243 lines
7.4 KiB
Plaintext
243 lines
7.4 KiB
Plaintext
<section xml:id="sec-Independent-Events" xmlns:xi="http://www.w3.org/2001/XInclude">
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<title>Independent Events</title>
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<p>
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The idea of conditional probability is that knowledge of one event can
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change our understanding of the probability of another.
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But it's also important to understand when this is not the case.
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</p>
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<definition xml:id="def-independent-events">
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<statement>
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<p>
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Events <m>A, B</m> are <term>independent</term> if
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<m>\Pr(A \mid B) = \Pr(A)</m>.
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</p>
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</statement>
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</definition>
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<p>
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The equation <m>\Pr(A\mid B) = \Pr(A)</m> very concretely says: knowledge
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that the event <m>B</m> has occurred does not change our understanding of
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<m>\Pr(A)</m>.
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Assuming <m>\Pr(A) \neq 0</m>, this is equivalent to
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<m>\Pr(A\cap B) = \Pr(A)\Pr(B)</m>, since:
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<md>
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<mrow> \Pr(A\mid B) \amp = \Pr(A) </mrow>
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<mrow> \frac{\Pr(A\cap B)}{\Pr(B)} \amp = \Pr(A) </mrow>
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<mrow> \Pr(A\cap B) \amp = \Pr(A)\Pr(B) </mrow>
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</md>
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This latter form of the equation is often more useful in calculations.
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</p>
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<example xml:id="example-rolls-independent">
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<statement>
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<p>
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In <xref ref="example-rolls-conditional"/>, we considered an experiment
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in which we rolled a fair 6-sided die twice.
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We defined events <m>A</m> that the sum of the rolls is at least 10 and
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<m>B</m> that the first roll is a 6, and we found that
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<m>\Pr(A \mid B) \neq \Pr(A)</m>, so <m>A</m> and <m>B</m> are not
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independent.
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</p>
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<p>
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Now consider the event <m>C</m> that the sum of the rolls is 7, which
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sounds very similar to the event <m>A</m>.
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We have:
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<md>
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<mrow> C \amp = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\} </mrow>
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<mrow> \Pr(C) \amp = \frac{6}{36} = \frac{1}{6} </mrow>
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<mrow> B\cap C \amp = \{(6, 1)\} </mrow>
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<mrow> \Pr(B\cap C) \amp = \frac{1}{36} </mrow>
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<mrow> \text{therefore: } \Pr(C\mid B) \amp = \frac{\Pr(C\cap B)}{\Pr(B)} = \frac{1/36}{1/6} = \frac{1}{6} = \Pr(C) </mrow>
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</md>
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So, even though the descriptions of events <m>A</m> and <m>C</m> are
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very similar, the event <m>C</m> is independent with <m>B</m>, while
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<m>A</m> is not.
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</p>
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</statement>
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</example>
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<exercises xml:id="exercises-Independent-Events">
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<exercise>
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<statement>
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<p>
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An experiment consists of rolling a fair die two times.
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Let <m>A</m> be the event that the sum is even, and let <m>B</m> be
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the event that the second roll is higher than the first.
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Are <m>A</m> and <m>B</m> independent?
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</p>
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</statement>
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<answer>
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<p>
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<m>A</m> and <m>B</m> are not independent.
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</p>
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</answer>
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<solution>
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<p>
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<md>
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<mrow> A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), </mrow>
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<mrow> \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), </mrow>
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<mrow> \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} </mrow>
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<mrow> B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), </mrow>
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<mrow> \amp (2, 3), (2, 4), (2, 5), (2, 6), </mrow>
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<mrow> \amp (3, 4), (3, 5), (3, 6), </mrow>
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<mrow> \amp (4, 5), (4, 6), </mrow>
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<mrow> \amp (5, 6)\} </mrow>
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<mrow> A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\} </mrow>
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</md>
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So <m>\Pr(A) = \frac{18}{36} = \frac{1}{2}</m>,
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<m>\Pr(B) = \frac{15}{36} = \frac{5}{12}</m>, and
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<m>\Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}</m>.
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Finally:
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<md>
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<mrow> \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} \neq \Pr(A), </mrow>
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</md>
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so <m>A</m> and <m>B</m> are not independent.
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</p>
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</solution>
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</exercise>
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<exercise>
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<statement>
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<p>
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An experiment consists of flipping a fair coin three times.
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Let <m>A</m> be the event that the first and second flips match.
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Let <m>B</m> be the event that there are at least two heads.
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Are <m>A</m> and <m>B</m> independent?
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</p>
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</statement>
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<answer>
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<p>
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<m>A</m> and <m>B</m> are independent.
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</p>
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</answer>
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<solution>
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<p>
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<md>
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<mrow> A \amp = \{ HHH, HHT, TTH, TTT \} </mrow>
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<mrow> B \amp = \{ HHH, HHT, HTH, THH \} </mrow>
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<mrow> A\cap B \amp = \{HHH, HHT\} </mrow>
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</md>
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So <m>\Pr(A) = \frac{4}{8} = \frac{1}{2}</m>,
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<m>\Pr(B) = \frac{4}{8} = \frac{1}{2}</m>, and
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<m>\Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}</m>.
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Finally:
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<md>
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<mrow> \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} = \Pr(A), </mrow>
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</md>
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so <m>A</m> and <m>B</m> are independent.
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</p>
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</solution>
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</exercise>
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<exercise>
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<statement>
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<p>
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Let <m>A = \{1, 2, 3\}</m> and <m>B = \{3, 4, 5\}</m> be events in the
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sample space <m>\Omega = \{1, 2, 3, 4, 5, 6\}</m>.
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Create a probability distribution for <m>\Omega</m> so that
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<m>A, B</m> are independent.
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</p>
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</statement>
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<answer>
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<table>
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<title>Example Distribution</title>
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<tabular halign="center">
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<row bottom="minor">
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<cell><m>x</m></cell>
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<cell><m>\Pr(x)</m></cell>
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</row>
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<row>
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<cell>1</cell>
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<cell>0.1</cell>
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</row>
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<row>
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<cell>2</cell>
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<cell>0.2</cell>
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</row>
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<row>
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<cell>3</cell>
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<cell>0.2</cell>
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</row>
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<row>
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<cell>4</cell>
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<cell>0.1</cell>
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</row>
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<row>
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<cell>5</cell>
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<cell>0.1</cell>
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</row>
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<row>
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<cell>6</cell>
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<cell>0.3</cell>
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</row>
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</tabular>
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</table>
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</answer>
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<solution>
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<table>
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<title>Example Distribution</title>
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<tabular halign="center">
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<row bottom="minor">
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<cell><m>x</m></cell>
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<cell><m>\Pr(x)</m></cell>
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</row>
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<row>
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<cell>1</cell>
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<cell>0.1</cell>
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</row>
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<row>
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<cell>2</cell>
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<cell>0.2</cell>
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</row>
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<row>
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<cell>3</cell>
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<cell>0.2</cell>
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</row>
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<row>
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<cell>4</cell>
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<cell>0.1</cell>
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</row>
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<row>
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<cell>5</cell>
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<cell>0.1</cell>
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</row>
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<row>
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<cell>6</cell>
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<cell>0.3</cell>
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</row>
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</tabular>
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</table>
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<p>
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Now <m>\Pr(A) = 0.5</m>, <m>\Pr(B) = 0.4</m>, and
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<md>
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<mrow>\Pr(A\cap B) = 0.2 = (0.5)(0.4) = \Pr(A)\Pr(B),</mrow>
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</md>
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so <m>A</m> and <m>B</m> are independent.
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</p>
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</solution>
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</exercise>
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</exercises>
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</section> |