373 lines
14 KiB
Plaintext
373 lines
14 KiB
Plaintext
<section xml:id="sec-Conditional-Probability" xmlns:xi="http://www.w3.org/2001/XInclude">
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<title>Conditional Probability</title>
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<subsection xml:id="subsec-conditional-probability">
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<title>Conditional Probability</title>
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<p>
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Sometimes events can interact with each other.
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We would like to have the language to talk about a scenario in which
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evidence that one event has occurred can alter our understanding of the
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probability of another event occurring.
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We would also like to develop the mathematical tools to quantify exactly
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how much that probability changes.
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</p>
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<definition xml:id="def-conditional-probability">
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<statement>
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<p>
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Let <m>A</m> and <m>B</m> be events.
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The <term>conditional probability</term> of <m>A</m> given <m>B</m> is
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<md>
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<mrow> \Pr(A \mid B) = \frac{\Pr(A\cap B)}{\Pr(B)}.</mrow>
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</md>
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</p>
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</statement>
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</definition>
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<example xml:id="example-rolls-conditional">
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<statement>
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<p>
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An experiment consists of rolling a fair 6-sided die two times.
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Let <m>A</m> be the event that the sum of the rolls is at least 10.
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To find <m>\Pr(A)</m>, we note that
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<m>A = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\}</m>, which
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has 6 elements.
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Since <m>|\Omega| = 36</m>, we have
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<md>
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<mrow>\Pr(A) = \frac{|A|}{|\Omega|} = \frac{6}{36} = \frac{1}{6}</mrow>
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</md>.
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</p>
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<p>
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Let <m>B</m> be the event that the first roll is 6.
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If we pause after the first die roll seeing the value of 6, we might
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be more inclined to expect a sum of at least 10.
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The evidence that we've already seen changes our understanding of the
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situation.
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Since <m>|B| = 6</m>, <m>\Pr(B) = \frac{6}{36} = \frac{1}{6}</m>.
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Also, <m>A\cap B = \{(6, 4), (6, 5), (6, 6)\}</m>, so
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<m>\Pr(A\cap B) = \frac{3}{36} = \frac{1}{12}</m>.
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Finally:
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<md>
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<mrow>\Pr(A \mid B) = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{1/12}{1/6} = \frac{1}{2}</mrow>
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</md>
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Before the experiment, we would have said there was only a
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<m>\frac{1}{6}</m> chance that the sum of the rolls is at least 10.
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However, with the additional knowledge of seeing the first roll of 6,
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we find the probability of a sum of at least 10 to be
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<m>\frac{1}{2}</m>, substantially more likely than before.
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</p>
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</statement>
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</example>
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</subsection>
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<subsection xml:id="subsec-diagnostic-testing">
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<title>Diagnostic Testing</title>
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<p>
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Diagnostic tests for diseases aren't perfect.
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When a test comes back positive or negative, a patient will want to
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understand the (conditional) probability that they have or don't have the
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disease based on the evidence (the diagnostic test result).
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</p>
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<definition xml:id="def-sensitivity-specificity">
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<statement>
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<p>
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The <term>sensitivity</term> of a diagnostic test is the probability
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that a patient who has the disease will see a positive test result.
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The <term>specificity</term> is the probability that a patient who
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does not have the disease will see a negative test result.
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</p>
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</statement>
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</definition>
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<p>
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Introducing event notation, let <m>D</m> be the event that a patient has
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the disease, and let <m>P</m> be the event that they receive a positive
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test result.
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Then the sensitivity of the diagnostic test is <m>\Pr(P \mid D)</m>, and
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the specificity is <m>\Pr(P^c \mid D^c)</m>.
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However, when the patient takes a diagnostic test, the conditional
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probabilities they would be most interested in would be
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<m>\Pr(D \mid P)</m> and <m>\Pr(D^c \mid P^c)</m>.
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</p>
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<p>
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Bayes' Theorem expresses the relationship between a conditional
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probability <m>\Pr(A \mid B)</m> and the flipped conditional probability
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<m>\Pr(B \mid A)</m>.
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</p>
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<theorem xml:id="thm-Bayes-v1">
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<title>Bayes' Theorem (v1)</title>
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<statement>
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<p>
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<md>
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<mrow> \Pr(B \mid A) = \frac{\Pr(A \mid B)\Pr(B)}{\Pr(A)} </mrow>
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</md>
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</p>
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</statement>
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</theorem>
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<p>
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For example, if a patient sees a positive diagnostic test result, they
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might try to calculate:
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<md>
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<mrow> \Pr(D \mid P) = \frac{\Pr(P \mid D)\Pr(D)}{\Pr(P)} </mrow>
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</md>
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It will take some work to be able to use this formula.
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We likely don't have direct access to <m>\Pr(P)</m> or <m>\Pr(D)</m>.
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</p>
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<p>
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Let's consider <m>\Pr(P)</m>, the probability of receiving a positive test
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result.
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The sensitivity <m>\Pr(P \mid D)</m> tells us this probability under the
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condition that the patient has the disease.
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For a patient who doesn't have the disease, the specificity isn't quite
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the number we're looking for.
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However, consider the complementary probability:
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<md>
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<mrow> \Pr(P \mid D^c) = 1 - \Pr(P^c \mid D^c) </mrow>
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</md>
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The total <m>\Pr(P)</m> can be divided into two categories: patients who
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have the disease and test positive, and patients who don't have the
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disease and test positive.
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So:
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<md>
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<mrow> \Pr(P) \amp = \Pr(P\cap D) + \Pr(P\cap D^c) </mrow>
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<mrow> \amp = \Pr(P\mid D)\Pr(D) + \Pr(P \mid D^c)\Pr(D^c) </mrow>
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<mrow> \amp = \Pr(P\mid D)\Pr(D) + (1 - \Pr(P^c \mid D^c))\Pr(D^c) </mrow>
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<mrow> \amp = (\text{sensitivity})\Pr(D) + (1 - \text{ specificity})\Pr(D^c) </mrow>
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</md>
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We can take this breakdown of <m>\Pr(P)</m> and write a new version of
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Bayes' Theorem:
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</p>
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<theorem xml:id="thm-Bayes-v2">
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<title>Bayes' Theorem (v2)</title>
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<statement>
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<p>
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<md>
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<mrow> \Pr(B \mid A) = \frac{\Pr(A \mid B)\Pr(B)}{\Pr(A \mid B)\Pr(B) + (1 - \Pr(A^c \mid B^c))\Pr(B^c)} </mrow>
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</md>
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</p>
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</statement>
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</theorem>
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<p>
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We're still missing a crucial piece of information: <m>\Pr(D)</m>, the
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probability (not conditioned on any evidence) that the patient has the
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disease.
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This is often referred to as the <term>prior</term>, as in, our prior
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understanding of the probability of something before we gained some new
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information from evidence.
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The conditional probability calculated using Bayes' Theorem is usually
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called the <term>posterior</term> (i.e., after taking evidence into
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account).
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</p>
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<p>
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There isn't always one single number that's reasonable to use as the prior
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probability.
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For example, in a diagnostic testing situation, the
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<term>prevalence</term> of the disease <mdash/> i.e., the proportion of
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the population who have the disease <mdash/> might feel like a natural
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number to use as the prior.
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However, what prevalence should you use?
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During the COVID-19 pandemic, the prevalence of COVID in a particular
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country, state, and city might be different.
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There's also the possibility of applying Bayes' Theorem multiple times to
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take into account multiple pieces of evidence, using the posterior
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probability from one application of Bayes' Theorem to play the role of the
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prior probability in the next.
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This idea would apply if, for example, a patient took a second diagnostic
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test to double-check.
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(Complicating the issue further, the developers of diagnostic tests often
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publish two or even more sets of sensitivity and specificity values,
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depending on whether a patient is already showing certain symptoms.)
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</p>
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<example>
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<statement>
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<p>
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A 50-year old woman with no symptoms is screened for breast cancer and
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tests positive.
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If the prevalence of breast cancer for women in her age group is 1%
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and the particular screening process used has a sensitivity of 90% and
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a specificity of 91%, what is the probability that the woman has
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breast cancer given her positive result?
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</p>
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</statement>
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<solution>
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<p>
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Let <m>P</m> be the event of testing positive and <m>D</m> the event
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of having the disease.
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Then the prevalence <m>\Pr(D)</m> is given as 1%, or 0.01.
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The sensitivity is <m>\Pr(P\mid D) = 0.9</m>, and the specificity is
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<m>\Pr(P^c\mid D^c) = 0.91</m>.
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So, according to Bayes' Theorem:
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<md>
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<mrow> \Pr(D\mid P) \amp = \frac{\Pr(P\mid D)\Pr(D)}{\Pr(P\mid D)\Pr(D) + (1 - \Pr(P^c\mid D^c))\Pr(D^c)} </mrow>
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<mrow> \amp = \frac{(0.9)(0.01)}{(0.9)(0.01) + (1 - 0.91)(1 - 0.01)} </mrow>
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<mrow> \amp \approx 0.092 </mrow>
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</md>
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This may seem like a surprising result.
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Despite sensitivity and specificty values around 90%, it turns out a
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positive test result only indicates a less than 10% chance of actually
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having the disease.
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Keep in mind that many ideas in probability and statistics can be
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highly counterintuitive.
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It's important to be very precise with statements and calculations.
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</p>
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</solution>
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</example>
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</subsection>
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<exercises xml:id="exercises-Conditional-Probability">
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<exercisegroup>
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<introduction>
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<p>
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In each of the following scenarios with given events <m>A</m> and
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<m>B</m>, calculate <m>\Pr(A), \Pr(B)</m>, <m>\Pr(A\cap B)</m>,
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<m>\Pr(A \mid B)</m>, and <m>\Pr(B \mid A)</m>.
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</p>
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</introduction>
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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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</p>
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</statement>
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<answer>
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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} </mrow>
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<mrow> \Pr(B \mid A) \amp = \frac{\Pr(B\cap A)}{\Pr(A)} = \frac{6/36}{18/36} = \frac{6}{18} = \frac{1}{3} </mrow>
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</md>
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</p>
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</answer>
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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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</p>
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</statement>
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<answer>
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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} </mrow>
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<mrow> \Pr(B \mid A) \amp = \frac{\Pr(B\cap A)}{\Pr(A)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} </mrow>
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</md>
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</p>
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</answer>
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</exercise>
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</exercisegroup>
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<exercise>
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<introduction>
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<p>
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A diagnostic test is developed to detect a disease present in 3.2% of
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the population.
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For a patient who has the disease, the test will accurately give a
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positive result 65% of the time.
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When the patient does not have the disease, the test will accurately
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give a negative result 99.9% of the time.
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</p>
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</introduction>
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<task>
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<statement>
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<p>
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For a patient who receives a positive test, what is the probability
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they have the disease?
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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>\approx 0.96</m>.
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</p>
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</answer>
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<solution>
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<p>
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Let <m>P</m> be the event of testing positive and <m>D</m> the event
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of having the disease.
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Then the prevalence <m>\Pr(D)</m> is given as 3.2%, or 0.032.
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The sensitivity is <m>\Pr(P\mid D) = 0.65</m>, and the specificity
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is <m>\Pr(P^c\mid D^c) = 0.999</m>.
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So, according to Bayes' Theorem:
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<md>
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<mrow> \Pr(D\mid P) \amp = \frac{\Pr(P\mid D)\Pr(D)}{\Pr(P\mid D)\Pr(D) + (1 - \Pr(P^c\mid D^c))\Pr(D^c)} </mrow>
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<mrow> \amp = \frac{(0.65)(0.032)}{(0.65)(0.032) + (1 - 0.999)(1 - 0.032)} </mrow>
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<mrow> \amp \approx 0.96 </mrow>
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</md>
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</p>
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</solution>
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</task>
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<task>
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<statement>
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<p>
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For a patient who receives a negative test, what is the probability
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they do not have the disease?
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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>\approx 0.99</m>.
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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> \Pr(D^c\mid P^c) \amp = \frac{\Pr(P^c\mid D^c)\Pr(D^c)}{\Pr(P^c\mid D^c)\Pr(D^c) + (1 - \Pr(P\mid D))\Pr(D)} </mrow>
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<mrow> \amp = \frac{(0.999)(1 - 0.032)}{(0.999)(1 - 0.032) + (1 - 0.65)(0.032)} </mrow>
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<mrow> \amp \approx 0.99 </mrow>
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</md>
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</p>
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</solution>
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</task>
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</exercise>
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</exercises>
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</section> |