Discrete and continuous distributions

1. Discrete and continuous distributions

2. Where does the binomial coefficient come from? Suppose I 7 blue and pink balls, each of them uniquely marked so that I can distinguish them A B C D E F G How many different samples can I draw containing the same balls but in a different order? 7! I have 7 choices for the first spot, 6 choices for the second (since I’ve picked 1 and now have only 6 to choose from), 5 choices for the third, etc. 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 G E C D B F A

3. Now if I am just counting the number of blue and pink balls, I don’t care about the order. So all possible arrangements (3!) of the pink balls look the same to me A B F D E C G A B G D E C F A B E D F C G A B D C A B G D F C E A B E D G C F A B F D G C E So instead of having 7! combinations, we have 7!/3! combinations, because where before we had 6 different possibilities of uniquely ordering different pink balls – they are equivalent

4. E F G The same goes for the blue balls, if we can’t tell them apart, we lose a factor of 4! number of ways of arranging n different things Binomial coefficient =C(n,k)= ----------------------------------------------------------------- (# of ways to arrange k things)*(# ways to arrange n-k things) n! = ----------------- k! (n-k)! Note that the binomial coefficient is symmetric – there are the same number of ways of choosing k or n-k things out of n

5. We’ve got the coefficient, what is the distribution about? • Suppose your sample of 7 is actually drawn from a very large population • (so large that it is basically unaffected by the removal of a measly 7 balls) • p = probability that ball is pink • (1-p) = probability that ball is not pink (blue) • The probability that you draw a sample with 3 pink balls and 4 blue balls in a particular order e.g. (two pink followed by 3 blues, followed by a pink followed by a blue) is prob(pink)*prob(pink)*prob(blue)*prob(blue)*prob(blue)*prob(pink)*prob(blue) = p3*(1-p)4

6. We’ve got the coefficient, what is the distribution about? • But the binomial distribution just tells us what the probability is of drawing e.g. 3 pink balls, not 3 pink balls at a particular point in the draw • The probability that you draw a sample with 3 pink balls and 4 blue balls in no particular order is = C(7,3) p3*(1-p)4 + ….

7. Probability distribution • A probability distribution lists all the possible outcomes and their probabilities • Outcomes are mutually exclusive • e.g. drawing 0, 1, 2, 3… pink balls • Outcome probabilities sum to one • e.g. when drawing 7 balls, the probability has to be one of {0,1,2,3,4,5,6,7} • Denote p(x) to mean P(X=x), that is the probability that the outcome is x

8. Binomial distribution • The binomial distribution tells us the probability of drawing k pink balls out of n • It depends on • n = the number of trials (draws) • k = the number of pink balls (successes) • p = the probability of drawing a pink ball (success)

9. the binomial distribution in R • dbinom(x, size, prob) • if blue and pink balls are equally likely > dbinom(3,7,0.5) [1] 0.2734375 >barplot(dbinom(0:7,7,0.5),names.arg=0:7)

10. what if p ≠ 0.5? • > barplot(dbinom(0:7,7,0.1),names.arg=0:7)

11. What is the mean? • mean of a binomial distribution is just n*p • in general  = E(X)= xp(x) probabilities that sum to 1 0.25 0.20 0.15 0.10 0.05 0 * + 1 * + 2 * + 3 * + 4 * + 5 * + 6 * + 7 * 0.00 m = 3.5

12. What is the variance? • variance of a binomial distribution is justn*p*(1-p) • in general s2 = E[(X-m)2]=  (x-m)2 p(x) (0.5)2 * (-0.5)2 * probabilities that sum to 1 0.25 0.20 (1.5)2 * 0.15 (-1.5)2 * 0.10 (-2.5)2 * (2.5)2 * 0.05 (-3.5)2 * (-3.5)2 * + + + + + + + 0.00

13. Which distribution has greater variance? p = 0.1 p = 0.5 var = n*p*(1-p) = 7*0.5*0.5 = 7*0.25 var = n*p*(1-p) = 7*0.1*0.9=7*0.09

14. briefly comparing an experiment to a distribution experiments = 1000 tosses = 7 for (i in 1:experiments) { x = sample(c("H","T"), tosses, replace = T) y[i] = sum(x=="H") } hist(y,breaks=-0.5:7.5) lines(0:7,dbinom(0:7,7,0.5)*1000) theoretical distribution result of 1000 trials Histogram of y 300 250 200 150 Frequency 100 50 0 0 2 4 6 y

15. cumulative distribution • aka CDF = cumulative density function • the probability that x is less than or equal to some value a

16. cumulative distribution P(X=x) P(X≤x) > barplot(dbinom(0:7,7,0.5),names.arg=0:7) > barplot(pbinom(0:7,7,0.5),names.arg=0:7)

17. cumulative distribution 1.0 1.0 0.8 0.8 0.6 0.6 cumulative distribution probability distribution 0.4 0.4 0.2 0.2 0.0 0.0 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 P(X=x) P(X≤x)

18. example: surfers on a website • Your site has a lot of visitors 45% of whom are female • You’ve created a new section on gardening • Out of the first 100 visitors, 55 are female. • What is the probability that this many or more of the visitors are female? • P(X≥55) = 1 – P(X≤54) = 1-pbinom(54,100,0.45)

19. another way to calculate cumulative probabilities • ?pbinom • P(X≤x) = pbinom(x, size, prob, lower.tail = T) • P(X>x) = pbinom(x, size, prob, lower.tail = F) > 1-pbinom(54,100,0.45) [1] 0.02839342 > pbinom(54,100,0.45,lower.tail=F) [1] 0.02839342

20. female surfers visiting a section of a website what is the area under the curve?

21. cumulative distribution > 1-pbinom(54,100,0.45) [1] 0.02839342 <3 %

22. Another discrete distribution: hypergeometric • randomly draw n elements without replacement from a set of N elements, r of which are S’s (successes) and (N-r) of which are F’s (failures) • hypergeometric random variable x is the number of S’s in the draw of n elements

23. hypergeometric example • fortune cookies • there are N = 20 fortune cookies • r = 18 have a fortune, N-r = 2 are empty • What is the probability that out of n = 5 cookies, s=5 have a fortune (that is we don’t notice that some cookies are empty) • > dhyper(5, 18, 2, 5) • [1] 0.5526316 • So there is a greater than 50% chance that we won’t notice.

24. hypergeometric and binomial • When the population N is (very) big, whether one samples with or without replacement is pretty much the same • 100 cookies, 10 of which are empty binomial hypergeometric number of full cookies out of 5

25. code aside > x = 1:5 > y1 = dhyper(1:5,90,10,5) > y2 = dbinom(1:5,5,0.9) > tmp = as.matrix(t(cbind(y1,y2))) > barplot(tmp,beside=T,names.arg=x) hypergeometric probability binomial probability

26. Poisson distribution • # of events in a given interval • e.g. number of light bulbs burning out in a building in a year • # of people arriving in a queue per minute • l = mean # of events in a given interval

27. Example: Poisson distribution • You got a box of 1,000 widgets. • The manufacturer says that the failure rate is 5 per box on average. • Your box contains 10 defective widgets. What are the odds? > ppois(9,5,lower.tail=F) [1] 0.03182806 • Less than 3%, maybe the manufacturer is not quite honest. • Or the distribution is not Poisson?

28. Poisson approximation to binomial • If n is large (e.g. > 100) and n*p is moderate (p should be small) (e.g. < 10), the Poisson is a good approximation to the binomial with l = n*p binomial Poisson 0.15 0.10 0.05 0.00 0 1 2 3 4 5 6 7 8 9 11 13 15

29. Continuous distributions • Normal distribution (aka “bell curve”) • fits many biological data well • e.g. height, weight • serves as an approximation to binomial, hypergeometric, Poisson • because of the Central Limit Theorem (more on this later) is important to inference problems

30. sampling from a normal distribution x <- rnorm(1000) h <- hist(x, plot=F) ylim <- range(0,h$density,dnorm(0)) hist(x,freq=F,ylim=ylim) curve(dnorm(x),add=T)

31. plotting on log axes • First of all, this is what a log function looks like > x = 1:1000 > y = log(x) > plot(x,y) y = log(x) is equivalent to x = exp(y) = ey

32. plotting the function y = e-x • > x = 1:20 • > y = exp(-x) • > plot(x,y) hard to tell what’s going on here, all the values are so close to 0

33. changing the axes just y on a log scale both x and y on a log scale > plot(x,y,log="y") > plot(x,y,log="xy")

34. from PS: CO2 levels over last ~ 50 years

35. CO2 levels over last ~ 400,000 years