#LyX 1.4.2svn created this file. For more info see http://www.lyx.org/ \lyxformat 245 \begin_document \begin_header \textclass literate-article \language english \inputencoding auto \fontscheme default \graphics default \paperfontsize default \spacing single \papersize default \use_geometry true \use_amsmath 1 \cite_engine basic \use_bibtopic false \paperorientation portrait \leftmargin 1in \topmargin 1in \rightmargin 1in \bottommargin 1in \secnumdepth 3 \tocdepth 3 \paragraph_separation indent \defskip medskip \quotes_language english \papercolumns 1 \papersides 1 \paperpagestyle default \tracking_changes false \output_changes true \end_header \begin_body \begin_layout Title Negative Binomial \end_layout \begin_layout Author Paul Johnson and Andrea Vieux \end_layout \begin_layout Date May1, 2006 \end_layout \begin_layout Section Introduction \end_layout \begin_layout Standard The Negative Binomial is discrete probability distribution, one that can be used for counts or other integer-valued variables. It gives probabilities for integer values from \begin_inset Formula $0$ \end_inset to \begin_inset Formula $\infty$ \end_inset . It is sometimes also known as the Pascal distribution or as Polya's distributio n. \end_layout \begin_layout Standard The Negative Binomial is best thought of as a variant of a Binomial distribution. The reader will recall that a \emph on Bernoulli trial \emph default is a \begin_inset Quotes eld \end_inset coin flip \begin_inset Quotes erd \end_inset experiment, one that returns \begin_inset Quotes eld \end_inset yes \begin_inset Quotes erd \end_inset or \begin_inset Quotes eld \end_inset no \begin_inset Quotes erd \end_inset , \begin_inset Quotes eld \end_inset failure \begin_inset Quotes erd \end_inset or \begin_inset Quotes eld \end_inset success. \begin_inset Quotes erd \end_inset The Binomial distribution describes the number of \begin_inset Quotes eld \end_inset successes \begin_inset Quotes erd \end_inset out of a given number of \begin_inset Quotes eld \end_inset Bernoulli trials \begin_inset Quotes erd \end_inset . As such, its values are defined from \begin_inset Formula $0$ \end_inset up to the number of trials. \end_layout \begin_layout Standard The Negative Binomial describes the same sequence of Bernoulli trials that is described by the Binomial distribution. The difference is \begin_inset Quotes eld \end_inset what \begin_inset Quotes erd \end_inset is being described. Whereas Bernoulli tracks the number of successes, the Negative Binomial represents the number of failures that occurs at the beginning of the sequence as we wait for a given number of successes to be achieved. \end_layout \begin_layout Standard The most frequent use of the Negative Binomial model has nothing to do with the property that it describes the chances of a string of failures. Rather, it is used to describe distributions that represent counts of events. It is a replacement for the commonly used Poisson distribution, which is often considered the default distribution for integer-valued count data. The Poisson has the property that its expected value equals its variance, a property that is not found in some empirically observed distributions. The Negative Binomial has a higher variance. \end_layout \begin_layout Section Mathematical Description \end_layout \begin_layout Standard Recall that the Binomial distribution gives the probability of observing \begin_inset Formula $r$ \end_inset successes out ot \begin_inset Formula $N$ \end_inset trials when the probability of observing a success is fixed at \begin_inset Formula $p$ \end_inset . \begin_inset Formula \begin{equation} B(r|N)=\left(\begin{array}{c} N\\ r\end{array}\right)p^{r}(1-p)^{N-r}=\frac{N!}{r!(N-r)!}p^{r}(1-p)^{N-r}\label{eq:Binomial}\end{equation} \end_inset \newline The symbol \begin_inset Formula $\left(\begin{array}{c} N\\ r\end{array}\right)$ \end_inset is the \begin_inset Quotes eld \end_inset binomial coefficient \begin_inset Quotes erd \end_inset (from intermediate algebra) and it is spoken out loud as \begin_inset Quotes eld \end_inset \begin_inset Formula $N$ \end_inset choose \begin_inset Formula $r$ \end_inset \begin_inset Quotes erd \end_inset , meaning the number of ways one can choose subsets of \begin_inset Formula $r$ \end_inset from a larger set of \begin_inset Formula $N$ \end_inset . \begin_inset Formula \begin{equation} \left(\begin{array}{c} N\\ r\end{array}\right)=\frac{N!}{N!(N-r)!}\label{eq:BinCoef}\end{equation} \end_inset \newline Thinking of one particular sequence the \begin_inset Formula $N$ \end_inset Bernoulli trials, one that has \begin_inset Formula $r$ \end_inset successes, we can easily calculate the probabilities. The chance of observing one particular sequence, \begin_inset Formula $S,S,F,S,F,\ldots S$ \end_inset , will be \begin_inset Formula $p\cdot p\cdot(1-p)\cdot p\cdot(1-p)\ldots p$ \end_inset . Regrouping indicates the probability of that particular sequence is \begin_inset Formula $p^{r}(1-p)^{N-r}$ \end_inset . That is the probability of one particular sequence that has \begin_inset Formula $r$ \end_inset successes. Put together the chances of all such sequences, of which there are \begin_inset Formula $\left(\begin{array}{c} N\\ r\end{array}\right)$ \end_inset , and the formula in \begin_inset LatexCommand \ref{eq:Binomial} \end_inset should be clearly understood. \end_layout \begin_layout Standard The Negative Binomial is defined with reference to the Binomial. The R help page for the rbinom function describes it as \begin_inset Quotes eld \end_inset the number of failures which occur in a sequence of Bernoulli trials before a target number of successes is reached \begin_inset Quotes erd \end_inset (R documentation). \end_layout \begin_layout Standard We found the description on the Math World web site ( \begin_inset LatexCommand \url{http://mathworld.wolfram.com/NegativeBinomialDistribution.html} \end_inset ) to have more intuitive appeal. The Negative Binomial Distribution gives \begin_inset Quotes eld \end_inset the probability of r-1 successes and x failures in x+r-1 trials, and success on the (x+r) th trial. \begin_inset Quotes erd \end_inset \end_layout \begin_layout Standard Take the first part of the definition, \begin_inset Formula $(r-1)$ \end_inset successes out of \begin_inset Formula $(x+r-1)$ \end_inset trials. By the Binomial distribution, that is \begin_inset Formula \begin{equation} Prob(r-1\mid x+r-1)=\frac{(x+r-1)!}{(r-1)!(x)!}p^{r-1}(1-p)^{x}\label{eq:Binr-1}\end{equation} \end_inset \newline The probability of observing a success on the last trial, the \begin_inset Formula $(x+r)$ \end_inset th trial, is \begin_inset Formula $p$ \end_inset . So, by the multiplication rule of probabilities, we multiply the previous equation \begin_inset LatexCommand \ref{eq:Binr-1} \end_inset by \begin_inset Formula $p$ \end_inset to obtain \begin_inset Formula \begin{equation} NB(x\mid r,p)=\frac{(x+r-1)!}{(r-1)!(x)!}p^{r}(1-p)^{x}\label{eq:Binr-2}\end{equation} \end_inset \newline Or, if you prefer to write it with the binomial coefficient, \end_layout \begin_layout Standard \begin_inset Formula \begin{equation} NB(x\mid r,p)=\left(\begin{array}{c} x+r-1\\ r-1\end{array}\right)p^{r}(1-p)^{x}\label{eq:Binr-3}\end{equation} \end_inset \end_layout \begin_layout Section Central Tendency and Dispersion \end_layout \begin_layout Standard If \begin_inset Formula $x$ \end_inset has a Negative Binomial distribution, then \begin_inset Formula $x$ \end_inset can take on integer values ranging from \begin_inset Formula $0$ \end_inset to \begin_inset Formula $\infty$ \end_inset . The upper limit is infinity because we are letting the process run until there are exactly \begin_inset Formula $(r-1)$ \end_inset failures and then we conduct one additional experiment on which we obtain success. \end_layout \begin_layout Standard The expected value and variance are: \end_layout \begin_layout Standard \begin_inset Formula \begin{equation} E[x]=\frac{r(1-p)}{p}\end{equation} \end_inset \end_layout \begin_layout Standard \begin_inset Formula \begin{equation} Var(x)=\frac{r(1-p)}{p^{2}}\end{equation} \end_inset \newline and the skewness and kurtosis are \begin_inset Formula \[ Skewness(x)=\frac{2-p}{\sqrt{(1-p)r}}\] \end_inset \end_layout \begin_layout Standard \begin_inset Formula \[ Kurtosis(x)=\frac{p^{2}-6p+6}{r(1-p)}\] \end_inset \end_layout \begin_layout Standard These are worth considering because of their dependence on \begin_inset Formula $r$ \end_inset . As one would expect, the expected number of successes is increasing in the number of observed failures. However, the important effects of \begin_inset Formula $r$ \end_inset are seen in the variance and skewness. As \begin_inset Formula $r$ \end_inset increases, the variance increases. But the skewness shrinks. Hence, for small values of \begin_inset Formula $r$ \end_inset , we should expect to see a very skewed distribution, whereas for higher values of \begin_inset Formula $r$ \end_inset , the distribution should be more symmetrical. \end_layout \begin_layout Section Illustrations \end_layout \begin_layout Standard The following code is for Figure \begin_inset LatexCommand \ref{cap:Negative-Binomial} \end_inset , which displays example distributions as a function of a target number of successes and probabilities for successes on individual trials. Recall that the probabilities indicate the chances of observing a given number of failures after observing a given number of successes. \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Standard <>= \end_layout \begin_layout Standard drawBinom <- function (S, p){ \end_layout \begin_layout Standard \end_layout \begin_layout Standard x<- seq(0,200) \end_layout \begin_layout Standard \end_layout \begin_layout Standard xprob <- dnbinom(x,size=S,prob=p) \end_layout \begin_layout Standard \end_layout \begin_layout Standard mytitle <- paste("x",S,p) \end_layout \begin_layout Standard \end_layout \begin_layout Standard plot(x, xprob,type="s",main=mytitle,xlab=paste("number of failures before ",S,"successes")) \end_layout \begin_layout Standard \end_layout \begin_layout Standard \end_layout \begin_layout Standard \end_layout \begin_layout Standard } \end_layout \begin_layout Standard \end_layout \begin_layout Standard \end_layout \begin_layout Standard \end_layout \begin_layout Standard par(mfcol=c(3,2)) \end_layout \begin_layout Standard \end_layout \begin_layout Standard sizes <- c(20,40,60) \end_layout \begin_layout Standard \end_layout \begin_layout Standard pvals <- c(0.33,0.67) \end_layout \begin_layout Standard \end_layout \begin_layout Standard for (i in 1:2){ \end_layout \begin_layout Standard \end_layout \begin_layout Standard for (j in 1:3){ \end_layout \begin_layout Standard \end_layout \begin_layout Standard drawBinom ( sizes[j], pvals[i] ) \end_layout \begin_layout Standard \end_layout \begin_layout Standard } \end_layout \begin_layout Standard \end_layout \begin_layout Standard } \end_layout \begin_layout Standard \end_layout \begin_layout Standard @ \end_layout \end_inset \end_layout \begin_layout Standard \begin_inset Float figure wide false sideways false status open \begin_layout Caption Negative Binomial \begin_inset LatexCommand \label{cap:Negative-Binomial} \end_inset \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Standard \backslash begin{center} \end_layout \begin_layout Standard \end_layout \begin_layout Standard <>= \end_layout \begin_layout Standard <> \end_layout \begin_layout Standard \end_layout \begin_layout Standard @ \end_layout \begin_layout Standard \end_layout \begin_layout Standard \backslash end{center} \end_layout \end_inset \end_layout \end_inset \end_layout \begin_layout Standard \paragraph_spacing double In the Figure \begin_inset LatexCommand \ref{cap:Negative-Binomial} \end_inset we can see how changes in the size and probability alter the Negative Binomial Distribution. The size moves the graph right to left, and the probability widens or shrinks the bell shape. For a graph which resembles the Normal Distribution, refer to Figure \begin_inset LatexCommand \ref{cap:Negative-Binomial-2} \end_inset . \end_layout \begin_layout Standard \begin_inset Float figure wide false sideways false status open \begin_layout Caption Negative Binomial 2 \begin_inset LatexCommand \label{cap:Negative-Binomial-2} \end_inset \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Standard \backslash begin{center} \end_layout \begin_layout Standard \end_layout \begin_layout Standard <>= \end_layout \begin_layout Standard x7<-seq(0,100) \end_layout \begin_layout Standard \end_layout \begin_layout Standard xprob7<-dnbinom(x7,size=50,prob=.5) \end_layout \begin_layout Standard \end_layout \begin_layout Standard plot(xprob7,type="s") \end_layout \begin_layout Standard \end_layout \begin_layout Standard @ \end_layout \begin_layout Standard \end_layout \begin_layout Standard \backslash end{center} \end_layout \end_inset \end_layout \end_inset \end_layout \begin_layout Section The Negative Binomial as a Mixture Distribution \end_layout \begin_layout Standard From J. Scott Long's book (Citation ??): \end_layout \begin_layout Standard \paragraph_spacing double The negative binomial distribution is a useful tool in the analysis of count data. It differs from the Poisson Distribution in that it allows for the conditional variance to exceed the conditional mean. The negative binomial distribution is the end result of a process which begins with a Poisson distribution with mean \begin_inset Formula $\lambda$ \end_inset . The fixed ( \begin_inset Formula $\lambda$ \end_inset ) is replaced with a random variable that has a mean of \begin_inset Formula $\lambda$ \end_inset , but has nonzero variance. With this new framework, we are representing the possibility that there is unobserved heterogeneity. \end_layout \begin_layout Standard Recall the definition of the Poisson gives the probability of observing \begin_inset Formula $x$ \end_inset events as a function of a parameter \begin_inset Formula $\lambda$ \end_inset . \begin_inset Formula \begin{equation} Poisson(x|\lambda)=\frac{\lambda^{x}e^{-\lambda}}{x!}\label{eq:ConditPoisson}\end{equation} \end_inset \end_layout \begin_layout Standard Instead of thinking of \begin_inset Formula $\lambda$ \end_inset as a constant, think of it as a random variable, representing the fact that the average count is likely to vary across experiments (counties, cities, etc). If we think of \begin_inset Formula $\lambda$ \end_inset as a variable, then the probability of observing \begin_inset Formula $x$ \end_inset failures has to be the result of a 2 part process. Part 1 chooses \begin_inset Formula $\lambda$ \end_inset and part 2 chooses \begin_inset Formula $x$ \end_inset . That means there are possibly many different routes through which we can observe a particular value \begin_inset Formula $x$ \end_inset , since \begin_inset Formula $\lambda$ \end_inset can vary from \begin_inset Formula $0$ \end_inset to \begin_inset Formula $\infty$ \end_inset , and for each possible value of \begin_inset Formula $\lambda$ \end_inset , there is \begin_inset Quotes eld \end_inset some chance \begin_inset Quotes erd \end_inset of observing any positive value. \begin_inset Formula \begin{equation} \int P(x|\lambda)p(\lambda)d\lambda\label{eq:mixture}\end{equation} \end_inset \end_layout \begin_layout Standard If one chooses \begin_inset Quotes eld \end_inset just the right \begin_inset Quotes erd \end_inset distribution for \begin_inset Formula $\lambda$ \end_inset , then the \begin_inset Quotes eld \end_inset mixture \begin_inset Quotes erd \end_inset distribution will be Negative Binomial. \end_layout \begin_layout Standard Suppose we take the lucky guess that \begin_inset Formula $\lambda$ \end_inset has a Gamma probability distribution \end_layout \begin_layout Standard \begin_inset Formula \begin{equation} Prob(\lambda)=\frac{1}{\beta^{\alpha}\Gamma(\alpha)}\lambda^{(\alpha-1)}e^{-\lambda/\beta}\end{equation} \end_inset \newline The \begin_inset Quotes eld \end_inset shape \begin_inset Quotes erd \end_inset parameter is \begin_inset Formula $\alpha$ \end_inset and the scale parameter is \begin_inset Formula $\beta$ \end_inset , and the expected value is \begin_inset Formula $\alpha\beta$ \end_inset and the variance is \begin_inset Formula $\alpha\beta^{2}$ \end_inset . \end_layout \begin_layout Standard Inserting the given distributions into \begin_inset LatexCommand \ref{eq:mixture} \end_inset , we are able to calculate the marginal distribution of \begin_inset Formula $x$ \end_inset . \end_layout \begin_layout Standard \begin_inset Formula \begin{equation} Prob(x)=\int_{0}^{\infty}\frac{e^{-\lambda}\lambda^{x}}{x!}\frac{1}{\beta^{\alpha}\Gamma(\alpha)}\lambda^{(\alpha-1)}e^{-\lambda/\beta}d\lambda\end{equation} \end_inset \begin_inset Formula \begin{equation} =\frac{1}{x!\beta^{\alpha}\Gamma(\alpha)}\int_{0}^{\infty}e^{-\lambda}\lambda^{x}\lambda^{(\alpha-1)}e^{-\lambda/\beta}d\lambda\end{equation} \end_inset \begin_inset Formula \begin{equation} =\frac{1}{x!\beta^{\alpha}\Gamma(\alpha)}\int_{0}^{\infty}\lambda^{x+\alpha-1}e^{-\lambda}e^{-\lambda/\beta}d\lambda\end{equation} \end_inset \end_layout \begin_layout Standard \begin_inset Formula \begin{equation} =\frac{1}{x!\beta^{\alpha}\Gamma(\alpha)}\int_{0}^{\infty}\lambda^{x+\alpha-1}e^{-\lambda(1+1/\beta)}d\lambda\label{eq:brute4}\end{equation} \end_inset \end_layout \begin_layout Standard I have tried to simplify this through brute force, but have been unable to do so. I feel sure there is a specialized result in calculus that will help to use the Gamma function, which would be adapted to this case as: \end_layout \begin_layout Standard \begin_inset Formula \begin{equation} \Gamma(x+\alpha)=\int_{0}^{\infty}\lambda^{x+\alpha-1}e^{-\lambda}d\lambda\end{equation} \end_inset \newline And the end result, the Negative Binomial distribution, is stated either as \end_layout \begin_layout Standard \begin_inset Formula \begin{equation} =\frac{\Gamma(\alpha+x)}{x!\Gamma(\alpha)}\left(\frac{1}{1+\beta}\right)^{\alpha}\left(\frac{\beta}{1+\beta}\right)^{x}\end{equation} \end_inset \newline or \begin_inset Formula \begin{equation} =\frac{\Gamma(\alpha+x)}{x!\Gamma(\alpha)}\left(\frac{1}{1+\beta}\right)^{\alpha}\left(1-\frac{1}{1+\beta}\right)^{x}\label{eq:mixture6}\end{equation} \end_inset \end_layout \begin_layout Standard This is the form of the Negative Binomial that was discussed earlier. \end_layout \begin_layout Standard I have been unable to find the brute force method to make the jump from \begin_inset LatexCommand \ref{eq:brute4} \end_inset to \begin_inset LatexCommand \ref{eq:mixture6} \end_inset , but in Gelman, Carlin, Stern, and Rubin's Bayesian Data Analysis, 2ed (p. 53), an alternative derivation is offered. Recall Bayes's theorem \begin_inset Formula \begin{equation} p(\lambda|x)=\frac{p(x|\lambda)p(\lambda)}{p(x)}\label{eq:Bayes1}\end{equation} \end_inset \newline Rearrange that \begin_inset Formula \begin{equation} p(x)=\frac{p(x|\lambda)p(\lambda)}{p(\lambda|x)}\label{eq:Bayes2}\end{equation} \end_inset \newline The left hand side is our target--the marginal distribution of \begin_inset Formula $x$ \end_inset . We already know formulae for the two components in the numerator, but the denominator requires a result from probability theory. \end_layout \begin_layout Standard Most statistics books--and all Bayesian statistics books--will have a result about updating from a Poisson distribution. If \begin_inset Formula $x$ \end_inset is a Poisson variable with parameter \begin_inset Formula $\lambda$ \end_inset , and the conjugate prior for \begin_inset Formula $\lambda$ \end_inset is the Gamma distribution, \begin_inset Formula $Gamma(\lambda|\alpha,\beta)$ \end_inset , then Bayes's theorem (expression \begin_inset LatexCommand \ref{eq:Bayes1} \end_inset ), leads to the conclusion that if \begin_inset Formula $x$ \end_inset \begin_inset Quotes eld \end_inset events \begin_inset Quotes erd \end_inset are observed, then \begin_inset Formula \begin{equation} p(\lambda|x)=Gamma(\lambda|\alpha+x,\beta+1)\end{equation} \end_inset \newline That is the last ingredient we need to solve expression \begin_inset LatexCommand \ref{eq:Bayes2} \end_inset . \end_layout \begin_layout Standard Gelman et al put it this way: \begin_inset Formula \begin{equation} p(x)=\frac{Poisson(x|\lambda)Gamma(\lambda|\alpha,\beta)}{Gamma(\lambda|\alpha+x,1|\beta)}\end{equation} \end_inset \newline Writing in the expressions obtained above, \begin_inset Formula \begin{equation} p(x)=\frac{\frac{\lambda^{x}e^{-\lambda}}{x!}\cdot\frac{1}{\beta^{\alpha}\Gamma(\alpha)}\lambda^{(\alpha-1)}e^{-\lambda/\beta}}{\frac{1}{(1+\beta)^{(\alpha+x)}\Gamma(\alpha+x)}\lambda^{(\alpha+x-1)}e^{-\frac{\lambda}{(1+\beta)}}}\end{equation} \end_inset \newline After cancellation and rearrangement, that simplifies to: \begin_inset Formula \begin{equation} p(x)=\frac{\Gamma(\alpha+x)}{\Gamma(\alpha)x!}\cdot\frac{\beta^{\alpha}}{(1+\beta)^{\alpha+x}}\end{equation} \end_inset \newline Using the definition of \begin_inset Formula $\Gamma(x)$ \end_inset and rearranging again, we find: \begin_inset Formula \begin{equation} =\frac{(\alpha+x-1)!}{(\alpha-1)!x!}\cdot\left(\frac{\beta}{1+\beta}\right)^{\alpha}\left(\frac{1}{1+\beta}\right)^{x}\end{equation} \end_inset \newline Recognizing that the first term on the left is the Binomial coefficient, the derivation is complete. \end_layout \begin_layout Standard We choose the shape and scale parameters so that the Gamma distributed value of \begin_inset Formula $\lambda$ \end_inset will allow us to simplify \begin_inset LatexCommand \ref{eq:mixture6} \end_inset . It appears obvious in this expression that the role of the probability of success on an individual trial is \begin_inset Formula $p=\frac{1}{1+\beta}$ \end_inset and that \begin_inset Formula $\alpha$ \end_inset is the \begin_inset Quotes eld \end_inset desired number of successes \begin_inset Quotes erd \end_inset in the Negative Binomial. That means that, if want to set the Gamma parameters equal to the proper levels, we choose \begin_inset Formula $\beta=(1-p)/p$ \end_inset and \begin_inset Formula $\alpha=n$ \end_inset (the number of desired successes). \end_layout \begin_layout Standard If \begin_inset Formula $n$ \end_inset represents the number of successes and the number of failures is \begin_inset Formula $x=N-n$ \end_inset \end_layout \begin_layout Standard \begin_inset Formula \[ Prob(x)=\frac{\Gamma(x+n)}{\Gamma(n)x!}p^{n}(1-p)^{x}\] \end_inset \end_layout \begin_layout Subsection* Another way to think of the mixture \end_layout \begin_layout Standard Instead of replacing \begin_inset Formula $\lambda$ \end_inset in the Poisson definition, suppose we think of multiplying it by a randomly chosen value \begin_inset Formula $\delta$ \end_inset . If \begin_inset Formula $\delta$ \end_inset has an expected value of \begin_inset Formula $1$ \end_inset , then \begin_inset Formula $E(\lambda\delta)=\lambda$ \end_inset , so, \begin_inset Quotes eld \end_inset on average, \begin_inset Quotes erd \end_inset the original \begin_inset Formula $\lambda$ \end_inset is reproduced. However, because \begin_inset Formula $\delta$ \end_inset might be higher or lower, then \begin_inset Formula $\lambda\delta$ \end_inset will have random variation, and so the number of observed events will fluctuate. Its average will still be \begin_inset Formula $\lambda,$ \end_inset but it will have greater variance. In such a way, one can see the Poisson in this way (replacing \begin_inset Formula $\lambda$ \end_inset by \begin_inset Formula $\lambda\delta$ \end_inset ). \end_layout \begin_layout Standard \begin_inset Formula \[ P(x\mid\delta)=\frac{e^{-(\lambda\delta)}(\lambda\delta)^{x}}{x!}\] \end_inset \end_layout \begin_layout Standard Here is the question: how can one select a mathematically workable model for \begin_inset Formula $\delta$ \end_inset so that \begin_inset Formula $E(\delta)=1$ \end_inset ? \end_layout \begin_layout Standard I've seen this done in several ways. Recall that the Gamma distribution has expected value \begin_inset Formula $\alpha\beta$ \end_inset and variance \begin_inset Formula $\alpha\beta^{2}$ \end_inset . So if we drew a variate \begin_inset Formula $y$ \end_inset from any Gamma distribution and then divided by \begin_inset Formula $\alpha\beta,$ \end_inset the result would be \begin_inset Formula \[ x/\alpha\beta\] \end_inset \newline and the expected value would be \begin_inset Formula $E(x/\alpha\beta)=E(y)/\alpha\beta=1$ \end_inset . \end_layout \begin_layout Standard More commonly, attention is focused on a subset of possible Gamma distributions, the ones for which the \begin_inset Formula $\beta=1/\alpha$ \end_inset . When \begin_inset Formula $\delta$ \end_inset follows a Gamma distribution \begin_inset Formula $Gamma(\delta|\alpha,1/\alpha)$ \end_inset , then it has an expected value of 1 and its variance is \begin_inset Formula $1/\alpha$ \end_inset . As \begin_inset Formula $\alpha$ \end_inset tends to 0, the variance tends toward infinity. \end_layout \begin_layout Standard Think of \begin_inset Formula $\lambda$ \end_inset as an \begin_inset Quotes eld \end_inset attribute \begin_inset Quotes erd \end_inset of an observation. If \begin_inset Formula $\delta$ \end_inset is a Gamma variate with mean 1 and variance \begin_inset Formula $1/\alpha$ \end_inset , then \begin_inset Formula $\lambda\delta$ \end_inset is also a Gamma variate, with mean \begin_inset Formula $\lambda$ \end_inset and variance \begin_inset Formula $\lambda^{2}/\alpha$ \end_inset . Maybe it is just a lucky guess, but it appears to me that \begin_inset Formula $\lambda\delta$ \end_inset would have the distribution \begin_inset Formula $Gamma(\alpha,\lambda/\alpha)$ \end_inset . \end_layout \end_body \end_document