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Use the prior probabilities in Example 2.3.8 for the events\({B_1},...,{{\bf{B}}_{11}}\). Let \({E_1}\)be the event that the first patient is a success. Compute the probability of \({E_1}\) and explain why it is so much less than the value computed in Example 2.3.7.

Short Answer

Expert verified

The probability of \({E_1}\)is 0.5.

Step by step solution

01

Given information

\({E_1}\) is the event that the first patient is a success.

02

Calculating the mean of the posterior distribution

Consider a clinical trial experiment for the study of treatment for depression.

Consider the decline as a failure and no decline as a success. We consider only patients of the imipramine treatment group. Let us consider pas the proportion of successes among all patients who might receive the treatment.

Let us select the 40 imipramine patients in this trial. Let us consider that there are 11 different possible compositions of the collection of patients.

Let the proportion of success for the 11 possible compositions are:

\(0,\frac{1}{{10}},...,\frac{9}{{10}},1\)

Let partition the sample space by the possible values of p.

Define the event\({B_j}\)represent the event that our sample was drawn from a collection with the proportion\(\frac{{\left( {j - 1} \right)}}{{10}}\)of successes.

Let’s consider the following prior probabilities for the event \({B_1},...,{B_{11}}\) as follows:

Let’s define the event\({E_1}\), representing the first patient is a success. Therefore, by using the prior probabilities of\({B_j}\)and the law of total probability, we compute the probability\({E_1}\)as follows

\(\begin{aligned}{c}\Pr \left( {{E_1}} \right) &= \sum\limits_{i = 1}^{11} {\Pr \left( {{B_j}} \right) \times \frac{{\left( {i - 1} \right)}}{{10}}} \\ &= 0.00\left( {\frac{0}{{10}}} \right) + 0.19\left( {\frac{1}{{10}}} \right) + ...\\ &= 0.356\end{aligned}\)

Now, compute the probability\({E_1}\)on the basis that the probabilities of\({B_j}\)is\(\frac{1}{{11}}\)for each j. In this situation, the probability\({E_1}\)as follows

\(\begin{aligned}{c}\Pr \left( {{E_1}} \right) &= \sum\limits_{i = 1}^{11} {\Pr \left( {{B_j}} \right) \times \frac{{\left( {i - 1} \right)}}{{10}}} \\ &= \sum\limits_{i = 1}^n {\frac{1}{{11}}} \times \frac{{\left( {i - 1} \right)}}{{10}}\\ &= 0.5\end{aligned}\)

It is observed that the probabilities of\({E_1}\)is less for prior probabilities of\({B_j}\)as compared to that for probabilities of\({B_j}\)is\(\frac{1}{{11}}\)for each j. Also, it is observed that prior probabilities are very smaller for the larger values of i.

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Most popular questions from this chapter

Suppose that a box contains one fair coin and one coin with a head on each side. Suppose that a coin is drawn at random from this box and that we begin to flip the coin. In Eqs. (2.3.4) and (2.3.5), we computed the conditional probability that the coin was fair, given that the first two flips both produce heads.

a. Suppose that the coin is flipped a third time and another head is obtained. Compute the probability that the coin is fair, given that all three flips produced

heads.

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For any two events A and B with Pr(B) > 0, prove that Pr(Ac|B) = 1 − Pr(A|B).

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