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Chapter 2: Q1E (page 84)

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

Expert verified

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

01

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 \(\Pr \left( {{B_1}|A} \right) < \Pr \left( {{B_1}} \right)\)

02

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

03

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,

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

04

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 \(i;\left( {i = 2, \ldots ,k} \right)\).

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Most popular questions from this chapter

Each time a shopper purchases a tube of toothpaste, he chooses either brand A or brand B. Suppose that for each purchase after the first, the probability is 1/3 that he will choose the same brand that he chose on his preceding purchase and the probability is 2/3 that he will switch brands. If he is equally likely to choose either brand A or brand B on his first purchase, what is the probability that both his first and second purchases will be brand A and both his third and fourth purchases will be brand B?

Suppose that a balanced die is rolled repeatedly until the same

number appears on two successive rolls, and let X denote the number of

rolls that are required. Determine the value of Pr(X= x), for x=2, 3,...

If S is the sample space of an experiment and A is any event in that space, what is the value of \({\bf{Pr}}\left( {{\bf{A}}\left| {\bf{S}} \right.} \right)\)?

A new test has been devised for detecting a particular type of cancer. If the test is applied to a person who has this type of cancer, the probability that the person will have a positive reaction is 0.95 and the probability that the person will have a negative reaction is 0.05. If the test is applied to a person who does not have this type of cancer, the probability that the person will have a positive reaction is 0.05 and the probability that the person will have a negative reaction is 0.95. Suppose that in the general population, one person out of every 100,000 people has this type of cancer. If a person selected at random has a positive reaction to the test, what is the probability that he has this type of cancer?

Consider a machine that produces items in sequence. Under normal operating conditions, the items are independent with a probability of 0.01 of being defective. However, it is possible for the machine to develop a 鈥渕emory鈥� in the following sense: After each defective item, and independent of anything that happened earlier, the probability that the next item is defective is\(\frac{{\bf{2}}}{{\bf{5}}}\). After each non-defective item, and independent of anything that happened earlier, the probability that the next item is defective is\(\frac{{\bf{1}}}{{{\bf{165}}}}\).

Please assume that the machine is operating normally for the whole time we observe or has a memory for the whole time we observe. LetBbe the event that the machine is operating normally, and assume that\({\bf{Pr}}\left( {\bf{B}} \right){\bf{ = }}\frac{{\bf{2}}}{{\bf{3}}}\). Let\({{\bf{D}}_{\bf{i}}}\)be the event that theith item inspected is defective. Assume that\({{\bf{D}}_{\bf{1}}}\)is independent ofB.

a. Prove that\({\bf{Pr}}\left( {{{\bf{D}}_{\bf{i}}}} \right){\bf{ = 0}}{\bf{.01}}\) for alli. Hint:Use induction.

b. Assume that we observe the first six items and the event that occurs is\({\bf{E = D}}_{\bf{1}}^{\bf{c}} \cap {\bf{D}}_{\bf{2}}^{\bf{c}} \cap {{\bf{D}}_3} \cap {{\bf{D}}_4} \cap {\bf{D}}_{\bf{5}}^{\bf{c}} \cap {\bf{D}}_{\bf{6}}^{\bf{c}}\). The third and fourth items are defective, but the other four are not. Compute\({\bf{Pr}}\left( {{\bf{B}}\left| {\bf{E}} \right.} \right)\).
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