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Bayes’ Theorem

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  • Partition of a sample space
  • Theorem of total probability


If `E_1, E_2 ,..., E_n` are n non empty events which constitute a partition of sample space S, i.e. `E_1, E_2 ,..., E_n` are pairwise disjoint and `E_1 ∪ E_2 ∪ ... ∪ E_n` = S and A is any event of nonzero probability, then

P(Ei|A) =`(P(E_i) P (A | E_i))/(  sum_(i=1)^n P(E_j) P(A|E _j ))`

P  for any i = 1, 2, 3, ..., n

Proof: By formula of conditional probability, we know that
`P(E_i|A) = (P(A ∩ E_i )) / (P(A))`
`= (P(E_i ) (P(A|E_i )))/ (P(A))` (by multiplication rule of probability) 

`= (P(E_i )P(A|E_i ))/ (sum_(j = 1)^n P(E _j)P(A|E_j)) ` (by the result of theorem of total probability)

Remark:  The following terminology is generally used when Bayes' theorem is applied. The events `E_1, E_2, ..., E_n` are  called hypotheses. 
The probability `P(E_i)` is called the priori probability of the hypothesis `E_i` 
The conditional probability `P(E_i |A)` is called a posteriori probability of the hypothesis `E_i`.
Bayes' theorem is also called the formula for the probability of "causes". Since the `E_i's` are a partition of the sample space S, one and only one of the events `E_i` occurs (i.e. one of the events `E_i` must occur and only one can occur). Hence, the above formula gives us the probability of a particular Ei (i.e. a "Cause"), given that the event A has occurred. 

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1) Partition of a sample space: 
A set of events `E_1, E_2, ..., E_n` is said to represent a partition of the sample space S if 

(a) `E_i ∩ E_j = φ, i ≠ j, i, j = 1, 2, 3, ..., n`

(b) `E_1 ∪ Ε_2 ∪ ... ∪ E_n= S`   and 

(c) `P(E_i) > 0  "for all"   i = 1, 2, ..., n.`
In other words, the events `E_1, E_2, ..., E_n` represent a partition of the sample space S if they are pairwise disjoint, exhaustive and have nonzero probabilities.
As an example, we see that any nonempty event E and its complement E′ form a partition of the sample space S since they satisfy E ∩ E′ = φ and E ∪ E′ = S.

2) Theorem of total probability:
Let   `{E_1, E_2,...,E_n}`  be a partition of the sample space S,  and suppose that each of the events `E_1, E_2,..., E_n` has nonzero probability of occurrence. Let A be any event associated with S, then
`P(A) = P(E_1) P(A|E_1) + P(E_2) P(A|E_2) + ... + P(E_n) P(A|E_n)` 
= ` sum _(j=1) ^ n P(E_j) P(A|E_j)`     

Proof :
 Given that `E_1, E_2,..., E_n` is a partition of the sample space S in following fig.

Therefore , S =` E_1 ∪ E_2 ∪ ... ∪ E_n`          ... (1) 
and `E_i ∩ E_j = φ, i ≠ j, i, j = 1, 2, ..., n`
Now, we know that for any event A, 
A = A ∩ S 
=` A ∩ (E_1 ∪ E_2 ∪ ... ∪ E_n)`
= `(A ∩ E_1) ∪ (A ∩ E_2) ∪ ...∪ (A ∩ E_n)`
Also A ∩ `E_i` and A ∩ `E_j` are respectively the subsets of `E_i` and `E_j`. We know that `E_i`  and `E_j` are disjoint, for i  ≠ j, therefore, `A  ∩ E_i` and `A  ∩ E_j`  are also disjoint for all i ≠ j,  i, j = 1, 2, ..., n.

`P(A) = P [(A ∩ E_1) ∪ (A ∩ E_2)∪ .....∪ (A ∩ E_n)]` 

= `P (A ∩ E_1) + P (A ∩ E_2) + ... + P (A ∩ E_n)`    
Now, by multiplication rule of probability, we have 

`P(A ∩ E_i) = P(E_i) P(A|E_i) as  P (E_i) ≠ 0 ∀ i = 1,2,..., n`

Therefore, P (A) = `P (E_1) P (A|E_1) + P (E_2) P (A|E_2) + ... + P (E_n)P(A|E_n)`

or `P(A) = sum_(j = 1)^n P(E_j) P(A|E_j)` 

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