Question

Question: Prove Theorem \(2\), the extended form of Bayes’ theorem. That is, suppose that \(E\) is an event from a sample space \(S\) and that \({F_1},{F_2},...,{F_n}\) are mutually exclusive events such that \(\bigcup\nolimits_{i = 1}^n {{F_i} = S} \). Assume that \(p\left( E \right) \ne 0\) and \(p\left( {{F_i}} \right) \ne 0\) for \(i = 1,2,...,n\). Show that

\(p\left( {{F_j}\left| E \right.} \right) = \frac{{p\left( {E\left| {{F_j}} \right.} \right)p\left( {{F_j}} \right)}}{{\sum\nolimits_{i = 1}^n {p\left( {E\left| {{F_i}} \right.} \right)p\left( {{F_i}} \right)} }}\)

(Hint: use the fact that \(E = \bigcup\nolimits_{i = 1}^n {\left( {E \cap {F_i}} \right)} \).)

Short answer

Answer

Hence, it is proved that \(p\left( {{F_j}\left| E \right.} \right) = \frac{{p\left( {E\left| {{F_j}} \right.} \right)p\left( {{F_j}} \right)}}{{\sum\nolimits_{i = 1}^n {p\left( {E\left| {{F_i}} \right.} \right)p\left( {{F_i}} \right)} }}\).

Step-by-step solution

Given Data

• \(E\)is an event from a sample space\(S\)and that\({F_1},{F_{}},...,{F_n}\)are mutually exclusive event such that\(\bigcup\nolimits_{i = 1}^n {{F_i} = S} \).
• To prove that\(p\left( {{F_j}\left| E \right.} \right) = \frac{{p\left( {E\left| {{F_j}} \right.} \right)p\left( {{F_j}} \right)}}{{\sum\nolimits_{i = 1}^n {p\left( {E\left| {{F_i}} \right.} \right)p\left( {{F_i}} \right)} }}\).

Proof

Consider that

\(\begin{array}{l}p\left( E \right) = p\left( {E \cap S} \right)\\{\rm{ }} = p\left( {E \cap \mathop \cup \limits_{i = 1}^n {F_i}} \right)\\{\rm{ }} = p\left( {\mathop \cup \limits_{i = 1}^n E \cap {F_i}} \right)\\{\rm{ }} = \sum\nolimits_{i = 1}^n {p\left( {E \cap {F_i}} \right)} \end{array}\)

Now, because\({F_1},{F_{}},...,{F_n}\)are mutually exclusive so-

\(\begin{array}{l}p\left( {{F_j}\left| E \right.} \right) = \frac{{p\left( {{F_i} \cap E} \right)}}{{p\left( E \right)}}\\{\rm{ }} = \frac{{p\left( {{F_j}\left| E \right.} \right) \cdot p\left( E \right)}}{{\sum\nolimits_{i = 1}^n {p\left( {E \cap {F_i}} \right)} }}\\{\rm{ }} = \frac{{p\left( {E\left| {{F_j}} \right.} \right)p\left( {{F_j}} \right)}}{{\sum\nolimits_{i = 1}^n {p\left( {E\left| {{F_i}} \right.} \right)p\left( {{F_i}} \right)} }}\end{array}\)

Hence, Proved.