# FND-STA-Probability_distribution.R # # Purpose: A Bioinformatics Course: # R code accompanying the FND-STA-Probability_distribution unit. # # Version: 1.0 # # Date: 2017 10 10 # Author: Boris Steipe (boris.steipe@utoronto.ca) # # Versions: # 1.0 First code live version # # TODO: # Add tasks # # == DO NOT SIMPLY source() THIS FILE! ======================================= # # If there are portions you don't understand, use R's help system, Google for an # answer, or ask your instructor. Don't continue if you don't understand what's # going on. That's not how it works ... # # ============================================================================== #TOC> ========================================================================== #TOC> #TOC> Section Title Line #TOC> ----------------------------------------------------------------------- #TOC> 1 Introduction 54 #TOC> 2 Three fundamental distributions 117 #TOC> 2.1 The Poisson Distribution 120 #TOC> 2.2 The uniform distribution 173 #TOC> 2.3 The Normal Distribution 193 #TOC> 3 quantile-quantile comparison 234 #TOC> 3.1 qqnorm() 244 #TOC> 3.2 qqplot() 304 #TOC> 4 Quantifying the difference 321 #TOC> 4.1 Chi2 test for discrete distributions 355 #TOC> 4.2 Kullback-Leibler divergence 446 #TOC> 4.2.1 An example from tossing dice 457 #TOC> 4.2.2 An example from lognormal distributions 579 #TOC> 4.3 Kolmogorov-Smirnov test for continuous distributions 620 #TOC> #TOC> ========================================================================== # = 1 Introduction ======================================================== # The space of possible outcomes of events is called a probability distribution # and the properties of probability distributions are crucial to our work. Many # distributions like the (discrete) Poisson, or the (continuous) Normal # distribution have been characterized in great detail, their behaviour is well # understood, and they are useful for countless applications - but we also need # to able to work with ad hoc distributions that have never been seen before. # Let's get a few facts about probability distributions out of the way: # The "support" of a probability distribution is the range of outcomes that have # a non-zero probability. The "domain" of a probability distribution is the # range of probabilities that the distribution can take over its support. Think # of this as the ranges on the x- and y-axis respectively. Thus the distribution # can be written as p = f(x). # The integral over a probability distribution is always 1. This means: the # distribution reflects the situation that an event does occur, any event, but # there is not "no event". # R's inbuilt probability functions always come in four flavours: # d... for "density": this is the probability density function (p.d.f.), # the value of f(x) at x. # p... for "probability": this is the cumulative distribution function # (c.d.f.). It is 0 at the left edge of the support, and 1 at # the right edge. # q... for "quantile": The quantile function returns the x value at which p... # takes a requested value. # r... for "random": produces random numbers that are distributed according # to the p.d.f. # To illustrate with the "Normal Distribution" (Gaussian distribution): # 1000 normally distributed values with default parameters: mean 0, sd 1. r <- rnorm(1000) # pastel green: histogram of 1000 random samples hist(r, freq = FALSE, breaks = 30, xlim = c(-4, 4), ylim = c(0, 1), main = "Normal Distribution", xlab = "x", ylab = "f(x)", col = "#E6FFF6") # 100 equally spaced point along x x <- seq(-4, 4, length.out = 100) # black: c. d. f. along x. Note that this asymptotically reaches 1 points(x, pnorm(x), type = "l") # dark red: p. d. f. along x points(x, dnorm(x), type = "l", lwd = 2, col="firebrick") # purple: 1% and 99% quantiles abline(v = qnorm(c(0.01, 0.99)), lwd = 0.5, col = "#CCAAFF") # Study this plot well and familiarize yourself with the terms. # = 2 Three fundamental distributions ===================================== # == 2.1 The Poisson Distribution ========================================== # The Poisson distribution is a discrete probability distribution that # characterizes how many events we expect among a given number of observations # if the mean probability of an event is known. Assume we know that there are # 256 transcription factors among the 6091 protein-coding genes of yeast, then # the probability of picking a transcription factor at random from all ORFs is # 256/6091 ~= 4.2%. How many do we expect if we look e.g. at 250 differentially # expressed genes? This means the mean number of transcription factors we would # expect in that sample of differentially expressed genes is (250 * 256)/6091. dpois(0, (250 * 256) / 6091) # Probability of seeing no TFs dpois(1, (250 * 256) / 6091) # Probability of seing one ... dpois(2, (250 * 256) / 6091) # Probability of seing two ... dpois(3:10, (250 * 256) / 6091) # Probability of seing from three to ten ... sum(dpois(0:4, (250 * 256) / 6091)) # Probability of seeing four or less ... # Lets plot this N <- 25 x <- dpois(0:N, (250 * 256) / 6091) names(x) <- as.character(0:N) midPoints <- barplot(x, col = "#E6FFF6", axes = TRUE, ylim = c(0, 0.15), xlab = "# of TF in set", ylab = "p") # Confirm that our understanding of dpois() is correct, by simulating actual # trials: N <- 1000 genes <- numeric(6091) # All genes are 0 genes[1:256] <- 1 # TFs are 1 x <- numeric(N) # initialize vector set.seed(112358) for (i in 1:N) { x[i] <- sum(sample(genes, 250)) # sum of TFs in our sample in this trial } (t <- table(x)/N) # Add these values to the plot y <- numeric(26) # initialize vector with 26 slots y[as.numeric(names(t)) + 1] <- t # put the tabled values there (index + 1) points(midPoints, y, pch = 21, cex = 0.7, bg = "firebrick") legend("topright", legend = c("poisson distribution", "samples"), pch = c(22, 21), pt.bg = c("#E6FFF6", "firebrick"), bty = "n") # == 2.2 The uniform distribution ========================================== # The uniform distribution has the same probability over its entire support. R's # runif() function takes the desired number, the min and the max as arguments. # Let's plot a histogram of a million values between -1 and 1 to demonstrate. hist(runif(1e6, -1, 1), breaks = 20, col = "#FFE6F6") abline(h = 1e6/20, col="firebrick") # One of the important uses of the uniform distribution is to write conditional # expressions that are TRUE with a given probability. For example, to get TRUE # values with a probability of 1/3, we pick a random number between 0 and 1, and # ask whether the number is smaller than 1/3. Example: runif(10) < 1/3 #confirm with a million trials sum(runif(1e6) < 1/3)/1e6 # should be close to 0.33333... # == 2.3 The Normal Distribution =========================================== # The king of probability distributions. Why? That's because of the Central # Limit Theorem (CLT) that essentially says that a process that is subject to # small fluctuations will tend to normally distributed outcomes ... and in # biology literally everything is subject to small fluctuations. # Lets simulate: Lets create a vector of 12345 numbers, choose samples of size # 77 from that vector and calculate the mean. We do that 9999 times. I am just # using odd numbers so it is clear in the code which number represents what. x <- runif(12345) v <- numeric(9999) for (i in 1:length(v)) { v[i] <- mean(sample(x, 77)) } hist(v, breaks = 20, col = "#F8DDFF") # Let's try this again, this time sampling from an exponential distribution ... x <- rexp(12345) v <- numeric(9999) for (i in 1:length(v)) { v[i] <- mean(sample(x, 77)) } hist(v, breaks = 20, col = "#F8DDFF") # ... or from a t-distribution ... x <- rt(12345, df = 3) v <- numeric(9999) for (i in 1:length(v)) { v[i] <- mean(sample(x, 77)) } hist(v, breaks = 20, col = "#F8DDFF") # The outcomes all give normal distributions, regardless what the details of our # original distribution were! # = 3 quantile-quantile comparison ======================================== # Once we have observed a distribution of outcomes, the next task is often to # quantify whether the values correspond to a known distribution. This can be # done with quantile-quantile plots that plot the quantile values of one # distribution against those of the other. Identically distributed quantiles lie # on a straight line. # == 3.1 qqnorm() ========================================================== # The functions qqnorm() and qqline() perform this # comparison with the normal distribution. set.seed(1112358) x <- rnorm(100, mean=0, sd=1) # 100 normally distributed balues qqnorm(x) qqline(x, col = "seagreen") # Our variables in x appear normally distribute - which is not surprising since # we produced them with rnorm(). What about the kind of sampling we did above? # Let's test whether samples from an exponential distribution are normally # distributed (previously we just visually inspected the histogram.) # Create a vector of sample means from the exponential distribution; use # only a few samples for the mean set.seed(112358) x <- rexp(12345) v <- numeric(999) for (i in 1:length(v)) { v[i] <- mean(sample(x, 12)) } qqnorm(v) qqline(v, col = "turquoise") # normal #try with more samples for (i in 1:length(v)) { v[i] <- mean(sample(x, 77)) } qqnorm(v) qqline(v, col = "steelblue") # normal #try with many samples for (i in 1:length(v)) { v[i] <- mean(sample(x, 374)) } qqnorm(v) qqline(v, col = "plum") # normal # Exactly as the CLT predicts, the more often we sample - we tried # 12, 77, and 374 samples - the more "normal" the distribution looks. # What does a distribution look like that is NOT normal? # Let's try simulating an "Extreme value distribution". This type of # distribution appears if we choose the max() of a sample set.seed(112358) rEVD <- numeric(9999) for (i in seq_along(rEVD)) { rEVD[i] <- max(rnorm(100)) } hist(rEVD, breaks = 20, col = "orchid") # Note the long tail on the right! qqnorm(rEVD) qqline(rEVD, col = "orchid") # normal # Definitely not "normal"! # == 3.2 qqplot() ========================================================== # qqplot() works like qqnorm(), except that we compare two arbitrary # distribtutions, rather than one distribution against the normal distribution # as we did before. Sample against sample. x <- seq(0, 4, length.out = 100) dl <- dlnorm(x) # log-normal distribution dg <- dgamma(x, shape=1.5) # gamma distribution plot(dl, type="l") points(dg, type="l", col = "maroon") qqplot(dl, dg) # clearly not equal # = 4 Quantifying the difference ========================================== # Quantile-quantile plots give us a visual estimate, but how do we quantify the # difference between distributions? Let's compare two types of extreme-value # distributions, a lognormal distribution with the same distribution that is # slightly shifted, and three gamma distributions with three different shape # parameters. Let's define and visualize the distributions first, to see what # we are comparing. x <- seq(0, 4, length.out = 100) set.seed(112358) dl1 <- dlnorm(x) # log-normal distribution dl2 <- dlnorm(x - 0.25) # log-normal distribution, shifted right (a bit) dg1.2 <- dgamma(x, shape=1.2) # three gamma distributions with... dg1.5 <- dgamma(x, shape=1.5) # ...wider, and wider... dg1.9 <- dgamma(x, shape=1.9) # ...peak myCols <- c("black", "grey", "maroon", "turquoise", "steelblue") plot(dl1, type="l", lwd=2) # visualize the distributions points(dl2, type="l", col = myCols[2]) points(dg1.2, type="l", col = myCols[3]) points(dg1.5, type="l", col = myCols[4]) points(dg1.9, type="l", col = myCols[5]) legend("topright", legend = c("dl1", "dl2", "dg1.2", "dg1.5", "dg1.9"), lty = 1, lwd = c(2, 1, 1, 1, 1), col = myCols, bty = "n") # == 4.1 Chi2 test for discrete distributions ============================== # The chi2 test can be used to compare discrete distributions - even though # it is not ideal for this purpose. # (https://stats.stackexchange.com/questions/113692/test-identicality-of-discrete-distributions) # # Let's draw 100 random variables according to our target distributions N <- 100 set.seed(112358) rL1 <- rlnorm(N) # log-normal distribution rL2 <- rlnorm(N, meanlog = 0.25) # log-normal distribution, shifted right rG1.2 <- rgamma(N, shape=1.2) # three gamma distributions with... rG1.5 <- rgamma(N, shape=1.5) # ...wider, and wider... rG1.9 <- rgamma(N, shape=1.9) # ...peak maxX <- max(c(rL1, rL2, rG1.2, rG1.5, rG1.9)) myBreaks <- seq(0, 5, length.out = 10) # 9 intervals from 0 to 5... myBreaks <- c(myBreaks, maxX) # ... and one that contains the outliers # It's easy to plot a histogram of one set of random deviates... hist(rG1.5, breaks = myBreaks, col = myCols[4]) # ... but basic R has no inbuilt function to stack histogram bars side-by-side. # We use the multhist() function in the plotrix package: check out the # package information - plotrix has _many_ useful utilities to enhance # plots or produce informative visualizations. if (! require(plotrix, quietly=TRUE)) { install.packages("plotrix") library(plotrix) } # Package information: # library(help = plotrix) # basic information # browseVignettes("plotrix") # available vignettes # data(package = "plotrix") # available datasets h <- multhist(list(rL1, rL2, rG1.2, rG1.5, rG1.9 ), breaks = myBreaks, col = myCols) legend("topright", legend = c("rL1", "rL2", "rG1.2", "rG1.5", "rG1.9"), pch=15, col = myCols, bty="n") # We have assigned the output of multihist to the variable h, which now # contains the densities, midpoints, breaks etc. of the histogram ... # as well as the matrix of counts: h[[2]] # But using hist() or multhist() for binning is a bit of a hack - even though # it works, of course. The "real" R function to bin data is cut(). cut() # sorts the values of a vector into bins, and returns the bin-labels as # integers. x <- cut(rL1, breaks = myBreaks, labels = FALSE) table(x) countsL1 <- table(cut(rL1 , breaks = myBreaks, labels = FALSE)) countsL2 <- table(cut(rL2 , breaks = myBreaks, labels = FALSE)) countsG1.2 <- table(cut(rG1.2, breaks = myBreaks, labels = FALSE)) countsG1.5 <- table(cut(rG1.5, breaks = myBreaks, labels = FALSE)) countsG1.9 <- table(cut(rG1.9, breaks = myBreaks, labels = FALSE)) chisq.test(countsL1, countsL2) # Note that this use of the chi2 test does not have very much power, since it # does not assume the values for the bins are ordered, but takes them to # be independent. Also, chisq.test() complains about the Chi-squared # approximation to be possibly incorrect because some of the counts are small. # In this case, rather than rely on the in-built chi-square table, we can do # an explicit simulation to estimate the p-value for the null hypothesis. # chisq.test(countsL1, countsL2, simulate.p.value = TRUE, B = 10000) # Note that the probability that the samples came from the same distribution is # quite high. We don't seem to be able to distinguish l1 and l2 with the chi2 # test. # Let's calculate this for the other distribution too: chisq.test(countsL1, countsG1.2, simulate.p.value = TRUE, B = 10000) chisq.test(countsL1, countsG1.5, simulate.p.value = TRUE, B = 10000) chisq.test(countsL1, countsG1.9, simulate.p.value = TRUE, B = 10000) # As a result, we would conclude that none of these distributions are # significantly different. # == 4.2 Kullback-Leibler divergence ======================================= # For discrete probability distributions, there is a much better statistic, the # Kullback-Leibler divergence (or relative entropy). It is based in information # theory, and evaluates how different the matched pairs of outcome categories # are. Its inputs are the probability mass functions (p.m.f.) of the two # functions to be compared. A probability mass function is the probability of # every outcome the process can have. Kullback-Leibler divergence therefore can # be applied to discrete distributions. But we need to talk a bit about # converting counts to p.m.f.'s. # === 4.2.1 An example from tossing dice # The p.m.f of an honest die is (1:1/6, 2:1/6, 3:1/6, 4:1/6, 5:1/6, 6:1/6). But # there is an issue when we convert sampled counts to frequencies, and estimate # probabilities with these frequencies. The problem is: by chance we might not # have observed a particular outcome! Consider this example: set.seed(47) N <- 20 (counts <- table(sample(1:6, N, replace = TRUE))) # We have not observed a "2"! # # Now, if we convert the counts to frequencies, and assume these to be # probabilities, we get an observed p.m.f: pmf <- numeric(6) for (i in seq_along(counts)) { outcome <- as.integer(names(counts)[i]) pmf[outcome] <- counts[i]/N } names(pmf) <- 1:6 pmf # This means we assign a probability of 0 to the outcome "2" - because we # haven't observed it. But how do we compare it then? What does it mean when we # should have 1/6 of "2"s but we have none whatsoever? Taken at face value it # means we are comparing apples to oranges - essentially a five-faced die with a # six-faced die - and therefore divergence is infinitely large. Obviously, that # was just a problem caused by our limited sampling, but the question remains: # how do we account for missing data? There are several solutions, for example, # for ordered data one could substitute the average values of the two bracketing # outcomes. But a simple and quite robust solution is to add "pseudocounts". # This is called adding a Laplace prior, or a Jeffreys prior: in our case, # simply add 0.5 to every value that the two functions don't share. # pmf of an honest die pmfHD <- rep(1/6, 6) names(pmfHD) <- 1:6 pmfHD # pmf, estimated from our sampled counts with and without pseudocounts pmfPC <- numeric(6) + 0.5 for (i in seq_along(counts)) { outcome <- as.integer(names(counts)[i]) pmfPC[outcome] <- counts[i] + pmfPC[outcome] } pmfPC <- pmfPC / sum(pmfPC) names(pmfPC) <- 1:6 pmf # before pmfPC # after # The definition of a function that takes samples and returns a pmf # with pseudocounts is straightforward, but requires a bit of juggling # the observed values into the right slots. Here's what this can look like: pmfPC <- function(cts, nam) { # Convert counts to a p.m.f, adding a pseudocount of 0.5 to all categories # of outcomes (Jeffreys prior). # Parameters: # cts num raw counts # nam names of the categories, converted to char # Value a probability mass function nam <- as.character(nam) if (!all(names(cts) %in% nam)) { stop("PANIC: names in \"cts\" do not match \"nam\"!") } pmf <- numeric(length(nam)) names(pmf) <- nam pmf[names(cts)] <- cts[names(cts)] pmf <- pmf + 0.5 # add pseudocounts return(pmf/sum(pmf)) } # The definition of the Kullback-Leibler divergence itself is quite simple # actually: for two distributions, p and q it is sum(p * log( p / q)) KLdiv <- function(p, q) { # p and q are two pmfs of discrete probability distributions # with the same outcomes, which are nowhere 0. # Value: Kullback-Leibler divergence sum(p * log( p / q))). if (length(p) != length(q)) { stop("PANIC: input vector lengths differ!") } if (any(c((p == 0), (q == 0)))) { stop("PANIC: 0's found in input vectors!") } return(sum(p * log( p / q ))) } # Now we can calculate KL-divergences for our six-faced die example: KLdiv(rep(1/6, 6), pmfPC(counts, 1:6)) # p.m.f. of an honest die vs. our # actual counts # Is that a lot? Let's look at the distribution of KL-divergences in our # six-sided die scenario: p.m.f. of 20 samples each versus the theoretical # p.m.f. N <- 1000 nSample <- 20 divs <- numeric(N) for (i in 1:N) { x <- table(sample(1:6, nSample, replace = TRUE)) divs[i] <- KLdiv(rep(1/6, 6), pmfPC(x, 1:6)) } hist(divs, col = "whitesmoke", xlab = "Kullback-Leibler divergence", main = sprintf("%d samples of honest die", nSample)) abline(v = KLdiv(rep(1/6, 6), pmfPC(counts, 1:6)), col="firebrick") # We see that our example sample (KL-divergence marked with the red line) is # somewhat but not drastically atypical. # === 4.2.2 An example from lognormal distributions # We had compared a set of lognormal and gamma distributions above, now we # can use KL-divergence to quantify their similarity: pmfL1 <- pmfPC(countsL1, nam = 1:10) pmfL2 <- pmfPC(countsL2, nam = 1:10) KLdiv(pmfL1, pmfL2) # 0.1087 # To evaluate what this number means, we can run a simple simulation: we create # random samples according to the rL1 distribution, calculate the Kullback # Leibler divergence with countsL1, and compare the distribution we get with the # value we observed as the difference with discL2. Essentially, this tells us # the probability that countsL2 is actually a sample from the L1 function. Here we # go: N <- 1000 divs <- numeric(N) set.seed(31416) for (i in 1:N) { x <- rlnorm(100) # get random samples from our reference distributions y <- table(cut(x, breaks = myBreaks, labels = FALSE)) # tabulate counts q <- pmfPC(y, nam = 1:10) # convert to p.m.f. with pseudocounts divs[i] <- KLdiv(pmfL1, q) # calculate Kullback-Leibler divergence } hist(divs, col = "thistle", xlab = "Kullback-Leibler divergence", main = sprintf("Samples from shifted lnorm()")) abline(v = KLdiv(pmfL1, pmfL2), col="firebrick") # How many KL-divergences were less than the difference we observed? sum(divs < KLdiv(pmfL1, pmfL2)) #933 # Therefore the probability that the samples came from the same distribution # is only 100 * (N - 933) / N (%) ... 6.7%. You see that this gives a much more # powerful statistical approach than the chi2 test we used above. # == 4.3 Kolmogorov-Smirnov test for continuous distributions ============== # The Kolmogorov-Smirnov (KS) test is meant for continuous distributions, i.e. # the probability it calculates assumes that the function values are all # different. In the case of random samples, that is a reasonable assumption. KS # is a two-sample goodness of fit test, i.e. it has a null-hypothesis that the # samples were taken from the same (unknown) distribution, and then asks whether # this is likely, given the actual values. It is sensitive both to differences # in location (position of the mean), and differences in shape. R has the # ks.test() function to calculate it. ks.test(dl1, dl2) # vs. shifted log-normal. ks.test(dl1, dg1.2) # vs. the three gamma distributions ks.test(dl1, dg1.5) ks.test(dl1, dg1.9) ks.test(dg1.5, dg1.9) # the two last gammas against each other # (The warnings about the presence of ties comes from the 0s in our function # values). The p-value that these distributions are samples of the same # probability distribution gets progressively smaller. # [END]