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6.2.1 An Example To illustrate Bayes rule, consider a medical diagnosis problem in which there are two alternative hypotheses: (1) that the patient has a particular form of cancer, and (2) that the patient does not. The available data is from a particular laboratory test with two possible outcomes: ? (positive) and ? (negative). We have prior knowledge that over the entire population of people only .008 have this disease. Furthermore, the lab test is only an imperfect indicator of the disease. The test returns a correct positive result in only 98% of the cases in which the disease is actually present and a correct negative result in only 97% of the cases in which the disease is not present. In other cases, the test returns the opposite result. The above situation can be summarized by the following probabilities: P(cancer) = .008, P(cancer) = .992 P(?|cancer) = .98, P(?|cancer) = .02 P(?|cancer) = .03, P(?|cancer) = .97 Suppose we now observe a new patient for whom the lab test returns a positive result. Should we diagnose the patient as having cancer or not? The maximum a posteriori hypothesis can be found using Equation (6.2): P(?|cancer)P(cancer) = (.98).008 = .0078 P(?|cancer)P(cancer) = (.03).992 = .0298 Thus, hMAP = cancer. The exact posterior probabilities can also be determined by normalizing the above quantities so that they sum to 1 (e.g., P(cancer|?) = .0078 / (.0078 + .0298) = .21). This step is warranted because Bayes theorem states that the posterior probabilities are just the above quantities divided by the probability of the data, P(?). Although P(?) was not provided directly as part of the problem statement, we can calculate it in this fashion because we know that P(cancer|?) and P(cancer|?) must sum to 1 (i.e., either the patient has cancer or they do not). Notice that while the posterior probability of cancer is significantly higher than its prior probability, the most probable hypothesis is still that the patient does not have cancer.

          6.2.1 An Example
To illustrate Bayes rule, consider a medical diagnosis problem in which there are two alternative hypotheses: (1) that the patient has a particular form of cancer, and (2) that the patient does not. The available data is from a particular laboratory test with two possible outcomes: ? (positive) and ? (negative). We have prior knowledge that over the entire population of people only .008 have this disease. Furthermore, the lab test is only an imperfect indicator of the disease. The test returns a correct positive result in only 98% of the cases in which the disease is actually present and a correct negative result in only 97% of the cases in which the disease is not present. In other cases, the test returns the opposite result. The above situation can be summarized by the following probabilities:
P(cancer) = .008, P(cancer) = .992
P(?|cancer) = .98, P(?|cancer) = .02
P(?|cancer) = .03, P(?|cancer) = .97
Suppose we now observe a new patient for whom the lab test returns a positive result. Should we diagnose the patient as having cancer or not? The maximum a posteriori hypothesis can be found using Equation (6.2):
P(?|cancer)P(cancer) = (.98).008 = .0078
P(?|cancer)P(cancer) = (.03).992 = .0298
Thus, hMAP = cancer. The exact posterior probabilities can also be determined by normalizing the above quantities so that they sum to 1 (e.g., P(cancer|?) = .0078 / (.0078 + .0298) = .21). This step is warranted because Bayes theorem states that the posterior probabilities are just the above quantities divided by the probability of the data, P(?). Although P(?) was not provided directly as part of the problem statement, we can calculate it in this fashion because we know that P(cancer|?) and P(cancer|?) must sum to 1 (i.e., either the patient has cancer or they do not). Notice that while the posterior probability of cancer is significantly higher than its prior probability, the most probable hypothesis is still that the patient does not have cancer.

6.2.1 An Example
To illustrate Bayes rule, consider a medical diagnosis problem in which there are two alternative hypotheses: (1) that the patient has a particular form of cancer, and (2) that the patient does not. The available data is from a particular laboratory test with two possible outcomes: ? (positive) and ? (negative). We have prior knowledge that over the entire population of people only .008 have this disease. Furthermore, the lab test is only an imperfect indicator of the disease. The test returns a correct positive result in only 98% of the cases in which the disease is actually present and a correct negative result in only 97% of the cases in which the disease is not present. In other cases, the test returns the opposite result. The above situation can be summarized by the following probabilities:
P(cancer) = .008, P(cancer) = .992
P(?|cancer) = .98, P(?|cancer) = .02
P(?|cancer) = .03, P(?|cancer) = .97
Suppose we now observe a new patient for whom the lab test returns a positive result. Should we diagnose the patient as having cancer or not? The maximum a posteriori hypothesis can be found using Equation (6.2):
P(?|cancer)P(cancer) = (.98).008 = .0078
P(?|cancer)P(cancer) = (.03).992 = .0298
Thus, hMAP = cancer. The exact posterior probabilities can also be determined by normalizing the above quantities so that they sum to 1 (e.g., P(cancer|?) = .0078 / (.0078 + .0298) = .21). This step is warranted because Bayes theorem states that the posterior probabilities are just the above quantities divided by the probability of the data, P(?). Although P(?) was not provided directly as part of the problem statement, we can calculate it in this fashion because we know that P(cancer|?) and P(cancer|?) must sum to 1 (i.e., either the patient has cancer or they do not). Notice that while the posterior probability of cancer is significantly higher than its prior probability, the most probable hypothesis is still that the patient does not have cancer.

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Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Ana Carolina Da Cruz

08/25/2022

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6.2.1 An Example To illustrate Bayes rule, consider a medical diagnosis problem in which there are two alternative hypotheses: (1) that the patient has a particular form of cancer, and (2) that the patient does not. The available data is from a particular laboratory test with two possible outcomes: ⊕ (positive) and ⊖ (negative). We have prior knowledge that over the entire population of people only .008 have this disease. Furthermore, the lab test is only an imperfect indicator of the disease. The test returns a correct positive result in only 98% of the cases in which the disease is actually present and a correct negative result in only 97% of the cases in which the disease is not present. In other cases, the test returns the opposite result. The above situation can be summarized by the following probabilities: P(cancer) = .008, P(cancer) = .992 P(⊕|cancer) = .98, P(⊖|cancer) = .02 P(⊕|cancer) = .03, P(⊖|cancer) = .97 Suppose we now observe a new patient for whom the lab test returns a positive result. Should we diagnose the patient as having cancer or not? The maximum a posteriori hypothesis can be found using Equation (6.2): P(⊕|cancer)P(cancer) = (.98).008 = .0078 P(⊕|cancer)P(cancer) = (.03).992 = .0298 Thus, hMAP = cancer. The exact posterior probabilities can also be determined by normalizing the above quantities so that they sum to 1 (e.g., P(cancer|⊕) = .0078 / (.0078 + .0298) = .21). This step is warranted because Bayes theorem states that the posterior probabilities are just the above quantities divided by the probability of the data, P(⊕). Although P(⊕) was not provided directly as part of the problem statement, we can calculate it in this fashion because we know that P(cancer|⊕) and P(cancer|⊕) must sum to 1 (i.e., either the patient has cancer or they do not). Notice that while the posterior probability of cancer is significantly higher than its prior probability, the most probable hypothesis is still that the patient does not have cancer.

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00:01 Let's say that now we have a new patient and we know that this patient, so we know, so given that the test was positive for this patient, we need to say if this patient has cancer or not in this case.

00:20 So we have to say it has cancer or not.

00:26 So how can we say this? we can compute this probability here for each one of the.

00:32 Of this case is cancer or not cancer.

00:35 So we need to compute the probability of not having cancer, but the test said that you have.

00:43 And we need to compute the probability of having cancer, given that your test was positive.

00:52 And if this probability is higher than this, we are going to say, so let's say like, let's say that this is one, and this is two.

01:04 So if you say the one is greater than two, this means that we are going to say that the patient does not have cancer.

01:13 But now if it is the opposite, this here is higher, we are saying that the patient has cancer.

01:20 Now we just need to compute this using the base rule.

01:24 Base rule.

01:26 So the formula he will be the probability, in this case, that we let's start with cancer given that your test was positive.

01:36 This is the same as the probability of your test being positive, but you have cancer, multiply by the probability that you have, that a person can have a cancer, divided by the probability that the test is positive.

01:52 And we also need to compute this with no cancer.

01:55 So the person does not have cancer, but the test is positive.

02:00 This is what we want to compute as well.

02:02 And we can compute this by calculating this probability here.

02:08 And again, this is coming from the base rule.

02:13 Now, we just need to compute the p positive here.

02:17 So the p positive here will be given by.

02:21 The test, let's just draw something here.

02:24 The test can be positive and negative for two cases.

02:29 If you don't have cancer, or if you have cancer, sorry, and if you don't have cancer.

02:35 And we have this probability, like the probability that if you have cancer, the test will be positive with 98%.

02:45 And if you don't have cancer, the test will be positive if you have with a probability 0 .03.

02:55 So this means that the test could be positive for these two cases when you don't have the cancer and when you have the cancer.

03:04 So this means that we need to consider these two cases that we have here...

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