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bayes-theorem.html
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<div id="content">
<h1 class="title">Bayes’ Theorem</h1>
<div id="outline-container-orgdd4ea6f" class="outline-2">
<h2 id="orgdd4ea6f">Preliminary</h2>
<div class="outline-text-2" id="text-orgdd4ea6f">
<p>
Necessary to understanding Bayes’ Theorem is an understanding of <b>conditional probability</b>.
</p>
<p>
Given the logical equivalence q → r ≡ ¬ q ∨ r, wouldn’t a <i>conditional</i>
probability just be the probability of the union of the effect (consequent)
with the complement of the cause (antecedent)?
</p>
<p>
So, \(p(E \rightarrow F) = p(\overline{E} \cup F)?\)
</p>
<p>
No! It doesn’t work that way. But we do want to ask, if \(E\) occurs, what is
the probability that \(F\) occurs also?
</p>
<p>
Here is how it works: if \(E\) and \(F\) are events in a probability space then
the conditional probability of \(F\) <i>given</i> \(E\), which is denoted \(p(F \mid
E)\), is the sum of probabilities of points in both \(E\) <b>and</b> \(F\) divided by
the sum of probabilities of points in just \(E\).
</p>
<ol class="org-ol">
<li>\(p(F \mid E) = p(F \cap E)\ /\ p(E)\)</li>
<li>\(p(E \mid F) = p(E \cap F)\ /\ p(F)\)</li>
</ol>
<p>
The obvious implication of the above two facts, together with the fact that
set intersection is commutative, is that
</p>
<p>
\(p(F \mid E) \cdot p(E) = p(E \mid F) \cdot p(F)\)
</p>
<p>
And if \(E\) and \(F\) are independent events, then
</p>
<p>
\(p(F \mid E) = p(F) \text{ and } p(E \mid F) = p(E)\).
</p>
</div>
<div id="outline-container-orga3ed32c" class="outline-3">
<h3 id="orga3ed32c">Example — Roll a Die Twice</h3>
<div class="outline-text-3" id="text-orga3ed32c">
<p>
What is the probability of a 3 on the first roll and a 4 on the second roll?
</p>
<p>
\(E =\) the event of a 3 on the first roll
</p>
<p>
Points in \(E\): [(3 . 1) (3 . 2) (3 . 3) (3 . 4) (3 . 5) (3 . 6)]
</p>
<p>
\(F =\) the event of a 4 on the second roll
</p>
<p>
Points in \(F\): [(1 . 4) (2 . 4) (3 . 4) (4 . 4) (5 . 4) (6 . 4)]
</p>
<p>
Points in \(F\) and \(E\): [(3 . 4)]
</p>
<p>
Turn it into a conditional — what is the probability of a 4 on the second
roll <i>given</i> a 3 on the first roll?
</p>
<p>
\(p(F \mid E) = p(F \cap E)\ /\ p(E) = \frac{1/36}{1/6} = \frac{1}{6}\)
</p>
</div>
</div>
<div id="outline-container-org6102c16" class="outline-3">
<h3 id="org6102c16"><b>Intuitive</b> versus <b>Calculated</b></h3>
<div class="outline-text-3" id="text-org6102c16">
<p>
Draw one card from a 52-card deck:
</p>
<p>
\(p(\clubsuit \mid Ace\ \clubsuit) = p(\clubsuit \cap Ace\ \clubsuit)\ /\ p(Ace\ \clubsuit) = \frac{1/52}{1/52} = 1\)
</p>
<p>
\(p(\spadesuit \mid Ace\ \heartsuit) = p(\spadesuit \cap Ace\ \heartsuit)\ /\ p(Ace\ \heartsuit) = \frac{0}{1/52} = 0\)
</p>
<p>
\(p(King\ \heartsuit \mid King) = p(King\ \heartsuit \cap King)\ /\ p(King) = \frac{1/52}{4/52} = \frac{1}{4}\)
</p>
<p>
\(p(\clubsuit \mid Black) = p(\clubsuit \cap Black)\ /\ p(Black) = \frac{13/52}{1/2} = \frac{1/4}{1/2} = \frac{1}{2}\)
</p>
</div>
</div>
<div id="outline-container-orga6a7be3" class="outline-3">
<h3 id="orga6a7be3">General Conjunction Rule</h3>
<div class="outline-text-3" id="text-orga6a7be3">
<p>
\(p(E \cap F) = p(F) \cdot p(E \mid F) = p(E) \cdot p(F \mid E) = p(F \cap E)\)
</p>
</div>
</div>
<div id="outline-container-org54c7d61" class="outline-3">
<h3 id="org54c7d61">More Examples</h3>
<div class="outline-text-3" id="text-org54c7d61">
<p>
Draw two cards in a row (without replacement):
</p>
<p>
\(p(Ace\ \spadesuit\ \mbox{first} \cap Ace\ \spadesuit\
\mbox{second}) = p(Ace\ \spadesuit\ \mbox{first}) \cdot p(Ace\
\spadesuit\ \mbox{second} \mid Ace\ \spadesuit\ \mbox{first}) =
\frac{1}{52} \cdot 0 = 0\)
</p>
<p>
\(p(\mbox{Red first} \cap \mbox{Red second}) = p(\mbox{Red first})
\cdot p(\mbox{Red second} \mid \mbox{Red first}) = \frac{1}{2} \cdot
\frac{25}{51} = \frac{25}{102} < \frac{1}{4} = \frac{1}{2} \cdot
\frac{1}{2}\)
</p>
<p>
\(p(\mbox{Ace first} \cap \mbox{Ace second}) = p(\mbox{Ace first})
\cdot p(\mbox{Ace second} \mid \mbox{Ace first}) = \frac{1}{13}
\cdot \frac{3}{51} = \frac{1}{13} \cdot \frac{1}{17} = \frac{1}{221}
< \frac{1}{169} = \frac{1}{13} \cdot \frac{1}{13}\)
</p>
</div>
</div>
<div id="outline-container-org92c222b" class="outline-3">
<h3 id="org92c222b">Independence and the Product Rule</h3>
<div class="outline-text-3" id="text-org92c222b">
<p>
Recall that independent experiments are when the outcome of one
experiment does not affect the outcome of others.
</p>
<p>
If \(p(F \mid E) = p(F)\) then \(E\) and \(F\) are independent.
</p>
<p>
If two independent experiments \(E\) and \(F\) are carried out in
sequence, then the probability of both outcomes is \(p(E) \cdot p(F)\)
</p>
<p>
\(p(F \mid E) = p(F)\)
</p>
<p>
\(p(F \mid E) = p(F \cap E)\ /\ p(E)\)
</p>
<p>
\(p(F) \cdot p(E) = p(F \cap E)\)
</p>
<p>
Recall the first example: What is the probability of rolling a 3 and
then rolling a 4 with a single die?
</p>
<p>
\(1/6 \cdot 1/6 = 1/36\). These events are independent.
</p>
<p>
<a href="http://setosa.io/conditional/">http://setosa.io/conditional/</a> is a <b>very</b> helpful site for grasping the
definition and significance of conditional probabilities and independent
events. See if you agree!
</p>
</div>
</div>
</div>
<div id="outline-container-org6c8d900" class="outline-2">
<h2 id="org6c8d900">Introduction</h2>
<div class="outline-text-2" id="text-org6c8d900">
<p>
If you read the guide at <a href="https://arbital.com/p/bayes_rule/?l=1zq">Arbital.com</a> you will get at least a glimmer of an
understanding that the key concept behind this mysterious thing called Bayes’
Theorem (also called Bayes’ Rule) is that extra information allows one to
derive a more realistic estimate of the probability that a particular event
occurs.
</p>
</div>
<div id="outline-container-org2468e49" class="outline-3">
<h3 id="org2468e49">The Theorem</h3>
<div class="outline-text-3" id="text-org2468e49">
<p>
If \(A\) and \(B\) are events such that \(p(A) \ne 0\) and \(p(B) \ne 0\),
then
</p>
<p>
\(p(A\ |\ B) = \frac{p(B\ |\ A)p(A)}{p(B\ |\ A)p(A) + p(B\ |\
\overline{A})p(\overline{A})}\).
</p>
</div>
</div>
<div id="outline-container-org47e2b09" class="outline-3">
<h3 id="org47e2b09">The Proof</h3>
<div class="outline-text-3" id="text-org47e2b09">
<p>
Bayes’ Theorem is just an equation — it expresses the equivalence between
something on the left-hand-side of the equals sign (LHS) and something else
on the right-hand-side (RHS).
</p>
<p>
Like most proofs of equations, this one is a straightforward application of
definitions, first principles (axioms) such as <i>quantities equal to the same
thing are equal to each other</i> and identities such as De Morgan’s laws or the
distributive laws (for sets).
</p>
<p>
From the definition of events and sample spaces we have \(B = B \cap
S\).
</p>
<p>
From the law of set complementarity we have \(S = A \cup
\overline{A}\).
</p>
<p>
Therefore, \(p(B) = p(B \cap S) = p(B \cap (A \cup \overline{A})) =
p((B \cap A) \cup (B \cap \overline{A}))\) (by the
distributive law) \(= p(B \cap A) + p(B \cap \overline{A}) - p(A
\cap \overline{A})\) (by the definition of probability of the
union of two events) \(= p(B \cap A) + p(B \cap \overline{A}) - 0\)
(by substituting equals for equals) \(= p(B \cap A) + p(B
\cap \overline{A})\).
</p>
<p>
Now, from the definition of conditional probability (the
probability that \(B\) will happen after it is known that \(A\)
happened) we have \(p(B\ |\ A) = p(B \cap A)\ /\ p(A)\).
</p>
<p>
From that, and because \(p(B \cap A) = p(A \cap B)\) (set
intersection is commutative), we have
</p>
<p>
\(p(B\ |\ A)p(A) = p(B \cap A) = p(A \cap B) = p(A\ |\ B)p(B),\
\text{ or }\ p(A\ |\ B) =\)
</p>
<p>
\(\frac{p(B\ |\ A)p(A)}{p(B)} = \frac{p(B\ |\ A)p(A)}{p(B \cap A) +
p(B \cap \overline{A})} = \frac{p(B\ |\ A)p(A)}{p(B\ |\ A)p(A) +
p(B\ |\ \overline{A})p(\overline{A})}\).
</p>
<p>
by dividing through by \(p(B)\), substituting what we saw \(p(B)\)
equals in the paragraph above that begins with the word
<i>Therefore</i>, and again using the definition of conditional
probability in the two terms of the denominator of the fraction to
substitute equals for equals.
</p>
<p>
Q.E.D.
</p>
</div>
</div>
</div>
<div id="outline-container-org912b9d4" class="outline-2">
<h2 id="org912b9d4">Example</h2>
<div class="outline-text-2" id="text-org912b9d4">
<p>
What is the probability that a random person who <i>tests positive</i> for a
disease <i>actually has</i> the disease, if we know that 1% of the population has
the disease, that 95% of those who have the disease test positive for it, and
2% of those who do not have the disease test positive for it?
</p>
<p>
Let \(H =\) a person Has the disease.
</p>
<p>
Let \(T =\) a person Tests positive for the disease.
</p>
<p>
We're given:
</p>
<ul class="org-ul">
<li>\(p(H) = 0.01\)</li>
<li>\(p(T \mid H) = 0.95\)</li>
<li>\(p(T \mid \overline{H}) = 0.02\)</li>
</ul>
<p>
We want \(p(H \mid T)\). Try it yourself, the answer is given below.
</p>
</div>
</div>
<div id="outline-container-orgae8ad10" class="outline-2">
<h2 id="orgae8ad10">Three Out of Four</h2>
<div class="outline-text-2" id="text-orgae8ad10">
<p>
The relationship between the four probabilities is good to know, because we
often have good probability estimates for three of the numbers, but we need to
compute the fourth.
</p>
<p>
For another example, suppose a doctor
</p>
<ul class="org-ul">
<li>knows that the disease meningitis gives a sufferer a stiff neck 70% of the
time.</li>
<li>knows some unconditional facts about meningitis and stiff necks, i.e.,
<ul class="org-ul">
<li>the prior probability of someone having meningitis is low, only .002% (1
out of 50,000 people);</li>
<li>the prior probability of someone having a stiff neck is 1%.</li>
</ul></li>
</ul>
<p>
Let S = has a stiff neck, M = has meningitis.
</p>
<table border="2" cellspacing="0" cellpadding="6" rules="groups" frame="hsides">
<colgroup>
<col class="org-left" />
<col class="org-left" />
<col class="org-right" />
</colgroup>
<tbody>
<tr>
<td class="org-left">p(S | M)</td>
<td class="org-left">=</td>
<td class="org-right">0.7</td>
</tr>
<tr>
<td class="org-left">p(M)</td>
<td class="org-left">=</td>
<td class="org-right">.00002</td>
</tr>
<tr>
<td class="org-left">p(S)</td>
<td class="org-left">=</td>
<td class="org-right">.01</td>
</tr>
<tr>
<td class="org-left">p(M | S)</td>
<td class="org-left">=</td>
<td class="org-right">p(S | M)p(M) / p(S) = (0.7)(.00002) / 0.01 = 0.0014</td>
</tr>
</tbody>
</table>
</div>
</div>
<div id="outline-container-org342248b" class="outline-2">
<h2 id="org342248b">Lowered Expectations</h2>
<div class="outline-text-2" id="text-org342248b">
<p>
So the doctor should expect fewer than 1 in 700 patients with a stiff
neck to have meningitis.
</p>
<p>
Note that even though meningitis strongly indicates a stiff neck
(with probability 0.7), the probability of meningitis stays small,
because the prior probability of stiff necks is so much higher than
that of meningitis.
</p>
<p>
The general idea is that we would like to determine, from our seeing
as evidence the <i>effect</i> of some unknown <i>cause</i>, what that <i>cause</i>
is.
</p>
</div>
</div>
<div id="outline-container-org7383370" class="outline-2">
<h2 id="org7383370">Cause and Effect</h2>
<div class="outline-text-2" id="text-org7383370">
<p>
In this context Bayes' Theorem says
</p>
<p>
p(cause | effect) = p(effect | cause) ⋅ p(cause) / p(effect)
</p>
<p>
Equivalently,
</p>
<p>
p(effect | cause) = p(cause | effect) ⋅ p(effect) / p(cause)
</p>
<p>
There are two directions here, <i>causal</i> and <i>diagnostic</i>:
</p>
<ul class="org-ul">
<li>The <i>causal</i> direction finds p(effect | cause).</li>
<li>The <i>diagnostic</i> direction finds p(cause | effect).</li>
</ul>
<p>
In medical diagnosis, estimates of the conditional probabilities on causal
relationships are often what a doctor has (i.e., p(symptoms | disease)), and
what the doctor needs is to derive a diagnosis (i.e., p(disease | symptoms)).
</p>
<p>
In a more general setting, <b>Truth</b> is to <b>cause</b> as <b>Test</b> is to <b>effect</b>.
</p>
</div>
</div>
<div id="outline-container-orgc177784" class="outline-2">
<h2 id="orgc177784">Truth versus Test</h2>
<div class="outline-text-2" id="text-orgc177784">
</div>
<div id="outline-container-orge52a0ae" class="outline-3">
<h3 id="orge52a0ae">Problem</h3>
<div class="outline-text-3" id="text-orge52a0ae">
<p>
What is the probability that a random person who tests positive for
a certain disease actually has the disease, if we know that 1% of
the population has the disease, that 95% of those who have the
disease test positive for it, and 2% of those who do not have the
disease test positive for it?
</p>
</div>
</div>
<div id="outline-container-org79dcec0" class="outline-3">
<h3 id="org79dcec0">Solution</h3>
<div class="outline-text-3" id="text-org79dcec0">
<p>
Let \(H\) be the event that a person has the disease.
</p>
<p>
Let \(T\) be the event that the person tests positive for it.
</p>
<p>
The problem statement gives \(p(T\ |\ H), p(T\ |\ \overline{H})\) and
\(p(H)\).
</p>
<p>
We know that \(p(T) = p(T\ |\ H) \cdot p(H) + p(T\ |\ \overline{H})
\cdot p(\overline{H})\) (question: how do we know that?)
</p>
<p>
What is asked for is \(p(H\ |\ T)\), which we can find using the
identity:
</p>
<p>
\(p(H\ |\ T) = p(T\ |\ H) p(H) / p(T)\).
</p>
<p>
\(p(H) = 0.01\)
</p>
<p>
\(p(\overline{H}) = 1 - p(H) = 1 - 0.01 = .99\)
</p>
<p>
\(p(T\ |\ H) = 0.95\)
</p>
<p>
\(p(T\ |\ \overline{H}) = 0.02\)
</p>
<p>
\(\therefore\)
</p>
<p>
\(p(T) = (.95)(.01) + (.02)(.99) = .0095 + .0198 = .0293\)
</p>
<p>
\(p(H\ |\ T) = (.95)(.01)/(.0293) = .0095/.0293 = .324 \approx 32\%\)
</p>
</div>
</div>
</div>
<div id="outline-container-orgacc6138" class="outline-2">
<h2 id="orgacc6138">Visualize</h2>
<div class="outline-text-2" id="text-orgacc6138">
<p>
It helps to visualize the four disjoint event subsets with their
probabilities in table form:
</p>
<p>
\(\begin{array}{|r|r|r|r|}
\hline
& & & \\
& p(\overline{H}) & p(H) & \\
\hline
& & & \\
p(T) & p(\overline{H} \cap T) = .0198 & p(H \cap T) = .0095 & .0293 \\
& & & \\
& = p(\overline{H}\ |\ T) \cdot p(T) & = p(H\ |\ T) \cdot p(T) & \\
& = p(T\ |\ \overline{H}) \cdot p(\overline{H}) & = p(T\ |\ H) \cdot p(H) & \\
& & & \\
\hline
& & & \\
p(\overline{T}) & p(\overline{H} \cap \overline{T}) = .9702 & p(H \cap \overline{T}) = .0005 & .9707 \\
& & & \\
& = p(\overline{H}\ |\ \overline{T}) \cdot p(\overline{T}) & = p(H\ |\ \overline{T}) \cdot p(\overline{T}) & \\
& = p(\overline{T}\ |\ \overline{H}) \cdot p(\overline{H}) & = p(\overline{T}\ |\ H) \cdot p(H) & \\
& & & \\
\hline
& & & \\
& .9900 & .0100 & 1.0 \\
\hline
\end{array}\)
</p>
</div>
</div>
</div>
</body>
</html>