Business Statistics, Can. ed. By Black, Chakrapani & Castillo

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1 Business Statistics, Can. ed. By Black, Chakrapani & Castillo Discrete Distributions Chapter 5 Discrete Distributions Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd. Prepared by Dr. Clarence S. Bayne, JMSB, Concordia University

2 Learning Objectives To distinguish between discrete random variables and continuous random variables. Learn how to determine and calculate the mean and variance of a discrete distribution. Identify the type of statistical experiments described by the binomial distribution When to use the Poisson distribution When and how to approximate the binomial distribution by the poisson distribution and solve problems Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

3 Discrete vs. Continuous Distributions Random Variable : a variable which contains the outcomes of a chance or random experiment Discrete Random Variable: the set of all possible values that is at the most a finite number; or an infinite number of all possible opportunities, but countable not measurable values Number of new subscribers to a magazine: outcome countable(person is subscriber or not subscriber), number of instances may be very large but finite Number of bad checks received by a restaurant: outcome countable (a bad check or not a bad check); instances or opportunities may be large, even infinite Number of absent employees on a given day: outcome only countable (an employee is absent or not absent): instances involve a finite number of employees Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

4 Continuous Random Variable Continuous random variables takes on values at every point over a given interval Continuous random variables have no gaps or unassumed values. It could be said that continuous random variables are generated by experiments in which things are measured, not counted. Examples of measures for which continuous random variables might be generated are time, height, weight, and volume Some cases: Current Ratio of a motorcycle distributorship Elapsed time between arrivals of bank customers Percent of the labour force that is unemployed Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

5 Discrete binomial Poisson Some Special Discrete and Continuous Distributions Continuous uniform normal exponential T Distribution chi square F Distribution Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

6 Discrete Distribution -- Example Distribution of Daily Crises Number of Probability Crises P r o b a b i l i t y Number of Crises Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

7 Requirements for a Discrete Probability Function Probabilities are between 0 and, inclusively 0 P( X) for all X Total of all probabilities equals PX ( ) over all x Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

8 Identification of a Discrete Probability Function : Examples X P(X) X P(X) X P(X) PROBABILITY DISTRIBUTION : YES NO NO Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

9 Expected Value or Mean of a Discrete Distribution EX XPX ( ) X P(X) X P( X) =.0 Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

10 Variance and Standard Deviation of a Discrete Distribution X P( X )... 0 X P(X) X ( X ) ( X ) P( X) Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd

11 Mean of Daily Crises Data Example EX XPX ( ) 5. X P(X) X P(X) P r o b a b i l i t y Number of Crises Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

12 Variance and Standard Deviation of Daily Crises Data Example X P( X ) X P(X) (X-µ) (X-µ) (X-µ) *P(x) Total s.00.4 Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

13 The Binomial Experiment and Applications The Binomial process The experiment involves n identical trials or selections Each trial has exactly two possible outcomes: success and failure Each trial is independent of the previous trials p is the probability of a success on any one trial q = ( p) is the probability of a failure on any one trial p and q are constant throughout the experiment X is the number of successes in the n trials Binomial Applications When sampling with replacement When sampling without replacement but n < 5% N Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

14 Binomial Distribution Probability function Mean value PX ( ) n! X! n X! n p X p q nx for 0 X n Variance and standard deviation npq n pq Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

15 Development of Binomial Distribution: Laying out the Problem Experiment : Randomly selecting with replacement, n = (two) families from two family types, residents of Tiny Town Success = Children in Household : p = 0.75 Failure = No Children in Household : q = p = 0.5 Let X be the number of families in the sample with Children in Household Family A B C D Children in Household Yes Yes No Yes Number of Automobiles 3 Listing of Sample Space (A,B), (A,C), (A,D), (A,A), (B,A), (B,B), (B,C), (B,D), (C,A), (C,B), (C,C), (C,D), (D,A), (D,B), (D,C), (D,D) Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

16 Development of Binomial Distribution: Counting Sample Points and Relative Frequencies Families A, B, and D have children in the household; family C does not Success is Children in Household: p = 0.75 Failure is No Children in Household: q = - p = 0.5 X is the number of families in the sample with Children in Household Listing of Sample Space (A,B), (A,C), (A,D), (A,A), (B,A), (B,B), (B,C), (B,D), (C,A), (C,B), (C,C), (C,D), (D,A), (D,B), (D,C), (D,D) P(outcome) X 0 Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

17 Developing the Binomial Distribution: Verifying Formula Listing of Sample Space (A,B), (A,C), (A,D), (A,A), (B,A), (B,B), (B,C), (B,D), (C,A), (C,B), (C,C), (C,D), (D,A), (D,B), (D,C), (D,D) P(outcome) X 0 X 0 P( X) P( X 0) P( X ) P( X ) P(X) 6/6 9/6 n! X! n X!! 0! 0 pq!!!!!!! x n x Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

18 Binomial Distribution: Deriving Formula Selecting groups of two types of families from four family types Families A, B, and D have children in the household; family C does not Success is Children in Household : p = 0.75 Failure is No Children in Household : q = p = 0.5 X = number of families in the sample with Children in Household Possible Sequences (F,F) (S,F) (F,S) (S,S) P(sequence) (. 5)(. 5) (. 75)(. 5) (. 5)(. 75) (. 75)(. 75) X 0 Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

19 Binomial Distribution: Formula Development Continued Possible Sequences P(sequence) X X P(X) (F,F) (S,F) (F,S) (S,S) P( X 0) P( X ) (. 5)(. 5) (. 5) (. 75)(. 5) (. 5)(. 75) (. 75)(. 75) (. 75)! 0! 0!!!! Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd. 0 P( X) (. 5)(. 5) (. 5) (. 5)(. 75) =0.375 (. 75)(. 75) (. 75) n! X! n X!! PX ( )!! =0.065 =0.565 pq x n x

20 Rationale for the Binomial Formula Note that the Binomial formula has two parts: The first part counts the number of different arrangement of events that have a particular structure (r successes and n r failures) in n selections with replacement from a group of size N; n<n. The count is derived as follows: n! r! n r! The second part is the probability of getting any given sequence of such n random trials, given as P r ( P) n r In the Family problem the event family with children and family without children has two (!/!!) possible sequences, either (S,F) or (F,S). Each has the identical probability 0.75 * 0.5 ( ), hence *0.75*0.5 = 0.35 Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

21 Binomial Distribution Application: Demonstration Problem 5.3 n 0 p. 06 q. 94 PX ( ) PX ( 0) PX ( ) PX ( ) P( X 0) 0! 0!(0 0)! ()()(. 90). 90 P( X ) 0!!( 0 )! ( 0)(. 06)(. 3086) P( X ) 0!!( 0 )! ( 90)(. 0036)(. 383). 46 Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

22 Binomial Table Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

23 n = 0 PROBABILITY X Using the Binomial Table Demonstration Problem 5.4 n 0 p PX ( 0) 0C0 07. Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

24 Binomial Distribution using Table: Demonstration Problem 5.3 n = 0 PROBABILITY X n p q. 94 P( X ) P( X 0) P( X ) P( X ) PX ( ) PX ( ) n p ( 0 )(. 06 ). 0 n pq ( 0 )(. 06 )(. 94 ) Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

25 Graphs of Selected Binomial Distributions n = 4 PROBABILITY X P(X) P = X P(X) P = X P(X) P = X Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

26 Normal Shape of Binomial Function for Large Values of n: Excel Graph Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

27 Poisson Distribution: The Law of Improbable Events Describes discrete occurrences over a continuum or interval or some opportunity A discrete distribution Describes rare or improbable events Each occurrence is independent of any other occurrences The number of occurrences in each interval can vary from zero to infinity The expected number of occurrences must hold constant throughout the experiment Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

28 Poisson Distribution: Applications Arrivals at queuing systems At airports: people, airplanes, automobiles, baggage At banks: people, automobiles, loan applications Number of cars per minute entering 407 Express Toll Route, Ontario coming from Highway 47 between 3 AM and 4 AM. Defects in goods, crises, epidemics Number of flaws per bolt of cloth Number of blemishes per square foot of painted surface Number of errors per typed page. Number of cases of a rare disease (HN, etc) Number of sewing flaws per pair of jeans during production When the events are occurring too frequently in an interval of time or opportunity the length of the interval can be reduced until the event becomes a rare occurrence Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

29 Poisson Distribution Poisson probability function PX ( ) where: e X e X! for X long run average 03,,,, (the base of natural logarithms) Mean value Variance Standard deviation Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

30 Poisson Distribution: Demonstration Problem customers / 4 minutes X = 0 customers / 8 minutes Adjusted = 64. customers / 8 minutes P(X) = X e X! PX ( = 0)= e 0! customers / 4 minutes X = 6 customers / 8 minutes Adjusted = 64. customers / 8 minutes P(X) = X e PX ( = 6)= X! e 6! Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

31 Poisson Distribution: Probability Table X Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

32 Poisson Distribution: Using the Poisson Tables X PX ( 4) Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

33 Poisson Distribution: Using the Poisson Tables X PX ( 5) PX ( 6) PX ( 7) PX ( 8) PX ( 9) Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

34 Poisson Distribution: Using the Poisson Tables X PX ( ) PX ( ) PX ( 0) PX ( ) Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

35 Poisson Distribution: Graphs Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

36 Discrete Distributions in R The discrete distributions of statistics are not continuous. Usually, they are constructed of a finite number of possible values for the random variable and each possibility is assigned a probability of occurrence. The Bernoulli Distribution One of the simplest discrete distributions is called the Bernoulli Distribution. This is a distribution with only two possible values. For example, consider the Bernoulli distribution in the table that follows: A Bernoulli Distribution x p In this case, there are only two possible values of the random variable, x = 0 or x =. When drawing numbers from this distribution, the probability of selecting x = 0 is 0.3, while the probability of selecting x = is 0.7. There are many applications that can be represented by Bernoulli Distributions. For example, x = 0and x = might represent the states of a lamp, with zero representing the state that the lamp is switched "off" and one representing the state that the lamp is switched "on." Or x = 0 and x = might represent the states of a biased coin, with zero representing "heads" and one representing "tails." Because the values of this distribution are discrete, the probability density density function will not be a continuous curve as in the Standard Normal Distribution or the Normal Distribution. The usual way to visualize a discrete distribution is with a sequence of "spikes." Use the following code to produce the image in Figure 3. x=c(0,) y=c(0.3,0.7) plot(x,y,type="h",xlim=c(-,),ylim=c(0,), lwd=,col="blue",ylab="p") points(x,y,pch=6,cex=,col="dark red") There are a few new constructs introduced in the above code:

37 . We use xlim and ylim to place "limits" on the horizontal and vertical axes, respectively.. In the points command, the pch argument determines the "plotting character" that is used. Experiment with the numbers through 6 to see a variety of different choices for the plotting character. Try?points to learn more. 3. In the points command, the cex argument determines the "character expansion." We've used it here to "scale" the size of the plotting character. Try values of or 3 and note the difference. Figure 3. The discrete Bernoulli probability density function. Note the "discrete" nature of the visualization.. On the x-axis, there are only two possibilities, x = 0 or x =.. On the vertical axis, the heights of the "spikes" represent probabilities. Note that the height of the spike at x = 0 is 0.3, the probability of selecting x = 0. The height of the spike at x = is 0.7, the probability of selecting x =. Key Idea: It is important to note that the sum of the probabilites equals, much as it did in thestandard Normal Distribution and the Normal Distribution.

38 Only this time, it is not the area under a curve that represents the total probability, but the sum of the probabilities represented by the heights of the "spikes." The Binomial Distribution The Binomial Distribution is best introduced with a simple example. Suppose that we have a "fair" coin, one in which there is an equal probability of flipping "heads" or "tails." Now, suppose that we toss the coin three times. Further, suppose that we define a random variable X that equals the number of heads obtained in three tosses of the fair coin. The outcomes and the value of the random variable are listed in the table that follows. Counting the Number of Heads Outcome X = # Heads HHH 3 HHT HTH HTT THH THT TTH TTT 0 Because the coin is "fair," each of the outcomes is equally likely, with each having a in 8 chance of occurring. Thus, the probability of tossing three heads (HHH) is /8. Next, there are three ways of obtaining two heads (HHT, HTH, and THH), each with a in 8 chance of occuring, so the probability of obtaining two heads in three tosses is 3/8. The remaining values of hte random variable and their probabilities are listed in the table that follows.

39 A Binomial Distribution x p 3 /8 = 0.5 3/8 = /8 = /8 = 0.5 Using dbinom It is readily apparent that this simple approach will rapidly become quite tedious if we increase the number of times that we toss the coin. Fortunately, this "binomial distribution" is easily calculated in R. To calculate the probability of obtaining three heads in three tosses of a "fair" coin, enter the following code: > dbinom(3,size=3,prob=/) [] 0.5 Note that this code gives a result that is identical to the first row in the table above. The general syntax dbinom(x, size=, prob = ) is fairly straightforward.. The argument x is the value of the random variable, the number of heads in the current example.. The size is the "number of trials," or the number of tosses in the current example. 3. The prob is the probability of success, or the probability of obtaining "heads" in the current example. We can produce the entire table with the following code: p=dbinom(0:3,size=3,prob=/) p []

40 In this case, the input for x is the vector 0:3, which produces upon expansion 0,,, and 3. We can create the "spikes" shown in Figure 4 with the following code: n=3 p=/ x=0:3 p=dbinom(x,size=n,prob=p) plot(x,p,type="h",xlim=c(-,4),ylim=c(0,),lwd=,col="blue",ylab="p") points(x,p,pch=6,cex=,col="dark red") Figure 4. The discrete binomial distribution describing the number of heads in three tosses of a "fair" coin. These commands become increasingly effective as we increase the number of trials, in this case the number of "tosses" of the fair coin. n=0 p=/ x=0:0 p=dbinom(x,size=n,prob=p) plot(x,p,type="h",xlim=c(-,),ylim=c(0,0.5),lwd=,col="blue",ylab="p") points(x,p,pch=6,cex=,col="dark red")

41 Figure 4. The discrete binomial distribution describing the number of heads in ten tosses of a "fair" coin. Using pbinom Now, suppose that we toss a "fair" coin ten times (the distribution shown in Figure 4) and we ask the following question: "What is the probbility of obtaining 4 or fewer heads?" One approach is to realize that this probability is the sum of the individual probabilities. That is, to obtain the probability of obtaining 4 or fewer heads, we would add the probabilities of obtaining 0,,, 3, or 4 heads. sum(dbinom(0:4,size=0,prob=/)) [] Alternatively, this sounds suspiciously similar to the questions we asked in the activities The Standard Normal Distribution and The Normal Distribution. In those activities we used thepnorm command. In similar fashion, we can use the pbinom command in the current situation. pbinom(4,size=0,prob=/) [] Note again that it's always the probability to the left of and including the number of successes.

42 Let's ask another question. Suppose that we want to know the probability of obtaining between 5 and 8 heads, inclusive. We could sum the probabilities of obtaining 5, 6, 7, or 8 heads. > sum(dbinom(5:8,size=0,prob=/)) [] pbinom(8,size=0,prob=/)-pbinom(4,size=0,prob=/) [] Using the binomial distribution in Figure 4, let's look at one final question and compute the probability that X is greater than 7. One obvious way to go is to compute the sum the probabilities that X is 8, 9, or 0. sum(dbinom(8:0,size=0,prob=/)) [] A second approach uses the pbinom command, which must always sum the probabilities to the left of or equal to a given number. In this case, the probability that X is greater that or equal to 8 (8, 9, or 0) is equal to minus the probability that X is less than or equal to 7 (0,,, 3, 4, 5, 6, or 7). -pbinom(7,size=0,prob=/) []

43 oisson Distribution he Poisson distribution is the probability distribution of independent event occurrences in n interval. If λ is the mean occurrence per interval, then the probability of having x ccurrences within a given interval is: roblem f there are twelve cars crossing a bridge per minute on average, find the probability of aving seventeen or more cars crossing the bridge in a particular minute. olution he probability of having sixteen or less cars crossing the bridge in a particular minute is iven by the function ppois. > ppois(6, lambda=) # lower tail [] ence the probability of having seventeen or more cars crossing the bridge in a minute is in he upper tail of the probability density function. > ppois(6, lambda=, lower=false) # upper tail [] 0.09 nswer f there are twelve cars crossing a bridge per minute on average, the probability of having eventeen or more cars crossing the bridge in a particular minute is 0.%.

44 Poisson {stats} R Documentation Description The Poisson Distribution Density, distribution function and random generation for the Poisson distribution with parameter lambda. Usage dpois(x, lambda) rpois(n, lambda) Arguments x vector of (non-negative integer) quantiles.. n lambda number of random values to return. vector of (non-negative) means. Details The Poisson distribution has density p(x) = λ^x exp(-λ)/x! for x = 0,,,. The mean and variance are E(X) = Var(X) = λ.. The quantile is right continuous: qpois(p, lambda) is the smallest integer x such that P(X x) p. Value

45 dpois gives the (log) density, ppois gives the (log) distribution function, qpois gives the quantile function, and rpois generates random deviates. Invalid lambda will result in return value NaN, with a warning. Examples par(mfrow = c(, )) x <- seq(-0.0, 5, 0.0) plot(x, ppois(x, ), type = "s", ylab = "F(x)", main = "Poisson() CDF") plot(x, pbinom(x, 00, 0.0), type = "s", ylab = "F(x)", main = "Binomial(00, 0.0) CDF") par(mfrow = c(, )) x <- seq(-0.0, 5, 0.0) plot(x, dpois(x, ), type = "s", ylab = "F(x)", main = "Poisson() probabilities") plot(x, dbinom(x, 00, 0.0), type = "s", ylab = "F(x)", main = "Binomial(00, 0.0) probabilitis")

46 Poisson Approximation of the Binomial Distribution Binomial probabilities are difficult to calculate when n is large. Under certain conditions binomial probabilities may be approximated by Poisson probabilities. If n 0 and n p 7, the approximation is acceptable. Poisson approximation Use n p. Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

47 Poisson Approximation of the Binomial Distribution X Poisson 5. Binomial Error n p X Poisson 30. Binomial n 0, 000 p Error Business Statistics, Can. Ed. 00 John Wiley & Sons Canada, Ltd.

48 Let X be a discrete random variable with values x = 0,, and probabilities P(X = 0) = 0.5, P(X = ) = 0.50, and P(X = ) = 0.5, respectively. (a) Find E(X). (b) Find E(X^). (c) Find var(x). (d) Find the expected value and variance of g(x) = 3X +. Answer (a) sum(x*p(x) = (b) sum(x^*p(x)) =.5 (c) var(x) = sum((x E(x)^)*p(x)) = 0.5 (d) E(g(x)) = E(3x+) = 3*E(x) + = 5 Var(g(x) =var(3x+) = 3^ var(x) = 4.5 Random digit dialing. When an opinion poll calls residential telephone numbers at random, only 0% of the calls reach a live person. You watch the random digit dialing machine make 5 calls. (a) What is the probability that exactly 3 calls reach a person? (b) What is the probability that at most 3 calls reach a person? (c) What is the probability that at least 3 calls reach a person? (d) What is the probability that fewer than 3 calls reach a person? (e) What is the probability that more than 3 calls reach a person? answer Let N be the number of live persons contacted among the 5 calls observed. Then N has the binomial distribution with n = 5 and p = 0.. (a) P(N = 3) = (5C3)(0.) 3 (0.8) = dbinom(3,5,0.) (b) P(N 3) = P(N = 0) + + P( N = 3) = sum(dbinom(0:3,5,0.)) (c) P(N 3) = P(N = 3) + + P(N = 5) = sum(dbinom(3:5,5,0.)) (d) P(N < 3) = P(N = 0) + + P(N = ) = sum(dbinom(0:,5,0.)) (e) P(N > 3) = P(N = 4) + + P(N = 5) = sum(dbinom(4:5,5,0.))

49 Binomial setting? In each situation below, is it reasonable to use a binomial distribution for the random variable X? Give reasons for your answer in each case. (a) An auto manufacturer chooses one car from each hour s production for a detailed quality inspection. One variable recorded is the count X of finish defects (dimples, ripples, etc.) in the car s paint. (b) The pool of potential jurors for a murder case contains 00 persons chosen at random from the adult residents of a large city. Each person in the pool is asked whether he or she opposes the death penalty; X is the number who say Yes. (c) Joe buys a ticket in his state s Pick 3 lottery game every week; X is the number of times in a year that he wins a prize. Answer (a) No: There is no fixed number of observations. (b) A binomial distribution is reasonable here; a large city will have a population much larger than 00 (the sample size), and each randomly selected juror has the same (unknown) probability p of opposing the death penalty. (c) In a Pick 3 game, Joe s chance of winning the lottery is the same every week, so assuming that a year consists of 5 weeks (observations), this would be binomial If Y is a binomial random variable with mean 6 and standard deviation, what are the values of n and p? p=/3 n=8 np=6 np(-p) = 4 6(-p)=4 -p =4/6 p=-4/6 = /3 np=6 n(/3)=6 n=8

50 .3. PARAMETRIC FAMILIES OF DISTRIBUTIONS 8 Quality control Defects occur along a strand of yarn. Genetics Mutations occur in a genome. Traffic flow Cars arrive at an intersection. Customer service Customers arrive at a service counter. Neurobiology Neurons fire. The rate at which events occur is often called λ; the number of events that occur in the domain of study is often called X; we write X Poi(λ). Important assumptions about Poisson observations are that two events cannot occur at exactly the same location in space or time, that the occurence of an event at location l does not influence whether an event occurs at any other location l, and the rate at which events arise does not vary over the domain of study. When a Poisson experiment is observed, X will turn out to be a nonnegative integer. The associated probabilities are given by Equation.6. P[X = k λ] = λk e λ. (.6) k! One of the main themes of statistics is the quantitative way in which data help us learn about the phenomenon we are studying. Example.4 shows how this works when we want to learn about the rate λ of a Poisson distribution. Example.4 (Seedlings in a Forest) Tree populations move by dispersing their seeds. Seeds become seedlings, seedlings become saplings, and saplings become adults which eventually produce more seeds. Over time, whole populations may migrate in response to climate change. One instance occurred at the end of the Ice Age when species that had been sequestered in the south were free to move north. Another instance may be occurring today in response to global warming. One critical feature of the migration is its speed. Some of the factors determining the speed are the typical distances of long range seed dispersal, the proportion of seeds that germinate and emerge from the forest floor to become seedlings, and the proportion of seedlings that survive each year. To learn about emergence and survival, ecologists return annually to forest quadrats (square meter sites) to count seedlings that have emerged since the previous year. One such study was reported in Lavine et al. [00]. A fundamental quantity of interest is the rate λ at which seedlings emerge. Suppose that, in one quadrat, three new seedlings are observed. What does that say about λ?

51 .3. PARAMETRIC FAMILIES OF DISTRIBUTIONS 9 Different values of λ yield different values of P[X = 3 λ]. To compare different values of λ we see how well each one explains the data X = 3; i.e., we compare P[X = 3 λ] for different values of λ. For example, P[X = 3 λ = ] = 3 e ! P[X = 3 λ = ] = 3 e 0.8 3! P[X = 3 λ = 3] = 33 e ! P[X = 3 λ = 4] = 43 e ! In other words, the value λ = 3 explains the data almost four times as well as the value λ = and just a little bit better than the values λ = and λ = 4. Figure.6 shows P[X = 3 λ] plotted as a function of λ. The figure suggests that P[X = 3 λ] is maximized by λ = 3. The suggestion can be verified by differentiating Equation.6 with respect to lambda, equating to 0, and solving. The figure also shows that any value of λ from about 0.5 to about 9 explains the data not too much worse than λ = 3. P[x=3] lambda Figure.6: P[X = 3 λ] as a function of λ

52 .3. PARAMETRIC FAMILIES OF DISTRIBUTIONS 0 Figure.6 was produced by the following snippet. lam <- seq ( 0, 0, length=50 ) y <- dpois ( 3, lam ) plot ( lam, y, xlab="lambda", ylab="p[x=3]", type="l" ) Note: seq stands for sequence. seq(0,0,length=50) produces a sequence of 50 numbers evenly spaced from 0 to 0. dpois calculates probabilities for Poisson distributions the way dbinom does for Binomial distributions. plot produces a plot. In the plot(...) command above, lam goes on the x-axis, y goes on the y-axis, xlab and ylab say how the axes are labelled, and type="l" says to plot a line instead of indvidual points. Making and interpreting plots is a big part of statistics. Figure.6 is a good example. Just by looking at the figure we were able to tell which values of λ are plausible and which are not. Most of the figures in this book were produced in R.