Oct 27 2009 LEC 10 ECON 240A 1 Probability Models L Phillips I Introduction So far we have used probability as a foundation for the binomial and normal distributions Then we have used these distributions as the models underlying our statistics such as the z variable normal with mean zero and variance one Interval estimation and hypothesis testing were based on such models Student s t distribution was based on a more elaborate model for a random variable that was the ratio of a z variable to a Chi Square variable But probability can be used directly for analysis We will begin by studying an application of Bayes Theorem using conditional probability joint probability and marginal probability to calculate the likelihood a patient has a disease given that his her test for the disease was positive II Bayes Theorem and Conditional Probability There are two underlying events in this example of epidemiology and statistics First an individual is sick has the disease event S or does not event S and second the individual tests positive for the disease event P or does not event P A two by two tableau is depicted in Table 1 showing the four joint probabilities such as the individual testing positive and having the disease p S P The marginal probabilities are shown along the right side and the bottom of this tableau The underlying facts are that 1 the incidence of the disease in the population is one in a thousand p S 0 001 2 the probability of testing positive for the disease given that you have it is ninety nine out of one hundred p P S 0 99 and 3 the probability of a false positive i e testing positive even if you do not have the disease is Oct 27 2009 LEC 10 ECON 240A 2 Probability Models L Phillips only two in a hundred i e p P S 0 02 Note that the test seems reasonably accurate or reliable The issue for the patient and the physician is what is the probability the patient has the disease if the test comes back positive i e what is p S P Table 1 Testing for a Disease Joint and Marginal Probabilities S Healthy S Diseased p S P p S P p P p S P p S P p P S p S p 1 P Test Positive P Test negative We use the facts to fill out the tableau as indicated in Table 2 Table 2 Testing for a Disease Joint and Marginal Probabilities Some Numbers P Test Positive P Test negative S Diseased p S P p S P p S 0 001 S Healthy p S P p S P p S 0 999 p P p P 1 We can use conditional probability and our facts to extend our knowledge p P S p S p P S 1 0 02 0 999 0 01998 2 and p P S p S p P S 3 0 99 0 001 0 00099 4 Oct 27 2009 LEC 10 ECON 240A 3 Probability Models L Phillips The calculation of these two joint probabilities can be used to fill in the tableau even more Table 3 Testing for a Disease Joint and Marginal Probabilities Sufficient Info P Test Positive P Test negative S Diseased p S P 0 00099 p S P p S 0 001 S Healthy p S P 0 01998 p S P p S 0 999 p P p P 1 There is now sufficient information to fill in all of the remaining blanks We are interested in the conditional probability p S P the probability we have the disease given that we tested positive p S P p P p S P 5 p S P 0 02097 0 00099 6 p S P 0 0472 7 So despite the apparent accuracy of the test with a conditional probability of testing positive if you have the disease of 0 99 and a conditional probability of testing positive if you are healthy of 0 02 the calculated conditional probability the patient has the disease given the test comes back positive is less than 5 So the test is not all that informative This is called the false positive paradox and is another good reason to have a good doctor and to be involved in your own health care Both the patient and the doctor need to conduct further tests and further diagnoses before settling on a course of treatment Oct 27 2009 LEC 10 ECON 240A 4 Probability Models L Phillips The tableau provided a cognitive and organized methodology for using the facts to derive their implications for decision making We could take a short cut by combining Eq s 3 and 5 p P S p S p P S p S P p P 8 p S P p P S p S p P 0 99 0 001 p P 9 and we still need p P to get the conditional probability that answers our question Looking at Table 1 we see p P p S P p S P p P S p S p P S p S 10 11 So by combining Eq s 9 and 11 we obtain p S P p P S p S p P S p S p P S p S 12 This is called Bayes Theorem III Duration Models Failure Times and the Exponential Distribution Duration studies or failure time analysis involve issues such as 1 the length of time an individual is unemployed before obtaining a job 2 the length of time an individual is on welfare before leaving the rolls 3 the duration of economic expansions or booms and 4 the length of life for a patient after receiving a therapy such as a heart bypass operation From the last example you can see that this type of statistics has a history in medicine as well as economics and terms such as number at risk carry over For illustration we will look at the length of expansions since World War II the number of months between the trough of the business cycle and its peak These turning points are established by the National Bureau of Economic Research and the information Oct 27 2009 LEC 10 ECON 240A 5 Probability Models L Phillips is published on occasion in Survey of Current Business published monthly by the U S Department of Commerce The timing of the post war business cycles is reproduced in Table 4 Table 4 Duration of Post War Economic Expansions in Months Trough October 1945 October 1949 May 1954 April 1958 February 1961 November 1970 March 1975 July 1980 November 1982 March 1991 November 2001 Peak November 1948 July 1953 August 1957 April 1960 December 1969 November 1973 January 1980 July 1981 July 1990 March 2001 not decided yet Sept 2007 Duration 37 45 39 24 106 36 58 12 92 120 70 so far For all continuous distributions the survivor function S t is defined as S t 1 F t 13 Where F t is the cumulative distribution function The hazard function is defined …
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