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Bayes' Theorem

by Maliha Hossain

 keyword: probability, Bayes' Theorem, Bayes' Rule 

INTRODUCTION

Bayes' Theorem (or Bayes' Rule) allows us to calculate P(A|B) from P(B|A) given that P(A) and P(B) are also known, where A and B are events. In this tutorial, we will derive Bayes' Theorem and illustrate it with a few examples.

Note that this tutorial assumes familiarity with conditional probability and the axioms of probability.

 Contents
- Bayes' Theorem
- Proof
- Example 1: Quality Control
- Example 2: The False Positive Paradox
- Example 3
- References

Bayes' Theorem

Let $ B_1, B_2, ..., B_n $ be a partition of the sample space $ S $, i.e. $ B_1, B_2, ..., B_n $ are mutually exclusive events whose union equals the sample space S. Suppose that the event $ A $ occurs. Then, by Bayes' Theorem, we have that

$ P[B_j|A] = \frac{P[A|B_j]P[B_j]}{P[A]}, j = 1, 2, . . . , n $

Bayes' Theorem is also often expressed in the following form:

$ P[B_j|A] = \frac{P[A|B_j]P[B_j]}{\sum_{k=1}^n P[A|B_k]P[B_k]} $

Proof

We will now derive Bayes'e Theorem as it is expressed in the second form, which simply takes the expression one step further than the first.

Let $ A $ and $ B_j $ be as defined above. By definition of the conditional probability, we have that

$ P[A|B_j] = \frac{P[A\cap B_j]}{P[B_j]} $

Multiplying both sides with $ B_j $, we get

$ P[A\cap B_j] = P[A|B_j]P[B_j] \ $

Using the same argument as above, we have that

$ \begin{align} P[B_j|A] & = \frac{P[B_j\cap A]}{P[A]} \\ \Rightarrow P[B_j\cap A] &= P[B_j|A]P[A] \end{align} $

Because of the commutativity property of intersection, we can say that

$ \begin{align} P[B_j|A]P[A] & = P[A|B_j]P[B_j] \\ \text{Dividing both sides by }P[A],\text{ we get } P[B_j|A] &= \frac{P[A|B_j]P[B_j]}{P[A]} \end{align} $

Finally, the denominator can be broken down further using the theorem of total probability so that we have the following expression

$ P[B_j|A] = \frac{P[A|B_j]P[B_j]}{\sum_{k=1}^n P[A|B_k]P[B_k]} $

Example 1: Quality Control

The following problem has been adapted from a few practice problems from chapter 2 of Probability, Statistics and Random Processes for Electrical Engineers by Alberto Leon-Garcia. The example illustrates how Bayes' Theorem plays a role in quality control.

A manufacturer produces a mix of "good" chips and "bad" chips. The proportion of good chips whose lifetime exceeds time $ t $ seconds decreases exponentially at the rate $ \alpha $. The proportion of bad chips whose lifetime exceeds t decreases much faster at a rate $ 1000\alpha $. Suppose that the fraction of bad chips is $ p $, and of good chips, $ 1 - p $

Let $ C $ be the event that the chip is functioning after $ t $ seconds. Let $ G $ be the event that the chip is good. Let $ B $ be the event that the chip is bad.

Here's what we can infer from the problem statement thus far:

the probability that the lifetime of a good chip exceeds $ t $: $ P[C|G] = e^{-\alpha t} $

the probability that the lifetime of a bad chip exceeds $ t $: $ P[C|B] = e^{-1000\alpha t} $

So by the theorem of total probability, we have that

$ P[C] = P[C|G]P[G] + P[C|B]P[B] $

$ = e^{-\alpha t}(1-p) + e^{-1000\alpha t}p $

Now suppose that in order to weed out the bad chips, every chip is tested for t seconds prior to leaving the factory. the chips that fail are discarded and the remaining chips are sent out to customers. Can you find the value of $ t $ for which 99% of the chips sent out to customers are good?

The problem requires that we find the value of $ t $ such that

$ P[G|C] = .99 $

We find $ P[G|C] $ by applying Bayes' Theorem

$ P[G|C] = \frac{P[C|G]P[G]}{P[C|G]P[G] + P[C|B]P[B]} $

$ = \frac{e^{-\alpha t}(1-p)}{e^{-\alpha t}(1-p) + e^{-1000\alpha t}} $

$ = \frac{1}{1 + \frac{pe^{-1000\alpha t}}{e^{-\alpha t}(1-p)}} = .99 $

The above equation can be solved for $ t $

$ t = \frac{1}{999\alpha}ln(\frac{99p}{1-p}) $

Example 2: The False Positive Paradox

The false positive paradox occurs when false positive tests are more probable than true positive tests. The fewer the number of incidents in the overall population, the higher the likelihood of a false positive test.

The following example illustrates how you would calculate the false positive rate for a test using Bayes' Theorem.

A manufacturer claims that its product can detect drug use among athletes 97% of the time (i.e. the test will show a positive 97% of the time given that the athletes used drugs). However, there is a 10% chance of a false alarm (i.e. non drug users will show positive results 10% of the time). Given that only 5% of the team actually use drugs, what is the probability that an athlete who tested positive is a non user?

Let $ D $ be the event that the athlete used drugs.

Therefore, $ D' $ is the event that the athlete did not use drugs.

Let $ Y $ be the event that the test result was positive

Therefore a negative result is described by the event $ Y' $

From the problem statement, we can infer the following.

$ P[D] = 0.05 $

$ P[D'] = 1 - P[D'] = 0.95 $

$ P[Y|D] = 0.97 $ (i.e. the probability of a positive test given the athlete used drugs)

$ P[Y'|D] = 1 - P[Y|D] = 0.03 $ (i.e. the probability of a negative test given the athlete used drugs)

$ P[Y|D'] = 0.1 $ (i.e. the probability that the test was positive given the athlete did not take drugs)

$ P[Y'|D'] = 1 - P[Y|D'] = 0.9 $ (i.e. the probability of a negative test given the athlete did not use drugs)

Now we need to find $ P[D'|Y] $, in other words, the probability that the athlete did not use drugs given the test was positive.

By Bayes' Theorem, we have that

$ P[D'|Y] = \frac{P[Y|D']P[D']}{P[Y|D]P[D] + P[Y|D']P[D']} $

$ = \frac{0.01\times 0.95}{0.97 \times 0.05 + 0.1 \times 0.95} = 0.662 $

So we see that a positive test is more likely to be a false alarm since the false positive rate is too high for this low incidence population.

Example 3: The Monty Hall Problem