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[[bayes_theorem_eg3_S13|Example 3: Monty Hall Problem]] | [[bayes_theorem_eg3_S13|Example 3: Monty Hall Problem]] | ||
+ | ---- | ||
+ | |||
+ | == References == | ||
+ | |||
+ | * Alberto Leon-Garcia, ''Probability, Statistics, and Random Processes for Electrical Engineering,'' Third Edition | ||
+ | |||
+ | * Mark Haddon, ''The Curious Incident of the Dog in the Night-Time'' |
Revision as of 14:20, 16 March 2013
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. In this tutorial, we will derive Bayes' Theorem and illustrate it with a few examples. After going over the examples, if you have any questions or if you find any mistakes please leave me a comment at the end of the relevant section.
Note that this tutorial assumes familiarity with conditional probability and the axioms of probability.
Contents - Bayes' Theorem - Proof - Example Problems - 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 Problems
Example 2: False Positive Paradox
References
- Alberto Leon-Garcia, Probability, Statistics, and Random Processes for Electrical Engineering, Third Edition
- Mark Haddon, The Curious Incident of the Dog in the Night-Time