3.3: Bayes' Rule
- Page ID
- 47234
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Consider a composite event \(M\) and a simple event \(A_i\). We know from conditional probability from the previous section that
\begin{align*} p(A_i|M) = \dfrac{p(A_i \cap M)}{p(M)} \\[4pt] p(M|A_i) = \dfrac{p(A_i \cap M)}{p(A_i)}, \end{align*}
and if we eliminate the denominator on the right-hand side, we find that
\begin{align*} p(M|A_i) = \dfrac{p(A_i|M) \, p(M)}{p(A_i)} \\[4pt] p(A_i|M) = \dfrac{p(M|A_i) \, p(A_i)}{p(M)}. \end{align*}
The second of these is most interesting - it gives the probability of a simple event, conditioned on the composite event, in terms of the composite event conditioned on the simple one! Recalling our above formula for \(p(M)\), we thus derive Bayes’ rule:
\[ p(A_i|M) = \dfrac{p(M|A_i) \, p(A_i)}{p(M|A_1) \, p(A_1) \, + \, . \, . \, . + \, p(M|A_n) \, p(A_n)}. \]
Here is an example of its use.
Consider a medical test that is 99% accurate - it gives a negative result for people who do not have the disease 99% of the time, and it gives a positive result for people who do have the disease 99% of the time. Only one percent of the population has this disease. Joe just got a positive test result: What is the probability that he has the disease?
Solution
The composite event \(M\) is that he has the disease, and the simple events are that he tested positive \((+)\) or he tested negative \((-)\). We apply
\begin{align*} p(M|+) &= \dfrac{p(+|M) \, p(M)}{p(+)} \\[4pt] &= \dfrac{p(+|M) \, p(M)} {p(+|M) \, p(M) + p(+|\bar{M}) \, p(\bar{M})} \\[4pt] &= \dfrac{0.99 \times 0.01} {0.99 \times 0.01 + 0.01 \times 0.99} \\[4pt] &= 1/2. \end{align*}
This example is not well appreciated by many healthcare consumers!
Here is another example, without so many symmetries.
Box A has nine red pillows in it and one white. Box B has six red pillows in it and nine white. Selecting a box at random and pulling out a pillow at random gives the result of a red pillow. What is the probability that it came from Box A?
Solution
\(M\) is the composite event that it came from Box A; the simple event is that a red pillow was collected \((R)\). We have
\begin{align*} p(M|R) &= \dfrac{p(R|M) \, p(M)}{p(R)} \\[4pt] &= \dfrac{p(R|M) \, p(M)} {p(R|M) \, p(M) + p(R|\bar{M}) \, p(\bar{M})} \\[4pt] &= \dfrac{0.9 \times 0.5} {0.9 \times 0.5 + 0.4 \times 0.5} \\[4pt] &= 0.692. \end{align*}