Question

Suppose thatkevents\({{\bf{B}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{B}}_{\bf{k}}}\)form a partition of the sample spaceS. For\({\bf{i = 1 \ldots k}}\), let\(\Pr \left( {{{\bf{B}}_{\bf{i}}}} \right)\)denote the prior probability of\({{\bf{B}}_{\bf{i}}}\). Also, for each eventAsuch that\({\bf{Pr}}\left( {\bf{A}} \right){\bf{ > 0}}\), let\({\bf{Pr}}\left( {{{\bf{B}}_{\bf{i}}}{\bf{|A}}} \right)\)denote the posterior probability of\({{\bf{B}}_{\bf{i}}}\)given that the eventAhas occurred. Prove that if\({\bf{Pr}}\left( {{{\bf{B}}_{\bf{1}}}{\bf{|A}}} \right){\bf{ < Pr}}\left( {{{\bf{B}}_{\bf{1}}}} \right)\)then\({\bf{Pr}}\left( {{{\bf{B}}_{\bf{i}}}{\bf{|A}}} \right){\bf{ > Pr}}\left( {{{\bf{B}}_{\bf{i}}}} \right)\)for at least one value of i\(\left( {{\bf{i = 2, \ldots ,k}}} \right)\).

Short answer

our assumption is wrong and conclude that if \(\Pr \left( {{B_1}|A} \right) < \Pr \left( {{B_1}} \right)\) then \(\Pr \left( {{B_i}|A} \right) > \Pr \left( {{B_i}} \right)\) for at least one value of \(i;\left( {i = 2, \ldots ,k} \right)\).

Step-by-step solution

Given information

Here we are given that there are k samples that are\({B_1},{B_2} \ldots {B_k}\)and\(\Pr \left( {{B_i}} \right)\)is the probability of\({B_i}\). The posterior probability of\({B_i}\)is\(\Pr \left( {{B_i}|A} \right)\)given that the event A has occurred.

And also given that \(\Pr \left( {{B_1}|A} \right) < \Pr \left( {{B_1}} \right)\)

State the properties

We know the properties, that are,

\(\sum\limits_{{\bf{i = 1}}}^{\bf{k}} {{\bf{Pr}}\left( {{{\bf{B}}_{\bf{i}}}} \right){\bf{ = 1}}} \) and \(\sum\limits_{{\bf{i = 1}}}^{\bf{k}} {{\bf{Pr}}\left( {{{\bf{B}}_{\bf{i}}}{\bf{|A}}} \right){\bf{ = 1}}} \)

Assume a condition

Let us assume that a condition that for all\(i;\left( {i = 2, \ldots ,k} \right)\)we have the posterior probabilities that is satisfies the condition,

\(\Pr \left( {{B_i}|A} \right) \le \Pr \left( {{B_i}} \right)\)

Proof

Now we can get the equation from the assumed condition.

\(\sum\limits_{i = 1}^k {\Pr \left( {{B_i}} \right) = \Pr \left( {{B_1}} \right) + \sum\limits_{i = 2}^k {\Pr \left( {{B_i}} \right)} } \)

Now by the properties and the given information, we get that,

\(\begin{aligned}{}\Pr \left( {{B_1}} \right) + \sum\limits_{i = 2}^k {\Pr \left( {{B_i}} \right)} > \Pr \left( {{B_1}|A} \right) + \sum\limits_{i = 2}^k {\Pr \left( {{B_i}} \right)} \\ > \Pr \left( {{B_1}|A} \right) + \sum\limits_{i = 2}^k {\Pr \left( {{B_i}|A} \right)} \\ > 1\end{aligned}\)

Here we noticed that 1>1. Which is a contradiction. So, our assumption is wrong and conclude that if \(\Pr \left( {{B_1}|A} \right) < \Pr \left( {{B_1}} \right)\) then \(\Pr \left( {{B_i}|A} \right) > \Pr \left( {{B_i}} \right)\) for at least one value of \(i;\left( {i = 2, \ldots ,k} \right)\).