MA : Introductory Probability

Transcription

1 MA : Introductory Probability David Murrugarra Department of Mathematics, University of Kentucky Spring 2017 David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

2 Negative Binomial Distribution Suppose we are given a coin which has probability p of coming up heads when it is tossed. We fix a positive integer k, and toss the coin until the k th head appears. We let X represent the number of tosses. When k = 1, X is geometrically distributed. David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

3 Negative Binomial Distribution Suppose we are given a coin which has probability p of coming up heads when it is tossed. We fix a positive integer k, and toss the coin until the k th head appears. We let X represent the number of tosses. When k = 1, X is geometrically distributed. For a general k, we say that X has a negative binomial distribution. David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

4 Negative Binomial Distribution Suppose we are given a coin which has probability p of coming up heads when it is tossed. We fix a positive integer k, and toss the coin until the k th head appears. We let X represent the number of tosses. When k = 1, X is geometrically distributed. For a general k, we say that X has a negative binomial distribution. We now calculate the probability distribution of X. If X = x, then it must be true that there were exactly k 1 heads thrown in the first x 1 tosses, and a head must have been thrown on the x th toss. David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

5 Negative Binomial Distribution Suppose we are given a coin which has probability p of coming up heads when it is tossed. We fix a positive integer k, and toss the coin until the k th head appears. We let X represent the number of tosses. When k = 1, X is geometrically distributed. For a general k, we say that X has a negative binomial distribution. We now calculate the probability distribution of X. If X = x, then it must be true that there were exactly k 1 heads thrown in the first x 1 tosses, and a head must have been thrown on the x th toss. There are ( ) x 1 k 1 sequences of length x with these properties, and each of them is assigned the same probability, namely David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

6 Negative Binomial Distribution Suppose we are given a coin which has probability p of coming up heads when it is tossed. We fix a positive integer k, and toss the coin until the k th head appears. We let X represent the number of tosses. When k = 1, X is geometrically distributed. For a general k, we say that X has a negative binomial distribution. We now calculate the probability distribution of X. If X = x, then it must be true that there were exactly k 1 heads thrown in the first x 1 tosses, and a head must have been thrown on the x th toss. There are ( ) x 1 k 1 sequences of length x with these properties, and each of them is assigned the same probability, namely p k 1 q x k David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

7 Negative Binomial Distribution Therefore, if we define Definition (Negative Binomial Distribution) If 0 < p < 1 and q = p 1, then the random variable X has a Negative Binomial Distribution if ( ) x 1 u(x, k, p) = P(X = x) = p k q x k for x = 1, 2, 3,.... k 1 We must show that P(X = x) = 1 x=1 David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

8 Negative Binomial Distribution Therefore, if we define Definition (Negative Binomial Distribution) If 0 < p < 1 and q = p 1, then the random variable X has a Negative Binomial Distribution if ( ) x 1 u(x, k, p) = P(X = x) = p k q x k for x = 1, 2, 3,.... k 1 We must show that P(X = x) = 1 x=1 This is true from the following formula ( ) k + j 1 (1 r) k = r j for 0 r < 1. j j=0 David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

9 Negative Binomial Distribution The last expression is equal to (1 r) k = For r = q, we have Then ( ) k + j 1 r j for 0 r < 1. k 1 j=0 1 (1 q) k = P(X = x) = x=1 x=0 j=0 ( k + j 1 k 1 ( ) x 1 p k q x k = k 1 ) q j for 0 r < 1. j=1 k ( k + j 1 k 1 ) p k q j. David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

10 Negative Binomial Distribution The last expression is equal to P(X = x) = x=1 ( k + j 1 k 1 j=0 ) p k q j = p k (1 q) k = 1. Thus u(x, k, p) = P(X = x) = 1. x=1 x=1 David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

11 Hypergeometric Distribution Suppose that we have a set of N balls, of which k are red and N k are blue. We choose n of these balls, without replacement, and define X to be the number of red balls in our sample. The distribution of X is called the hypergeometric distribution. There are ( ) N n different samples of size n, and the number of such samples with exactly x red balls is obtained by multiplying the number of ways of choosing x red balls from the set of k red balls and the number of ways of choosing n x blue balls from the set of N k blue balls. David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

12 Hypergeometric Distribution Hence, we define Definition (Hypergeometric Distribution) h(n, k, n, x) = P(X = x) = ( k N k ) x)( n x ( N. n) We must show that P(X = x) = 1 x=1 David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

13 Example (Fisher s Exact Test) It is often of interest to consider two traits, such as eye color and hair color, and to ask whether there is an association between the two traits. Two traits are associated if knowing the value of one of the traits for a given person allows us to predict the value of the other trait for that person. In this example, we will use the traits of gender and political party, and we will assume that there are only two possible genders, female and male, and only two possible political parties, Democratic and Republican. Suppose that we have collected data concerning these traits. To test whether there is an association between the traits, we first assume that there is no association between the two traits. This gives rise to an expected" data set, in which knowledge of the value of one trait is of no help in predicting the value of the other trait. David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

14 Example (Fisher s Exact Test) Our collected data set usually differs from this expected data set. If it differs by quite a bit", then we would tend to reject the assumption of independence of the traits. To nail down what is meant by quite a bit", we decide which possible data sets differ from the expected data set by at least as much as ours does, and then we compute the probability that any of these data sets would occur under the assumption of independence of traits. If this probability is small, then it is unlikely that the difference between our collected data set and the expected data set is due entirely to chance. Democrat Republican Female male David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

15 Example (Fisher s Exact Test) A general data table will look as follows, Democrat Republican Female s 11 s 12 t 11 male s 21 s 22 t 12 t 21 t 22 N Table: General data table. The row and column sums are called marginal totals, or marginals. We will denote the row sums by t 11 and t 12, and the column sums by t 21 and t 22. The (i, j) entry in the table will be denoted by s ij. Finally, the size of the data set will be denoted by N. David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

16 Example (Fisher s Exact Test) Democrat Republican Female s 11 s 12 t 11 male s 21 s 22 t 12 t 21 t 22 N Table: General data table. We then put t 21 yellow balls and t 22 green balls, corresponding to the Democratic and Republican marginals, into an urn. We draw t 11 balls, without replacement, from the urn, and call these balls females. The t 12 balls remaining in the urn are called males. David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

17 Example (Fisher s Exact Test) Democrat Republican Female male Table: Observed data. The probability of getting the actual data under this model is given by h(50, 32, 28, 24) = P(X = 24) = ( 32 )( ( ) ). David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

18 Example (Fisher s Exact Test) Democrat Republican Female male Table: Expected data. To construct the expected data set, If we choose 28 balls out of 50, we should expect to see, on the average, the same percentage of yellow balls in our sample as in the urn. Thus, we should expect to see, on the average, 28(32/50) = yellow balls in our sample. The other expected values are computed in exactly the same way. We note that the value of s 11 determines the other three values in the table, since the marginals are all fixed. David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

19 Example (Fisher s Exact Test) Democrat Republican Female male Table: Expected data. Thus, in considering the possible data sets that could appear in this model, it is enough to consider the various possible values of s 11. What is the probability of drawing exactly a" yellow balls, i.e., what is the probability that s 11 = a? It is h(50, 32, 28, a) = P(X = a) = ( 32 )( 18 a ( a ) ). (1) David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

20 Example (Fisher s Exact Test) We are now ready to decide whether our actual data differs from the expected data set by an amount which is greater than could be reasonably attributed to chance alone. David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

21 Example (Fisher s Exact Test) We are now ready to decide whether our actual data differs from the expected data set by an amount which is greater than could be reasonably attributed to chance alone. We note that the expected number of female Democrats is 18, but the actual number in our data is 24. The other data sets which differ from the expected data set by more than ours correspond to those where the number of female Democrats equals 25, 26, 27, or 28. David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

22 Example (Fisher s Exact Test) We are now ready to decide whether our actual data differs from the expected data set by an amount which is greater than could be reasonably attributed to chance alone. We note that the expected number of female Democrats is 18, but the actual number in our data is 24. The other data sets which differ from the expected data set by more than ours correspond to those where the number of female Democrats equals 25, 26, 27, or 28. Thus, to obtain the required probability, we sum the expression in Equation 1, ( )( 18 ) a 28 a a=24 ( ) = David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15

23 Example (Fisher s Exact Test) We are now ready to decide whether our actual data differs from the expected data set by an amount which is greater than could be reasonably attributed to chance alone. We note that the expected number of female Democrats is 18, but the actual number in our data is 24. The other data sets which differ from the expected data set by more than ours correspond to those where the number of female Democrats equals 25, 26, 27, or 28. Thus, to obtain the required probability, we sum the expression in Equation 1, ( )( 18 ) a 28 a a=24 ( ) = Thus, we should reject the hypothesis that the two traits are independent. David Murrugarra (University of Kentucky) MA 320: Section 5.1 Spring / 15