grb <- read.table ("http://astrostatistics.psu.edu/datasets/GRB_afterglow.dat", header=T, skip=1) flux <- grb[,2] hist(flux)The histogram suggests that the univariate distribution has roughly the shape of an exponential distribution (we'll speak more about what this means later). Let us replot these data in a particular (and particularly common) manner besides the histogram that is also suggestive of an exponential distribution.

As a first step, let us calculate something akin to the (x,y) coordinates of the empirical distribution function -- the function that has a jump of size 1/n at every one of the sorted data points.

n <- length(flux) xx <- sort(flux) yy <- (1:n)/nWe could now obtain the empirical cdf by connecting the (xx,yy) points using a stairstep pattern. However, we'll look at these points slightly differently.

The exponential distribution has a distribution function given by F(x) = 1-exp(-x/mu) for positive x, where mu>0 is a scalar parameter equal to the mean of the distribution. This implies among other things that log(1-F(x)) = -x/mu is a linear function of x in which the slope is the negative reciprocal of the mean. Let us then look for the characteristic linear pattern if we plot log(1-F(x)) against x using the empirical distribution function for F:

plot(xx, log(1-yy+1/n), xlab="flux", ylab="log(1-F(flux))")You may recall from the EDA and regression tutorial that this plot looks like the plot of time vs. flux that we produced as part of a regression analysis. This is only a coincidence;

The plot certainly looks linear, so let us proceed on the assumption that the flux data are a sample from an exponential distribution with unknown parameter mu.

The overriding question of this section is this: How shall we estimate mu?

As mentioned above, mu is equal to the mean of this population. For a quick refresher on some probability theory, let us recall why this is so: The first step in going from the distribution function F(x) = 1 - exp(-x/mu) to the mean, or expectation, is to obtain the density function by differentiating: f(x) = exp(-x/mu)/mu. Notice that we typically use F(x) to denote the distribution function and f(x) to denote the density function. Next, we integrate x*f(x) over the interval 0 to infinity, which gives the mean, mu.

Since mu is the population mean, it is intuitively appealing to simply estimate mu using the sample mean. This method, in which we match the population moments to the sample moments and then solve for the parameter estimators, is called the method of moments. Though it is a well-known procedure, we focus instead on a much more widely used method (for good reason) called maximum likelihood estimation.

The first step in maximum likelihood estimation is to write down the likelihood function, which is nothing but the joint density of the dataset viewed as a function of the parameters. Next, we typically take the log, giving what is commonly called the loglikelihood function. Remember that all logs are natural logs unless specified otherwise.

The loglikelihood function in this case is (with apologies for the awkward notation)

l(mu) = -n log(mu) - x

A bit of calculus reveals that l(mu) is therefore maximized at the sample mean. Thus, the sample mean is not only the method of moments estimator in this case but the maximum likelihood estimate as well.

In practice, however, it is sometimes the case that the linear-looking plot produced earlier is used to estimate mu. As we remarked, the negative reciprocal of the slope should give mu, so there is a temptation to fit a straight line using, say, least-squares regression, then use the resulting slope to estimate mu.

mean (flux) # This is the MLE m1 <- lm(log(1-yy+1/n) ~ xx) m1 -1/m1$coef[2] # An alternative estimatorThere is a possible third method that I am told is sometimes used for some kinds of distributions. We start with a histogram, which may be viewed as a rough approximation of the density:

h <- hist(flux)All of the information used to produce the histogram is now stored in the h object, including the midpoints of the bins and their heights on a density scale (i.e., a scale such that the total area of the histogram equals one).

To see how to use this information, note that the logarithm of the density function is log f(x) = -log(mu) - x/mu, which is a linear function of x. Thus, plotting the logarithm of the density against x might be expected to give a line.

counts <- h$counts dens <- h$density[counts>0] midpts <- h$mids[counts>0] plot(midpts, log(dens))When using linear regression to estimate the slope of the linear pattern just produced, I am told that it is standard to weight each point by the number of observations it represents, which is proportional to the reciprocal of the variance of the estimated proportion of the number of points in that bin. We can obtain both the weighted and unweighted versions here. We can then obtain an estimate of mu using either the intercept, which is -log(mu), or the slope, which is -1/mu:

m1 <- lm(log(dens) ~ midpts) m2 <- lm(log(dens) ~ midpts, weights=counts[counts>0]) exp(-m1$coef[1]) # This is one estimate -1/m1$coef[2] # This is another exp(-m2$coef[1]) # Yet another -1/m2$coef[2] # And anotherWe have thus produced no fewer than six different estimators of mu (actually seven, except that the MLE and the method of moments estimator are the same in this case). How should we choose one of them?

There are a couple of ways to answer this question. One is to appeal to statistical theory. The method of maximum likelihood estimation is backed by a vast statistical literature that shows it has certain properties that may be considered optimal. The method of moments is also a well-established method, but arguably with less general theory behind it than the method of maximum likelihood. The regression-based methods, on the other hand, are all essentially ad hoc.

A second way to choose among estimators is to run a simulation study in which we repeatedly simulate datasets (whose parameters are then known to us) and test the estimators to see which seems to perform best. In order to do this, we will need to be able to generate random numbers, which is the next topic in this tutorial.

In the previous example, the log-density function is -log(mu) - x/mu. The second derivative with respect to the parameter is 1/mu^2 - 2x/mu^3. To find the Fisher information, consider x to be a random variable; we know its expectation is mu, so the expectation of the second derivative equals -1/mu^2. We conclude that the Fisher information equals n/mu^2. Intuitively, this means that we get more "information" about the true value of mu when this value is close to zero than when it is far from zero. For the exponential distribution with mean mu, this makes sense.

Earlier we calculated the MLE. Let's give it a name:

mu.hat <- mean (flux) # The MLE found earlierThe standard error of mu.hat, based on the Fisher information, is (square root of) the inverse of the F.I. evaluated at mu.hat:

sqrt(mu.hat^2/n) # SE based on (expected) information mu.hat + 1.96 * c(-1,1) * sqrt(mu.hat^2/n) # approx. 95% CIThe F.I. calculated above is sometimes called the

-n * log(mu.hat) - sum(flux) / mu.hatand the negative second derivative of this function, evaluated at mu.hat, is

-n / mu.hat^2 + 2*sum(flux) / mu.hat^3 # observed information at MLENotice in this case (though not in every model) that the observed information evaluated at the MLE is equal to the expected information evaluated at the MLE:

n / mu.hat^2 # expected information at MLEDo you see why?

R handles many common distributions easily. To see a list, type

help.search("distribution",package="stats")Let's consider the well-known normal distribution as an example:

?NormalThe four functions 'rnorm', 'dnorm', 'pnorm', and 'qnorm' give random normals, the normal density (sometimes called the differential distribution function), the normal cumulative distribution function (CDF), and the inverse of the normal CDF (also called the quantile function), respectively. Almost all of the other distributions have similar sets of four functions. The 'r' versions are rbeta, rbinom, rcauchy, rchisq, rexp, rf, rgamma, rgeom, rhyper, rlogis, rlnorm, rmultinom, rnbinom, rnorm, rpois, rsignrank, rt, runif, rweibull, and rwilcox (there is no rtukey because generally only ptukey and qtukey are needed).

As an example, suppose we wish to simulate a vector of 10 independent, standard (i.e., mean 0 and standard deviation 1) normal random variables. We use the rnorm function for this purpose, and its defaults are mean=0 and standard deviation=1. Thus, we may simply type

rnorm(10)Suppose we wish to simulate a large number of normal random variables with mean 10 and standard deviation 3, then check a histogram against two normal density functions, one based on the true parameters and one based on estimates, to see how it looks. We'll use 'col=2, lty=2, lwd=3' to make the curve based on the true parameters red (color=2), dashed (line type=2), and wider than normal (line width=3). Also note that we are requesting 100 bins in the histogram (nclass=100) and putting it on the same vertical scale as the density functions (freq=FALSE).

z <- rnorm(200000, mean=10, sd=3) hist(z,freq=FALSE,nclass=100) x <- seq(min(z),max(z),len=200) lines(x,dnorm(x, mean=10, sd=3),col=2, lty=2, lwd=3) lines(x,dnorm(x,mean=mean(z),sd=sqrt(var(z))))We can find out what proportion of the deviates lie outside 3 standard deviations from the true mean, a common cutoff used by physical scientists. We can also see the true theoretical proportion:

sum(abs((z-10)/3)>3)/length(z) 2*pnorm(-3)In the first line above, we are using sum to count the number of TRUE's in the logical vector (abs((z-10)/3)>3). This works because logical values are coerced to 0's and 1's when necessary.

The function dnorm has a closed form: With mean=0 and sd=1, dnorm(x) equals exp(-x

pnorm(1:3)-pnorm(-(1:3)) qnorm(c(.05,.95))The first line above summarizes the well-known 68, 95, 99.7 rule for normal distributions (these are the approximate proportions lying within 1, 2, and 3 standard deviations from the mean). The second line gives the critical values used to construct a 90% confidence interval for a parameter when its estimator is approximately normally distributed.

Let us now briefly consider an example of a

k <- 0:10 dpois(k,lambda=2.5) # or equivalently, exp(-2.5)*2.5^k/factorial(k)Next, simulate some Poisson variables:

x <- rpois(10000,lambda=2.5) table(x) mean(x) var(x)

The density function for the Pareto is f(x)=ab

As another example, consider the Salpeter function, the simple but widely known expression of the initial mass function (IMF), in which the mass of a randomly selected newly formed star has a Pareto distribution with parameter a=1.35.

It turns out that a Pareto random variable is simply b*exp(X), where X is an exponential random variable with rate=a (i.e., with mean=1/a). However, rather than exploiting this simple relationship, we wish to build functions for the Pareto distribution from scratch. Our default values, which may be changed by the user, will be a=0.5 and b=1.

dpareto <- function(x, a=0.5, b=1) a*b^a/x^(a+1)Next, we integrate the density function to obtain the distribution function, which is F(x)=1-(b/x)

ppareto <- function(x, a=0.5, b=1) (x > b)*(1-(b/x)^a)Note that (x > b) in the above function is coerced to numeric, either 0 or 1.

Inverting the distribution function gives the quantile function. The following simplistic function is wrong unless 0<u<1, so a better-designed function should do some error-checking.

qpareto <- function(u, a=0.5, b=1) b/(1-u)^(1/a)Finally, to simulate random Pareto random variables, we use the fact that whenever the quantile function is applied to a uniform random variable, the result is a random variable with the desired distribution:

rpareto <- function(n, a=0.5, b=1) qpareto(runif(n),a,b)Creating functions in R, as illustrated above, is a common procedure. Note that each of the arguments of a function may be given a default value, which is used whenever the user calls the function without specifying the value of this parameter. Also note that each of the above functions consists of only a single line; however, longer functions may be created by enclosing them inside curly braces { }.

par(mfrow=c(2,2)) x <- seq(1,50,len=200) plot(x,dpareto(x),type="l") plot(x,ppareto(x),type="l",lty=2) u <- seq(.005,.9,len=200) plot(u,qpareto(u),type="l",col=3) z <- rpareto(200) dotchart(log10(z), main="200 random logged Pareto deviates", cex.main=.7) par(mfrow=c(1,1))The above commands illustrate some of the many plotting capabilities of R. The par function sets many graphical parameters, for instance, 'mfrow=c(2,2)', which divides the plotting window into a matrix of plots, set here to two rows and two columns. In the plot commands, 'type' is set here to "l" for a line plot; other common options are "p" for points (the default), "b" for connected dots, and "n" for nothing (to create axes only). Other options used: 'lty' sets the line type (1=solid, 2=dashed, etc.), 'col' sets color (1=black, 2=red, 3=green, etc.), 'main' puts a string into the plot title, and 'cex.main' sets the text magnification.

Type '? par' to see a list of the many plotting parameters and options.

To evaluate the estimators, we will look at their mean squared error (MSE), which is just what it sounds like: The average of the squared distances from the estimates to the true parameter value of a=1.35.

To illustrate the estimators we'll evaluate, let's start by simulating a single dataset of size 100:

d <- rpareto(100, a=1.35)Here are the estimators we'll consider:

- The
**maximum likelihood**estimator. Since the density with b=1 is given by f(x) = a/x^(a+1), the loglikelihood function is

l(a) = n log(a) - (a+1)(log x_{1}+ log x_{2}+ ... + log x_{n}).

The maximizer may be found using calculus to equal n/(log x_{1}+ ... + log x_{n}). For our dataset, this may be found as follows:

1/mean(log(d))

We used the sum of logarithms above where we could have used the equivalent mathematical expression given by the log of the product. Sometimes the former method gives more numerically stable answers for very large samples, though in this case "100/log(prod(d))" gives exactly the same answer.

- The
**method of moments**estimator. By integrating, we find that the mean of the Pareto distribution with b=1 is equal to a/(a-1). (This fact requires that a be greater than 1.) Setting a/(a-1) equal to the sample mean and solving for a gives 1/(1-1/samplemean) as the estimator.

1/(1-1/mean(d))

- What we'll call the
**EDF (empirical distribution function) estimator**. Since log(1-F(x)) equals -a log(x) when b=1, by plotting the sorted values of log(d) against log(n/n), log((n-1)/n), ..., log(1/n), we should observe roughly a straight line. We may then use least-squares regression to find the slope of the line, which is our estimate of -a:

lsd <- log(sort(d)) lseq <- log((100:1)/100) plot(lsd, lseq) tmp <- lm(lseq ~ lsd) abline(tmp,col=2) -tmp$coef[2]

- What we'll call the
**unweighted histogram estimator**. Since log f(x) equals log(a) - (a+1) log(x) when b=1, if we plot the values of log(d) against histogram-based estimates of the log-density function, we should observe roughly a straight line with slope -(a+1) and intercept log(a). Let's use only the slope, since that is the feature that is most often the focus of a plot that is supposed to illustrate a power-law relationship.hd <- hist(d,nclass=20,plot=F) counts <- hd$counts ldens <- log(hd$density[counts>0]) lmidpts <- log(hd$mids[counts>0]) plot(lmidpts, ldens) tmp <- lm(ldens~lmidpts) abline(tmp,col=2) -1-as.numeric(tmp$coef[2])

- What we'll call the
**weighted histogram estimator**. Exactly the same as the unweighted histogram estimator, but we'll estimate the slope using weighted least squares instead of ordinary least squares. The weights should be proportional to the bin counts.plot(lmidpts, ldens) tmp <- lm(ldens~lmidpts, weights=counts[counts>0]) abline(tmp,col=2) -1-as.numeric(tmp$coef[2])

five <- function(d) { lsd <- log(sort(d)) n <- length(d) lseq <- log((n:1)/n) m1 <- lm(lseq ~ lsd)$coef hd <- hist(d,nclass=n/5,plot=F) counts <- hd$counts ldens <- log(hd$density[counts>0]) lmidpts <- log(hd$mids[counts>0]) m2 <- lm(ldens~lmidpts)$coef m3 <- lm(ldens~lmidpts, weights=counts[counts>0])$coef out <- c(max.lik=1/mean(log(d)), meth.mom=1/(1-1/mean(d)), EDF=-as.numeric(m1[2]), unwt.hist=-1-as.numeric(m2[2]), wt.hist=-1-as.numeric(m3[2])) return(out) }The very last line of the function, "out", is the value that will be returned. (We could also have written "return(out)".) Let's test this function on our dataset:

five(d)There is no good way to compare these estimators based on a single sample like this. We now need to simulate multiple samples. Let's begin by taking n=100.

n.100 <- NULL for(i in 1:250) { dd <- rpareto(100, a=1.35) n.100 <- rbind(n.100, five(dd)) }Now we can get estimates of the biases of the estimators (their expectations minus the true parameter) and their variances. Note that we'll use the biased formula for the variance (i.e., the one that uses n instead of n-1 in the denominator) for a technical reason explained below.

bias.100 <- apply(n.100,2,mean) - 1.35 var.100 <- apply(n.100,2,var) * (249/250)It is a mathematical identity that the mean squared error (MSE) equals the square of the bias plus the variance, as we may check numerically for (say) the first column of n.100. However, the identity only works if we use the biased formula for the variance, which is why we used the multiplier (249/250) above.

mean((n.100[,1]-1.35)^2) bias.100[1]^2 + var.100[1]Thus, we can construct the MSEs and view the results as follows:

mse.100 <- bias.100^2 + var.100 rbind(bias.100, var.100, mse.100)Finally, let's repeat the whole experiment using samples of size 200.

n.200 <- NULL for(i in 1:250) { dd <- rpareto(200, a=1.35) n.200 <- rbind(n.200, five(dd)) } bias.200 <- apply(n.200,2,mean) - 1.35 var.200 <- apply(n.200,2,var) * (249/250) mse.200 <- bias.200^2 + var.200 rbind(bias.200, var.200, mse.200)

The name "EM algorithm" has its genesis in a seminal 1977 paper by Dempster, Laird, and Rubin in the Journal of the Royal Statistical Society, Series B. Many distinct algorithms published prior to 1977 were examples of EM, including the Lucy-Richardson algorithm for image deconvolution that is apparently quite well known in astronomy. The major contribution of Dempster et al was to unify these algorithms and prove certain facts about them. Interesting historical note: They even "proved" an untrue fact that was refuted six years (!) later (even thirty years ago, publications in statistics churned through the pipeline at a snail's pace).

We'll derive a simple EM algorithm on a toy example: We'll pretend that some of the gamma ray burst flux measurements were right-censored, as follows:

cflux <- flux cflux[flux>=60] <- 60 n <- length(cflux) yy <- (1:n)/n plot(sort(cflux),log(1-yy+1/n))The situation may be viewed like this: The complete dataset is a set of n observations from an exponential distribution with unknown mean mu, say, X

m <- sum(flux>=60) s <- sum(cflux) loglik <- function(mu) -(n-m)*log(mu)-s/muAs it turns out, this loglikelihood function can be maximized explicitly:

mle <- s/(n-m)However, we will construct an EM algorithm anyway for two reasons: First, it is instructive to see how the EM operates. Second, not all censoring problems admit a closed-form solution like this one does!

We start by writing down the complete-data loglikelihood for the sample. This is straightforward because the complete data are simply a random sample from an exponential distribution with mean mu. Next, we pick a starting value of mu, say mu

Let's start with mu

mu <- 20 loglik(mu) mu <- s/n + m*mu/n; loglik(mu) mu <- s/n + m*mu/n; loglik(mu) # repeat the last line a few timesNotice that the value of the (observed data) loglikelihood increases at each iteration. This is the fundamental property of any EM algorithm! In fact, it is very helpful when debugging computer code, since there must be a bug somewhere whenever the loglikelihood is ever observed to decrease. Notice also that the value of mu has converged to the true maximum likelihood estimator after a few iterations.

qso <- scan("http://www.astrostatistics.psu.edu/datasets/QSO_absorb.txt", skip=1, nlines=104)[2*(1:104)]Here is a function to implement an EM algorithm for a two-component mixture of normals, where the variances of the two components are assumed to be the same. The function may be modified to allow for different variances by using the commented-out lines rather than the ones preceding them:

twocomponentnormalmixtureEM <- function(mu1, mu2, sigsqrd, lambda, x) #twocomponentnormalmixtureEM <- function(mu1, mu2, sigsqrd1, sigsqrd2, lambda, x) { mx<-mean(x) nover2<-length(x)/2 dl <- 1 l<--Inf iterations<-0 a1<-(x-mu1)^2; b1<-lambda*exp(-a1/2/sigsqrd) # a1<-(x-mu1)^2; b1<-lambda/sqrt(sigsqrd1)*exp(-a1/2/sigsqrd1) a2<-(x-mu2)^2; b2<-(1-lambda)*exp(-a2/2/sigsqrd) # a2<-(x-mu2)^2; b2<-(1-lambda)/sqrt(sigsqrd2)*exp(-a2/2/sigsqrd2) while (dl>1e-4) { iterations<-iterations+1 postprobs <- b1/(b1+b2) lambda<-mean(postprobs) mu1<-mean(postprobs*x)/lambda mu2<-(mx-lambda*mu1)/(1-lambda) sigsqrd<-mean(postprobs*a1+(1-postprobs)*a2) # sigsqrd1<-mean(postprobs*a1); sigsqrd2<-mean((1-postprobs)*a2) oldl<-l a1<-(x-mu1)^2; b1<-lambda*exp(-a1/2/sigsqrd) # a1<-(x-mu1)^2; b1<-lambda/sqrt(sigsqrd1)*exp(-a1/2/sigsqrd1) a2<-(x-mu2)^2; b2<-(1-lambda)*exp(-a2/2/sigsqrd) # a2<-(x-mu2)^2; b2<-(1-lambda)/sqrt(sigsqrd2)*exp(-a2/2/sigsqrd2) l<-sum(log(b1+b2)) - nover2*log(sigsqrd) # l<-sum(log(b1+b2)) dl<-l-oldl } list(mu=c(mu1,mu2), variance=sigsqrd, # list(mu=c(mu1,mu2), variance=c(sigsqrd1, sigsqrd2), lambda=lambda, loglik=l, iterations=iterations) }Now let's try it:

hist(qso, nclass=20) twocomponentnormalmixtureEM (.6, 1, .1, .1, qso)This seems to converge to a sensible solution. However, try some other starting values and see what happens. If you find a different solution (i.e., a different local maximum), how does its loglikelihood value compare to that of the first solution?