In Chapter 17, I delve into probability in a semiformal way, and introduce distributions

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1 IN THIS CHAPTER»» Understanding the beta version»» Pursuing Poisson»» Grappling with gamma»» Speaking eponentially Appendi A More About Probability In Chapter 7, I delve into probability in a semiformal way, and introduce distributions of random variables. The binomial distribution is the starting point. In this chapter, I eamine additional distributions. One of the symbols on the pages of this book (and other books in the For Dummies series) lets you know that technical stuff follows. It might have been a good idea to hang that symbol above this chapter s title. So here s a small note of caution: Some mathematics follows. I put the math in to help you understand what you re doing when you work with the arguments of the R functions in this chapter. Are these functions on the esoteric side? Well... yes. Will you ever have occasion to use them? Well... you just might. Discovering Beta This distribution connects with the binomial distribution, which I discuss in Chapter 7. The beta distribution (not to be confused with beta, the probability of a Type 2 error) is a sort of chameleon in the world of distributions. It takes on a wide variety of appearances, depending on the circumstances. I won t give you all the mathematics behind the beta distribution, because the full treatment involves calculus. APPENDIX A More About Probability INDD Trim size: in 9.25 in February 4, 207 3:4 AM

2 The connection with the binomial is this: In the binomial, the random variable is the number of successes in N trials with p as the probability of a success. N and p are constants. In the beta distribution, the random variable is the probability of a success, with N and the number of successes as constants. Why is this useful? In the real world, you usually don t know the value of p, and you re trying to find it. Typically, you conduct a study, find the number of successes in a set of trials, and then you have to estimate p. Beta shows you the likelihood of possible values of p for the number of trials and successes in your study. Some of the math is complicated, but I can at least show you the rule that generates the density function for N trials with r successes, when N and r are whole numbers: f, r N N! r! N r! r N r The vertical bar in the parentheses on the left means given that. So this density function is for specific values of N and r. Calculus enters the picture when N and r aren t whole numbers. (Density function? Given that? See Chapter 7.) To give you an idea of what this function looks like, I used R s dbeta() function to generate and graph the density function for four successes in ten trials. The dbeta() function can take five arguments, of which only three concern us here: dbeta(, shape, shape2) The first argument is in the equation. The documentation defines shape and shape2 as non-negative parameters of the beta distribution. In English, that means the number of successes (shape) and the number of failures (shape2) in the eample I m working through. For this eample, then, it s dbeta(, 4, 6) once I specify what is. Mathematicians, in fact, write the equation for the density function in terms of shape and shape2, but they refer to shape as α and shape2 as β, which makes the density function! f!! 2 Statistical Analysis with R For Dummies INDD 2 Trim size: in 9.25 in February 4, 207 3:4 AM

3 To graph this member of the beta distribution family I first create a vector of -values.values <- seq(0,.95,.0) The ggplot code is ggplot(null,aes(=.values,y=dbeta(.values,4,6)))+ geom_line() Figure A- shows the graph. Each value on the -ais, remember is a possible value for the probability of a success. The curve shows probability density. As I point out in Chapters 8 and 7, probability density is what makes the area under the curve correspond to probability. A glance at the graph shows that the curve s maimum point is at.4, which is what you would epect for four successes in ten trials. FIGURE A-: The Beta Density function for four successes in ten trials. Suppose I toss a die (one of a pair of dice), and I define a success as any toss that results in a 3. I assume I m tossing a fair die, so I assume that p pr 3 / 6. Suppose I toss a die ten times and get four 3s. How good does that fair-die assumption look? The graph in Figure A- gives you a hint: The area to the left of.6667 (the decimal equivalent of /6) is a pretty small proportion of the total area, meaning that the probability that p is /6 or less is pretty low. APPENDIX A More About Probability INDD 3 Trim size: in 9.25 in February 4, 207 3:4 AM

4 Now, if you have to go to all the trouble of creating a graph, and then guesstimate proportions of area to come with an answer like pretty low, you re doing a whole lot of work for very little return. Fortunately, R has a better way: the pbeta() function. Supply p, the number of successes, the number of failures, and lower.tail = TRUE, and here s what you get: > pbeta(/6,4,6,lower.tail=true) [] If you ve obtained four successes in ten tosses and would like to know the 95 per cent confidence limits of p, you d use qbeta(): > qbeta(c(.025,.975),4,6) [] The first argument is the vector of probabilities for the lower and upper limits, the second argument is the number of successes and the third is the number of failures. If you had to generate, say, three random numbers from this beta distribution (although I don t know why you would): > rbeta(3,4,6) [] Poisson If you have the kind of process that produces a binomial distribution, and you have an etremely large number of trials and a very small number of successes, the Poisson distribution approimates the binomial. The equation of the Poisson is pr( ) e! In the numerator, μ is the mean number of successes in the trials, and e is (and infinitely more decimal places), a constant near and dear to the hearts of mathematicians. (See Chapter 6.) Here s an eample. FarKlempt Robotics, Inc., produces a universal joint for its robots elbows. The production process is under strict computer control, so that the probability a joint is defective is.00. What is the probability that, in a sample 4 Statistical Analysis with R For Dummies INDD 4 Trim size: in 9.25 in February 4, 207 3:4 AM

5 of,000, one joint is defective? What s the probability that two are defective? Three? Named after the 9th-century mathematician Siméon-Denis Poisson, this distribution is computationally easier than the binomial or at least it was back when mathematicians had no computational aids. With R, you can easily use dbinom() to do the binomial calculations. Then why bother to bring up this distribution at all? Because it s an easy segué from the binomial to this important distribution, which I discuss in greater detail in Chapter 8. First, I apply the Poisson distribution to the FarKlempt eample. If N 000, the mean is.00 and N (See Chapter 7 for an eplanation of N.) Now for the Poisson. The probability that one joint in a sample of,000 is defective is: pr() e! !. 368 For two defective joints in,000, it s pr( 2) e! ! And for three defective joints in,000: pr( 3) e! ! As you read this section, it may seem odd to refer to a defective item as a success it s just a way of labeling a specific event. R s dpois() function does all this in one fell swoop: dpois(c(,2,3),) Or if you want to look really cool: dpois(:3,) In either case, the first argument is the -values and the second argument is μ. APPENDIX A More About Probability INDD 5 Trim size: in 9.25 in February 4, 207 3:4 AM

6 Applying either format yields > dpois(:3,) [] In the R documentation for dpois() and the other Poisson functions, the second argument is called lambda, the Greek letter (λ) that many mathematicians use for that component of the Poisson distribution. So how close is the Poisson approimation to the binomial for this eample? > dbinom(:3,000,.00) [] Pretty close! Although the Poisson s usefulness as an approimation is outdated, it has taken on a life of its own. Phenomena as widely disparate as reaction times in psychology eperiments, degeneration of radioactive substances, and scores in professional hockey games seem to fit Poisson distributions. This is why business analysts and scientific researchers like to base models on this distribution. ( Base models on? What does that mean? I tell you all about modeling in Chapter 8.) Working with Gamma You might recall from Chapter 8 that the number of ways of arranging N objects in a sequence is N! ( N factorial ). You might also recall that N! N N N 2 2. Obviously, the factorial only works for whole numbers, right? The gamma function Not so fast. Mathematicians (some pretty famous ones) have etended the factorial concept to include non-integers and even negative numbers (which gets very hairy). This etension is called the gamma function. When gamma s argument is a positive whole number let s call it N the result is N!. Otherwise, gamma returns the result of a calculus-based equation. Rather than go into all the calculus, I ll just give you an eample: 4! 24 and 5! 20. So the factorial of 4.3 (whatever that would mean) should be somewhere 6 Statistical Analysis with R For Dummies INDD 6 Trim size: in 9.25 in February 4, 207 3:4 AM

7 between 24 and 20. Because of the N I just mentioned, you d find this factorial by letting gamma loose on 5.3 (rather than 4.3). And gamma(5.3) = You can verify this in R: > gamma(5.3) [] The gamma distribution All the discussion in the preceding section is mostly within the realm of theoretical mathematics. Things get more interesting (and more useful) when you tie gamma to a probability distribution. This marriage is called the gamma distribution. The gamma distribution is related to the Poisson distribution in the same way the negative binomial distribution is related to the binomial. The negative binomial tells you the number of trials until a specified number of successes in a binomial distribution. The gamma distribution tells you how many samples you go through to find a specified number of successes in a Poisson distribution. Each sample can be a set of objects (as in the FarKlempt Robotics universal joint eample), a physical area, or a time interval. The probability density function for the gamma distribution is f e! Again, this works when α is a whole number. If it s not, you guessed it calculus. (By the way, when this function has only whole-number values of α, it s called the Erlang distribution, just in case that ever comes up in conversation.) The letter e, once again, is the constant I mention earlier. Don t worry about the eotic-looking math. As long as you understand what each symbol means, you re in business. R does the heavy lifting for you. So here s what the symbols mean. For the FarKlempt Robotics eample, α is the number of successes and β corresponds to μ in the Poisson distribution. The variable tracks the number of samples. Thus, if you re interested in the probability density associated with finding the second success in the third sample, is 3, α is 2, and β is if the average number of successes per sample (of,000) is. (Where does come from, again? That s,000 universal joints per sample multiplied by.00, the probability of producing a defective one.) APPENDIX A More About Probability INDD 7 Trim size: in 9.25 in February 4, 207 3:4 AM

8 To determine probability, you have to work with area under the density function. This brings me to the R function for the gamma distribution s density function. If you re thinking it s called dgamma(), you re right. Like the functions for the beta distribution, this one takes five arguments, of which I ll just consider the first three. To help you visualize what s going on, I use dgamma() to plot the density function. Specifically, I show the density function for finding the second success in a range of samples with an average of success per sample. I begin with a vector for the range of samples:.values <- seq(,5) To calculate the densities for these values dgamma(.values,2,) The first argument is the values I m finding the densities for, the second is the number of successes I m interested in, and the third is the average number of successes per sample. That s the function I put inside ggplot for the aesthetic mapping to y: ggplot(null, aes(=.values,y=dgamma(.values,2,))) + geom_line()+ scale continuous(breaks=.values) The first argument to ggplot, NULL, indicates that I m not using a data frame. The third statement adds the vector values to the -ais. The result is Figure A-2. In real life you work with probabilities, not densities. So the probability of finding the second success by the third sample is the area underneath the density function to the left of 3. Figure A-2 shows that s quite a bit of area. Eactly how much? That s up to pgamma(): > pgamma(3,2,) [] The first argument is the number of samples, the second is the number of successes, and the third is the average number of successes per sample. 8 Statistical Analysis with R For Dummies INDD 8 Trim size: in 9.25 in February 4, 207 3:4 AM

9 FIGURE A-2: The density function for gamma, with number of successes = 2 and average of successes per sample =. The result indicates about an 80 percent chance of finding the second defective joint (that s a success, remember) by the third sample, with an average of one defective joint per sample. Eponential In the gamma distribution, if you have the eponential distribution. This gives the probability that it takes a specified number of samples to get to the first success. What does the density function look like? Ecuse me... I m about to go mathematical on you for a moment. Here, once again, is the density function for gamma: f e! If, it looks like this: f e R provides a set of functions for dealing with the eponential distribution (dep(), pep(), qep(), and rep()). I d use dep(), the.values vector and ggplot() to visualize the density function for you, but it looks quite a bit (although not eactly) like Figure A-2. I leave that as an eercise for you. APPENDIX A More About Probability INDD 9 Trim size: in 9.25 in February 4, 207 3:4 AM

10 If you want to do that eercise, work with dep(.values,) Continuing with the universal joint eample, I use pep() to calculate the probability of finding the first defective joint by the third sample. This function can take four arguments, of which I discuss just the first three: pep(q, rate =, lower.tail = TRUE) In the contet of our eample, the first argument is the number of samples. The second argument corresponds to in the density function. This means that if the average number of successes per sample is two (instead of one, as in this eample), the rate is 0.5. The third argument is the default that specifies returning the area to the left of under the eponential distribution s density function. Because I m working with the default I can omit lower.tail, and the probability is > pep(3,) [] The default for the second argument also just happens to be the value in this eample, so I could have done it this way: > pep(3) [] but that s only for this eample. So that s a 95 percent chance of finding the first defective joint by the third sample, if the average number of successes per sample is one. 0 Statistical Analysis with R For Dummies INDD 0 Trim size: in 9.25 in February 4, 207 3:4 AM