Discrete Probability Distributions

Transcription

1 Chapter 4 Discrete Probability Distributions 4.1 Random variable A random variable is a function that assigns values to different events in a sample space. Example Consider the experiment of rolling two dice together. Let X denote the sum of the two numbers. The sample 1

2 space is given by, S = (4.1.1) Then X can take values 2, 3,..., 12 with probabilities 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36, and 1/36 respectively. Thus X is a random variable. Discrete random variable Random variable whose values can be listed or counted are called discrete random variable. Example Suppose a physician agrees to use a new antihypertensive drug on a trial basis on the first four untreated hypertensive patients she encounters in her practice. Let X denote the number Chapter 4 2

3 of patients out of 4 whose hypertension are brought under control. then X is a discrete random variable with possible values 0, 1, 2, 3, and 4. Discrete random variables does not always have to assume integer values. A pathologist, while grading liver biopsies, categorized the amount of fat in the liver in the following ways: 0: no fat (< 5%), 1: 5%-30%, 2: 30%-70%, and 3: >70%. After finishing the grading he realized that he should have defined the category 0 as absolutely no fat (0%). Thy pathologist cleverly split the zero category into two to add a category >0-5% which he denoted by.5. Thus the whole set of values can be listed as: {0,.5, 1, 2, and 3}. Continuous random variable Random variable whose values cannot be listed or counted are called discrete random variable. Continuous random variables have uncountable sets as their support (set of values of the random variable). Example Height, Weight, BMI, Time to events, total amount Chapter 4 3

4 of drug taken, lab values such as creatinine clearance, drug half life are examples of continuous random variables. 4.2 Probability mass function The rule or function that expresses the probabilities associated with the values of a random variable in terms of its values is called probability mass function or probability distribution. If x is a possible value of a discrete random variable X, then the probability mass function assigns the probability Pr(X = x) to the value x. Often the relationship cannot be given by a single equation or formula and in such cases a tabular representation of the values and the probabilities form the probability mass function. Example Consider the experiment of rolling two dice together. Let X denote the sum of the two numbers. Then X can take values 2, 3,..., 12 with probabilities 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36, and 1/36 respectively. Thus one can Chapter 4 4

5 write the probability distribution of X as follows: Table 4.1: Probability distribution of sum of two numbers when two dice are rolled together. x Pr(X = x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 Or, one can write the probability mass function using a mathematical function Pr(X = x) = 6 7 x, x = 2, 3,..., 12. (4.2.1) 12 Example Example 4.6 (FOB). Suppose a physician agrees to use a new antihypertensive drug on a trial basis on the first four untreated hypertensive patients she encounters in her practice. Let X denote the number of patients out of 4 whose hypertension are brought under control. then X is a discrete random variable with possible values 0, 1, 2, 3, and 4. Suppose from the study conducted by the drug company we know that when a patient is treated with this drug, there is a 70% probability of response. Assuming that the Chapter 4 5

6 four patients are independent, Pr(X = 0) = Pr(No Response on all) = = Pr(X = 1) = Pr(Response on the first but not on the rest) +Pr(Response on the 2nd but not on the rest) +Pr(Response on the 3rd but not on the rest) +Pr(Response on the 4th but not on the rest) = 4( ) = Pr(X = 2) = Pr(Response on first and second but not on the rest) +Pr(Response on 1st and third but not on the rest) +. +Pr(Response on 3rd and 4th but not on the rest) = 6( ) = Chapter 4 6

7 Pr(X = 3) = Pr(No response on the first but on the rest) +Pr(No response on the 2nd but on the rest) +Pr(No response on the 3rd but on the rest) +Pr(No response on the 4th but on the rest) = 4( ) = And, similarly, Pr(X = 4) = Pr(Response on all) = = In tabular representation, Table 4.2: Probability distribution of number of Success in hypertension control. x Pr(X = x) In functional form, Pr(X = x) = 4 x (0.7) x (0.3) 4 x, x = 0, 1, 2, 3, 4, which is so-called Binomial distribution. Probability mass function satisfies: Chapter 4 7

8 (i)0 Pr(X = x) 1, and (ii) x Pr(X = x) = 1. Probability Distribution and Frequency Distribution Probability distribution of a random variable describes how frequently the values of the random variable are expected to occur in an infinite number of experiments. Whereas the relative frequency distribution gives a snapshot of the same observed in a finite number of experiments. Example Example 4.8 (FOB). Suppose the drug company provided the drug to 100 physicians and asked them to treat first four of their untreated hypertensive patients with it. Out of 100 physicians, 19 were able to bring all the four patients under control, 48 brought 3 patients under control, 24 brought 2 patients under control and the remaining 9 brought only 1 patient under control. Here is the frequency distribution: Chapter 4 8

9 Table 4.3: Frequency distribution of number of Success in hypertension control. x Total Frequency Relative Frequency 0/100 9/100 24/100 48/100 19/ , Compare this with the Probability distribution given in Table Table 4.4: Frequency and probability distribution of number of Success in hypertension control. x Total Relative Frequency Pr(X = x) In practice, one would like to know if the claim made by the company is true or not. To do that one needs to compare the probability distribution with the frequency distribution and see how close they are. This is done by so-called goodness-of-fit test in statistics which compares a theoretical probability model to an observed one. Chapter 4 9

10 4.2.1 Expected value/population mean Let us continue the hypertension example. What is the mean (average) number of patients brought under control out of 4 patients by 100 physicians? x = 100 = 0(0/100) + 1(9/100) + 2(24/100) + 3(48/100) + 4(19/100) = (4.2.2) Thus on average each physician brought 2.8 patients under control out of 4. Notice that x is being calculated based on the relative frequencies from the frequency distribution. Similarly, one can think of what the expected number of patients out of 4 would be brought under control if the probability model provided by the drug company is correct. This number is called the expected value of the random variable Chapter 4 10

11 number of Success in hypertension control, and is calculated as µ = 0(0.008) + 1(0.076) + 2(0.265) + 3(0.411) + 4(0.240) = (4.2.3) Thus, if the company s claim of 70% response were true, then on average, we would expect 2.8 out of 4 patients to be under control when treated with this drug. This expected number of responders in samples of 4 patients is close to the observed average. Expected value of a discrete random variable µ = E(X) = x x Pr(X = x) Chapter 4 11

12 Example Example 4.10 (FOB). The probability mass function of X,the number of episodes of otitis media in the first two years of life is given by Table 4.5: Probability mass function X= the number of episodes of otitis media in the first two years of life. x Pr(X = x) What is the expected number of episodes of otitis media in the first two years of life? µ = E(X) = 0(0.129)+1(0.264)+2(0.271)+3(0.185)+4(0.095)+ 5(0.039) + 6(0.017) = Thus on average a child would be expected to have about 2 episodes of otitis media in the first two years of life. Chapter 4 12

13 4.2.2 Variance of a discrete random variable Sample variance, as introduced in chapter 2 describes how the observations are spread over the whole range. Variance of a random variable, or the population variance measures the spread relative to the expected value. Variance of a discrete random variable σ 2 = V ar(x) = x (x µ)2 Pr(X = x), or σ 2 = V ar(x) = E(X 2 ) µ 2, σ 2 = V ar(x) = x x2 Pr(X = x) µ 2. Example Example 4.12 (FOB). The probability mass function of X,the number of episodes of otitis media in the first two years of life is given by Table 4.6: Probability mass function X= the number of episodes of otitis media in the first two years of life. x Pr(X = x) Chapter 4 13

14 To calculate the variance, we will use the second formula. First let us extend the above table to include a x 2 row. x x Pr(X = x) E(X 2 ) = 0(0.129) + 1(0.264) + 4(0.271) + 9(0.185) + 16(0.095) + 25(0.039) + 36(0.017) = Then, σ 2 = E(X 2 ) µ 2 = 6.12 (2.04) 2 = Corresponding population standard deviation, σ = σ 2 = 1.4. Chapter 4 14

15 4.3 Cumulative distribution function/distribution function For certain real number x, the cumulative distribution function F(x) gives the probability that the random variable X assumes a value less than or equal to x. Cumulative distribution function (c.d.f) F(x) = Pr(X x). Example Example 4.14 (FOB). The cumulative distribution function of the random variable X, the number of episodes of otitis media in the first two years of life. x Pr(X = x) F(x) =Pr(X x) Chapter 4 15

16 The c.d.f F(x) is defined for all real numbers. In the above example, we can calculate, for example, F( 2) = 0, F(2.1) = F(2) = 0.664, F(8) = F(6) = Factorial, permutations and combinations Factorial How many ways can you order n objects? Or, how many ways n individuals can sit in n chairs? Start with n = 2. n = 2 : AB, BA 2 = 2 1 = 2! n = 3 : ABC, ACB, BAC, BCA, CAB, CBA 6 = = 3! Chapter 4 16

17 n = 4 : ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, DCBA 24 = = 4! For general n, n! = n(n 1)(n 2) ! = Permutations How many ways can you choose k objects from n( k) objects? Or, how many ways n individuals can sit in k chairs? The first chair can be filled in n possible ways. The 2nd chair can be filled in n 1 possible ways.. The kth chair can be filled in n k + 1 possible ways. Chapter 4 17

18 Thus the total number of ways k chairs can be occupied by n individuals is n Pk = n (n 1)... (n k + 2) (n k + 1) n! = (n k)!. (4.4.1) Combinations Sometimes order of selection does not matter. For example, how many ways can you choose 2 individuals from 4 who volunteered to be in a clinical trial? Note that choosing A and B is same as choosing B and A. The possible ways are AB, AC, AD, BC, BD, CD. We write, 4 C2 = 4 2 = 6. In general, Chapter 4 18

19 n Ck = n k = n! k!(n k)!. Simple way to calculate combinations = 1, 1 1 = 1 = 1, = 2, = = 1, 3 1 = 3, 3 2 = 3, 3 3 = 1, etc. Chapter 4 19

20 4.5 Binomial Distribution Let us revisit the antihypertensive drug example (Example in this chapter). With a 70% chance of response, if a physician tries the drug on 4 patients, what is the probability that 2 of the four patients hypertension will be under control? Note that which patient responds is not important, we only look for two responders out of four. How many ways can 2 responders can be chosen out of 4? Obviously, 4 2 = 6 ways (RRNN,RNRN,RNNR,NRRN,N Now what is the probability that exactly one of theses sequences will occur? Pr(RRNN) = (.7)(.7)(.3)(.3) = (.7) 2 (.3) 2 = Chapter 4 20

21 Notice that every sequence has the same number of responders and non-responders. Pr(exactly 2 out of 4 will respond) = 4 2 (.7) 2 (.3) 2 = In this example, we had Two possible outcomes (Response/No response) A fixed number (4) of trials (patients) A constant probability (0.70) of success (response) for each trial, and The trials (patients) are independent. Under these conditions, the probability distribution of the number of successes is said to follow a binomial distribution. For n independent trials with each trial having probability of success p, the probability distribution of the number of successes X is given by Binomial distribution Chapter 4 21

22 Pr(X = x) = n x p x (1 p) n x, x = 0, 1, 2,..., n. Using the above distribution, we can easily calculate the probability of any number of success in any number of trials. For instance, for the antihypertensive drug example Pr(3 out of 4 will respond) = (1.7) 4 3 = Example Example 4.25 (FOB). What is the probability of obtaining 2 boys out of 5 children if the probability of a boy is 0.51 at each birth and the sexes of successive children are considered to be independent of each other? Here, n = 5, p = Pr(X = 2) = (1.51) 5 2 = Example What is the probability of obtaining at least 2 boys out of 5 children if the probability of a boy is 0.51 at each birth and the sexes of successive children are considered to be independent Chapter 4 22

23 of each other? Here, n = 5, p = Pr(X 2) = 1 Pr(X 1) = 1 {Pr(X = 0) + Pr(X = 1)} = (1.51) (1.51) 5 1 = 1 { } = = The above example shows that if n becomes even moderately large, it is time consuming, if not difficult to compute the probabilities. Realizing this, statisticians have tabulated the binomial probabilities for moderately large sample sizes (Table 1, Appendix, FOB) and some specific values of p. Example What is the probability of obtaining exactly 3 boys out of 7 children if the probability of a boy is 0.51 at each birth Chapter 4 23

24 and the sexes of successive children are considered to be independent of each other? Here, n = 7, p = From Table 1, Appendix Pr(X = 3) = The exact probability is (1.51) 7 3 = To calculate this probability using Microsoft excel, use the formula =Binomdist(3,7,.51,false). Example What is the probability of obtaining more than 3 boys out of 7 children if the probability of a boy is 0.51 at each birth and the sexes of successive children are considered to be independent of each other? Chapter 4 24

25 Here, n = 7, p = Pr(X > 3) = 1 Pr(X 3) = 1 {Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3) = 1 { } = = 0.5. To calculate this probability using Microsoft Excel, use the formula =1 - Binomdist(3,7,.51,true). The exact probability calculated using excel is Example Example (FOB). Compute (i) the probability of obtaining exactly 75 cases of chronic bronchitis and (ii) the probability of obtaining at least 75 cases of chronic bronchitis in the first year of life among 1500 families, where both parents are chronic bronchitics, if the underlying incidence rate of chronic bronchitis in the first year of life is Here, n = 1500, p = 0.05, and X=cases of chronic bronchitis in the Chapter 4 25

26 first year of life among 1500 families. (i) Pr(X = 75) = 1500 (.05) 7 5(1.05) (ii) Pr(X 75) = = Binomdist(1500, 75,.05, false) (In MS Excel) = x= x (.05) x (1.05) 1500 x = 1 Binomdist(1500, 74,.05, true) (In MS Excel) = = Chapter 4 26

27 Expected value (Mean) and Variance of the Binomial Distribution If we conduct a Bernoulli trial (a trial that results in only two outcomes - success and failure) with probability of success p n times, what would be the expected number of successes? Mean and variance of a binomial distribution E(X) = np, V ar(x) = np(1 p). Now that we know the formula, we can easily calculate the expected number of successes in four hypertensive patients when treated with an antihypertensive drug with a response rate of 70%. The answer is: 4*.70=2.8 (same as what we found in equation on Page 11). Parameters of binomial distribution n and p are the two parameters of Binomial distribution. Chapter 4 27

28 4.6 Poisson Distribution In the binomial distribution, we had a fixed number of trials. However, in many situations the number of trials might not be fixed. Here are some situations: Number of Lexus brand cars crossing a particular intersection within a fixed time interval (theoretically it could be 0, 1, 2,...,.) Number of deaths caused by typhoid fever in 20 years Number of bacterial colonies growing on a 100 cm 2 agar plate Number of goals scored by a team in a 20-minute game In all these cases, there is no fixed number of trials. But what if we want to calculate, for instance, the probability that the team scores just one goal in a 20-minute period? Chapter 4 28

29 Assumptions of Poisson Distribution 1. The probability of an event in an infinitesimal time interval is very small 2. The number of events in two distinct time intervals are independent, and 3. The rate of occurrence depends only on the length of time (proportional to the length), but not on where the interval starts or ends. Under the above assumptions, the distribution of the number of events within a specific period of time is said to follow a Poisson distribution. Suppose the rate at which the events occur in an interval is µ. Then the probability mass function for X, the number of events in that interval is: Probability mass function of a Poisson random variable Pr(X = x) = e µ µ x x!, x = 0, 1, 2,... Chapter 4 29

30 Parameters of Poisson distribution µ is the only parameter of the Poisson distribution. Example Suppose that the number of deaths from typhoid fever over a 1-year period is distributed as a Poisson random variable with parameter µ = What is the probability mass function of the number of deaths in 6-months period? Since µ = 4.6/year, µ = 2.3/6-month. Let X = the number of deaths within 6-month period. Then the distribution of the number of deaths within 6-month period is Poisson with mass function: Pr(X = x) = e 2.3 (2.3) x, x = 0, 1, 2,.... x! 2. What is the probability that 3 deaths occur in 6-months period? Pr(X = 3) = e 2.3 (2.3) 3 3! = Chapter 4 30

31 3. What is the probability of more than 3 deaths in 6-months period? Pr(X > 3) = 1 {Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3)} = 1 {e e 2.3 (2.3) + e 2.3 (2.3) 2 + e 2.3 (2.3) 3 } 2! 3! = 1 { } = (4.6.1) 4. What is the probability of no more than 2 deaths in 3-months period? Let Y = the number of deaths within 6-month period. Then the distribution of the number of deaths within 3-month period is Poisson with mass function: Pr(Y = y) = e 2.3 (1.15) y, y = 0, 1, 2,.... y! Pr(Y 2) = Pr(Y = 0) + Pr(Y = 1) + Pr(Y = 2) = e e 1.15 (1.15) + e 1.15 (1.15) 2 = = (4.6.2) 2! Chapter 4 31