Page 1 Module 10 Reasoning with Uncertainty - Probabilistic reasoning Version 2 CSE IIT, Kharagpur Page 2 Module 10 Reasoning with Uncertainty - Probabilistic reasoning Version 2 CSE IIT, Kharagpur Lesson 27 Probabilistic Inference Version 2 CSE IIT, Kharagpur Page 3 Module 10 Reasoning with Uncertainty - Probabilistic reasoning Version 2 CSE IIT, Kharagpur Lesson 27 Probabilistic Inference Version 2 CSE IIT, Kharagpur 10.4 Probabilistic Inference Rules Two rules in probability theory are important for inferencing, namely, the product rule and the Bayes' rule. Here is a simple example, of application of Bayes' rule. Suppose you have been tested positive for a disease; what is the probability that you actually have the disease? It depends on the accuracy and sensitivity of the test, and on the background (prior) probability of the disease. Let P(Test=+ve | Disease=true) = 0.95, so the false negative rate, P(Test=-ve | Disease=true), is 5%. Let P(Test=+ve | Disease=false) = 0.05, so the false positive rate is also 5%. Suppose the disease is rare: P(Disease=true) = 0.01 (1%). Let D denote Disease and "T=+ve" denote the positive Tes. Then, P(T=+ve|D=true) * P(D=true) P(D=true|T=+ve) = ------------------------------------------------------------ P(T=+ve|D=true) * P(D=true)+ P(T=+ve|D=false) * P(D=false) Version 2 CSE IIT, Kharagpur Page 4 Module 10 Reasoning with Uncertainty - Probabilistic reasoning Version 2 CSE IIT, Kharagpur Lesson 27 Probabilistic Inference Version 2 CSE IIT, Kharagpur 10.4 Probabilistic Inference Rules Two rules in probability theory are important for inferencing, namely, the product rule and the Bayes' rule. Here is a simple example, of application of Bayes' rule. Suppose you have been tested positive for a disease; what is the probability that you actually have the disease? It depends on the accuracy and sensitivity of the test, and on the background (prior) probability of the disease. Let P(Test=+ve | Disease=true) = 0.95, so the false negative rate, P(Test=-ve | Disease=true), is 5%. Let P(Test=+ve | Disease=false) = 0.05, so the false positive rate is also 5%. Suppose the disease is rare: P(Disease=true) = 0.01 (1%). Let D denote Disease and "T=+ve" denote the positive Tes. Then, P(T=+ve|D=true) * P(D=true) P(D=true|T=+ve) = ------------------------------------------------------------ P(T=+ve|D=true) * P(D=true)+ P(T=+ve|D=false) * P(D=false) Version 2 CSE IIT, Kharagpur 0.95 * 0.01 = -------------------------------- = 0.161 0.95*0.01 + 0.05*0.99 So the probability of having the disease given that you tested positive is just 16%. This seems too low, but here is an intuitive argument to support it. Of 100 people, we expect only 1 to have the disease, but we expect about 5% of those (5 people) to test positive. So of the 6 people who test positive, we only expect 1 of them to actually have the disease; and indeed 1/6 is approximately 0.16. In other words, the reason the number is so small is that you believed that this is a rare disease; the test has made it 16 times more likely you have the disease, but it is still unlikely in absolute terms. If you want to be "objective", you can set the prior to uniform (i.e. effectively ignore the prior), and then get P(T=+ve|D=true) * P(D=true) P(D=true|T=+ve) = ------------------------------------------------------------ P(T=+ve) 0.95 * 0.5 0.475 = -------------------------- = ------- = 0.95 0.95*0.5 + 0.05*0.5 0.5 This, of course, is just the true positive rate of the test. However, this conclusion relies on your belief that, if you did not conduct the test, half the people in the world have the disease, which does not seem reasonable. A better approach is to use a plausible prior (eg P(D=true)=0.01), but then conduct multiple independent tests; if they all show up positive, then the posterior will increase. For example, if we conduct two (conditionally independent) tests T1, T2 with the same reliability, and they are both positive, we get P(T1=+ve|D=true) * P(T2=+ve|D=true) * P(D=true) P(D=true|T1=+ve,T2=+ve) = ------------------------------------------------------------ P(T1=+ve, T2=+ve) 0.95 * 0.95 * 0.01 0.009 = ----------------------------- = ------- = 0.7826 0.95*0.95*0.01 + 0.05*0.05*0.99 0.0115 The assumption that the pieces of evidence are conditionally independent is called the naive Bayes assumption. This model has been successfully used for mainly application including classifying email as spam (D=true) or not (D=false) given the presence of various key words (Ti=+ve if word i is in the text, else Ti=-ve). It is clear that the words are not independent, even conditioned on spam/not-spam, but the model works surprisingly well nonetheless. Version 2 CSE IIT, Kharagpur Page 5 Module 10 Reasoning with Uncertainty - Probabilistic reasoning Version 2 CSE IIT, Kharagpur Lesson 27 Probabilistic Inference Version 2 CSE IIT, Kharagpur 10.4 Probabilistic Inference Rules Two rules in probability theory are important for inferencing, namely, the product rule and the Bayes' rule. Here is a simple example, of application of Bayes' rule. Suppose you have been tested positive for a disease; what is the probability that you actually have the disease? It depends on the accuracy and sensitivity of the test, and on the background (prior) probability of the disease. Let P(Test=+ve | Disease=true) = 0.95, so the false negative rate, P(Test=-ve | Disease=true), is 5%. Let P(Test=+ve | Disease=false) = 0.05, so the false positive rate is also 5%. Suppose the disease is rare: P(Disease=true) = 0.01 (1%). Let D denote Disease and "T=+ve" denote the positive Tes. Then, P(T=+ve|D=true) * P(D=true) P(D=true|T=+ve) = ------------------------------------------------------------ P(T=+ve|D=true) * P(D=true)+ P(T=+ve|D=false) * P(D=false) Version 2 CSE IIT, Kharagpur 0.95 * 0.01 = -------------------------------- = 0.161 0.95*0.01 + 0.05*0.99 So the probability of having the disease given that you tested positive is just 16%. This seems too low, but here is an intuitive argument to support it. Of 100 people, we expect only 1 to have the disease, but we expect about 5% of those (5 people) to test positive. So of the 6 people who test positive, we only expect 1 of them to actually have the disease; and indeed 1/6 is approximately 0.16. In other words, the reason the number is so small is that you believed that this is a rare disease; the test has made it 16 times more likely you have the disease, but it is still unlikely in absolute terms. If you want to be "objective", you can set the prior to uniform (i.e. effectively ignore the prior), and then get P(T=+ve|D=true) * P(D=true) P(D=true|T=+ve) = ------------------------------------------------------------ P(T=+ve) 0.95 * 0.5 0.475 = -------------------------- = ------- = 0.95 0.95*0.5 + 0.05*0.5 0.5 This, of course, is just the true positive rate of the test. However, this conclusion relies on your belief that, if you did not conduct the test, half the people in the world have the disease, which does not seem reasonable. A better approach is to use a plausible prior (eg P(D=true)=0.01), but then conduct multiple independent tests; if they all show up positive, then the posterior will increase. For example, if we conduct two (conditionally independent) tests T1, T2 with the same reliability, and they are both positive, we get P(T1=+ve|D=true) * P(T2=+ve|D=true) * P(D=true) P(D=true|T1=+ve,T2=+ve) = ------------------------------------------------------------ P(T1=+ve, T2=+ve) 0.95 * 0.95 * 0.01 0.009 = ----------------------------- = ------- = 0.7826 0.95*0.95*0.01 + 0.05*0.05*0.99 0.0115 The assumption that the pieces of evidence are conditionally independent is called the naive Bayes assumption. This model has been successfully used for mainly application including classifying email as spam (D=true) or not (D=false) given the presence of various key words (Ti=+ve if word i is in the text, else Ti=-ve). It is clear that the words are not independent, even conditioned on spam/not-spam, but the model works surprisingly well nonetheless. Version 2 CSE IIT, Kharagpur In many problems, complete independence of variables do not exist. Though many of them are conditionally independent. X and Y are conditionally independent given Z iff In full: X and Y are conditionally independent given Z iff for any instantiation x, y, z of X, Y,Z we have n example of conditional independence: onsider the following three Boolean random variables: eaveBy8, GetTrain, OnTime uppose we can assume that: (OnTime | GetTrain, LeaveBy8) = P(OnTime | GetTrain) ut NOT P(OnTime | LeaveBy8) = P(OnTime) hen, OnTime is dependent on LeaveBy8, but independent of LeaveBy8 given GetTrain. e can represent P(OnTime | GetTrain, LeaveBy8) = P(OnTime | GetTrain) raphically by: LeaveBy8 -> GetTrain -> OnTime A C L S P b T W g Version 2 CSE IIT, KharagpurRead More

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