C150&amp;ssl=1 150w, https://i2.wp.com/yutsumura.com/wp-content/uploads/2020/01/Event_f_definition.jpg?resize=160%2C160&amp;ssl=1 160w, https://i2.wp.com/yutsumura.com/wp-content/uploads/2020/01/Event_f_definition.jpg?resize=320%2C320&amp;ssl=1 320w" sizes="(max-width: 600px) 100vw, 600px" data-recalc-dims="1" /></p>
<p>		From the figure, we can see that the events $\{F_n\}_{n \geq 1}$ are mutually exclusive, that is, $F_iF_j = \emptyset$ for any $i \neq j$.<br />
		Furthermore, note that<br />
		\[\bigcup_{i=1}^n F_i = \bigcup_{i=1}^n E_i \tag{*}\]
		for any $n$, and<br />
		\[\bigcup_{i=1}^{\infty} F_i = \bigcup_{i=1}^{\infty} E_i. \tag{**}\]
<p>		Now we proceed to the main part and prove the equality<br />
		\[\lim_{n \to \infty} P(E_n) = P \left(\lim_{n \to \infty} E_n \right).\]
		We start with the right-hand-side and show that it is equal to the left-hand-side.<br />
		\begin{align*}<br />
		P\left(\lim_{n \to \infty} E_i \right) &#038;= P \left(\bigcup_{i=1}^{\infty} E_n \right) &#038;&#038; \text{ definition of } \lim_{n \to \infty} E_n\\[6pt]
		&#038;=  P \left(\bigcup_{i=1}^{\infty} F_i \right) &#038;&#038; \text{ by } (**) \\[6pt]
		&#038;= \sum_{i=1}^{\infty} P(F_i) &#038;&#038; \text{ by one of the axioms of probability} \\<br />
		&#038; &#038;&#038; \text{ with mutually exclusive events } \{F_n\}\\[6pt]
		&#038;= \lim_{n \to \infty} \sum_{i=1}^n P(F_i) \\[6pt]
		&#038;= \lim_{n \to \infty} P\left( \bigcup_{i=1}^n F_i \right) &#038;&#038; \text{ by additivity with mutually exclusive events } \{F_n\}\\[6pt]
		&#038;= \lim_{n \to \infty} P\left(\bigcup_{i=1}^n E_i \right) &#038;&#038; \text{ by } (*)\\[6pt]
		&#038;= \lim_{n \to \infty} P(E_n) &#038;&#038; \text{as $\{E_n\}$ is increasing}.<br />
		\end{align*}<br />
		This proves the desired equality.</p>
<h3>Proof of (2)</h3>
<p>Now, we consider a decreasing sequence of events $\{E_n\}_{n\geq 1}$.<br />
	As $\{E_n\}_{n\geq 1}$ is decreasing, its complement $\{E_n^c\}_{n \geq 1}$ is increasing. Thus, by the result of part (1), we see that<br />
	\[\lim_{n \to \infty} P(E_n^c) = P \left(\lim_{n \to \infty} E_n^c \right). \tag{***}\]
	By definition, we have<br />
	\[\lim_{n \to \infty} E_n^c = \bigcup_{n=1}^{\infty} E_n^c = \left(\bigcap_{n=1}^{\infty} E_n \right)^c.\]
<p>	It follows that the right-hand-side of the equality (***) becomes<br />
	\begin{align*}<br />
	P \left(\lim_{n \to \infty} E_n^c \right) &#038;= P\left(\left(\bigcap_{n=1}^{\infty} E_n \right)^c \right)\\[6pt]
	&#038;= 1 &#8211; P\left(\bigcap_{n=1}^{\infty} E_n \right)\\[6pt]
	&#038;= 1 &#8211; P\left(\lim_{n \to \infty} E_n \right).<br />
	\end{align*}<br />
	On the other hand, the left-hand-side of the equality (***) becomes<br />
	\begin{align*}<br />
		\lim_{n \to \infty} P(E_n^c) &#038;= \lim_{n \to \infty} \left( 1 &#8211; P(E_n) \right)\\[6pt]
		&#038;= 1 &#8211; \lim_{n \to \infty} P(E_n).<br />
	\end{align*}</p>
<p>	Combining these results yields<br />
	\[1 &#8211; \lim_{n \to \infty} P(E_n) = 1 &#8211; P\left(\lim_{n \to \infty} E_n \right).\]
	Equivalently, we have<br />
	\[\lim_{n \to \infty} P(F_n) = P\left(\lim_{n \to \infty} E_n \right),\]
	which is the desired equality. This completes the proof.</p>
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