06_StatisticalInference/01_05_ConditionalProbability/index.html

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    <h1>Conditional Probability</h1>
    <h2>Statistical Inference</h2>
    <p>Brian Caffo, Jeff Leek, Roger Peng<br/>Johns Hopkins Bloomberg School of Public Health</p>
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  <article></article>  
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    <h2>Conditional probability, motivation</h2>
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  <article data-timings="">
    <ul>
<li>The probability of getting a one when rolling a (standard) die
is usually assumed to be one sixth</li>
<li>Suppose you were given the extra information that the die roll
was an odd number (hence 1, 3 or 5)</li>
<li><em>conditional on this new information</em>, the probability of a
one is now one third</li>
</ul>

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    <h2>Conditional probability, definition</h2>
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    <ul>
<li>Let \(B\) be an event so that \(P(B) > 0\)</li>
<li>Then the conditional probability of an event \(A\) given that \(B\) has occurred is
\[
P(A ~|~ B) = \frac{P(A \cap B)}{P(B)}
\]</li>
<li>Notice that if \(A\) and \(B\) are independent, then
\[
P(A ~|~ B) = \frac{P(A) P(B)}{P(B)} = P(A)
\]</li>
</ul>

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    <h2>Example</h2>
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    <ul>
<li>Consider our die roll example</li>
<li>\(B = \{1, 3, 5\}\)</li>
<li>\(A = \{1\}\)
\[
\begin{eqnarray*}
P(\mbox{one given that roll is odd})  & = & P(A ~|~ B) \\ \\
& = & \frac{P(A \cap B)}{P(B)} \\ \\
& = & \frac{P(A)}{P(B)} \\ \\ 
& = & \frac{1/6}{3/6} = \frac{1}{3}
\end{eqnarray*}
\]</li>
</ul>

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    <h2>Bayes&#39; rule</h2>
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  <article data-timings="">
    <p>\[
P(B ~|~ A) = \frac{P(A ~|~ B) P(B)}{P(A ~|~ B) P(B) + P(A ~|~ B^c)P(B^c)}.
\]</p>

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    <h2>Diagnostic tests</h2>
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    <ul>
<li>Let \(+\) and \(-\) be the events that the result of a diagnostic test is positive or negative respectively</li>
<li>Let \(D\) and \(D^c\) be the event that the subject of the test has or does not have the disease respectively </li>
<li>The <strong>sensitivity</strong> is the probability that the test is positive given that the subject actually has the disease, \(P(+ ~|~ D)\)</li>
<li>The <strong>specificity</strong> is the probability that the test is negative given that the subject does not have the disease, \(P(- ~|~ D^c)\)</li>
</ul>

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    <h2>More definitions</h2>
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    <ul>
<li>The <strong>positive predictive value</strong> is the probability that the subject has the  disease given that the test is positive, \(P(D ~|~ +)\)</li>
<li>The <strong>negative predictive value</strong> is the probability that the subject does not have the disease given that the test is negative, \(P(D^c ~|~ -)\)</li>
<li>The <strong>prevalence of the disease</strong> is the marginal probability of disease, \(P(D)\)</li>
</ul>

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    <h2>More definitions</h2>
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    <ul>
<li>The <strong>diagnostic likelihood ratio of a positive test</strong>, labeled \(DLR_+\), is \(P(+ ~|~ D) / P(+ ~|~ D^c)\), which is the \[sensitivity / (1 - specificity)\]</li>
<li>The <strong>diagnostic likelihood ratio of a negative test</strong>, labeled \(DLR_-\), is \(P(- ~|~ D) / P(- ~|~ D^c)\), which is the \[(1 - sensitivity) / specificity\]</li>
</ul>

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    <h2>Example</h2>
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    <ul>
<li>A study comparing the efficacy of HIV tests, reports on an experiment which concluded that HIV antibody tests have a sensitivity of 99.7% and a specificity of 98.5%</li>
<li>Suppose that a subject, from a population with a .1% prevalence of HIV, receives a positive test result. What is the probability that this subject has HIV?</li>
<li>Mathematically, we want \(P(D ~|~ +)\) given the sensitivity, \(P(+ ~|~ D) = .997\), the specificity, \(P(- ~|~ D^c) =.985\), and the prevalence \(P(D) = .001\)</li>
</ul>

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    <h2>Using Bayes&#39; formula</h2>
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  <article data-timings="">
    <p>\[
\begin{eqnarray*}
  P(D ~|~ +) & = &\frac{P(+~|~D)P(D)}{P(+~|~D)P(D) + P(+~|~D^c)P(D^c)}\\ \\
 & = & \frac{P(+~|~D)P(D)}{P(+~|~D)P(D) + \{1-P(-~|~D^c)\}\{1 - P(D)\}} \\ \\
 & = & \frac{.997\times .001}{.997 \times .001 + .015 \times .999}\\ \\
 & = & .062
\end{eqnarray*}
\]</p>

<ul>
<li>In this population a positive test result only suggests a 6% probability that the subject has the disease </li>
<li>(The positive predictive value is 6% for this test)</li>
</ul>

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    <h2>More on this example</h2>
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    <ul>
<li>The low positive predictive value is due to low prevalence of disease and the somewhat modest specificity</li>
<li>Suppose it was known that the subject was an intravenous drug user and routinely had intercourse with an HIV infected partner</li>
<li>Notice that the evidence implied by a positive test result does not change because of the prevalence of disease in the subject&#39;s population, only our interpretation of that evidence changes</li>
</ul>

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  <hgroup>
    <h2>Likelihood ratios</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li>Using Bayes rule, we have
\[
P(D ~|~ +) = \frac{P(+~|~D)P(D)}{P(+~|~D)P(D) + P(+~|~D^c)P(D^c)} 
\]
and
\[
P(D^c ~|~ +) = \frac{P(+~|~D^c)P(D^c)}{P(+~|~D)P(D) + P(+~|~D^c)P(D^c)}.
\]</li>
</ul>

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  <hgroup>
    <h2>Likelihood ratios</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li>Therefore
\[
\frac{P(D ~|~ +)}{P(D^c ~|~ +)} = \frac{P(+~|~D)}{P(+~|~D^c)}\times \frac{P(D)}{P(D^c)}
\]
ie
\[
\mbox{post-test odds of }D = DLR_+\times\mbox{pre-test odds of }D
\]</li>
<li>Similarly, \(DLR_-\) relates the decrease in the odds of the
disease after a negative test result to the odds of disease prior to
the test.</li>
</ul>

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  <hgroup>
    <h2>HIV example revisited</h2>
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  <article data-timings="">
    <ul>
<li>Suppose a subject has a positive HIV test</li>
<li>\(DLR_+ = .997 / (1 - .985) \approx 66\)</li>
<li>The result of the positive test is that the odds of disease is now 66 times the pretest odds</li>
<li>Or, equivalently, the hypothesis of disease is 66 times more supported by the data than the hypothesis of no disease</li>
</ul>

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  <hgroup>
    <h2>HIV example revisited</h2>
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  <article data-timings="">
    <ul>
<li>Suppose that a subject has a negative test result </li>
<li>\(DLR_- = (1 - .997) / .985  \approx .003\)</li>
<li>Therefore, the post-test odds of disease is now \(.3\%\) of the pretest odds given the negative test.</li>
<li>Or, the hypothesis of disease is supported \(.003\) times that of the hypothesis of absence of disease given the negative test result</li>
</ul>

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