MSH3 Generalized linear model Ch. 6 Count data models

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1 Contents MSH3 Generalized linear model Ch. 6 Count data models 6 Count data model Introduction: The Children Ever Born Data The Poisson Distribution Log-Linear Models ML estimation Goodness of fit and hypothesis test Children ever born example Grouped Data and the Offset The Deviance Table The Additive Model Negative binomial distribution Model for excess zero Zero-inflated Poisson model Hurdle model Model selection example SydU MSH3 GLM (2015) First semester Dr. J. Chan 207

2 6 Count data model MSH3 Generalized linear model Ch. 6 Count data models In this chapter we study log-linear models for count data with a Poisson error structure and a log link function. 6.1 Introduction: The Children Ever Born Data The following table, adapted from Little (1979), comes from the Fiji Fertility Survey, typical in the reports of the World Fertility Survey. Each cell in the table shows the mean, the variance and the number of observations. The unit of analysis is the individual woman, the response is the number of children she has borne, and the 3 discrete predictors are the duration since her first marriage, the type of place of residence (Suva the capital, other urban and rural), and her educational level (none, lower primary, upper primary, and secondary or higher). Data such as these have traditionally been analyzed using ordinary linear models with normal errors. The key concern, however, is not the normality of the errors but rather the assumption of constant variance. log(var) log(mean) Figure 6.1: The Mean-variance Relationship for the CEB Data SydU MSH3 GLM (2015) First semester Dr. J. Chan 208

3 The plot of variance against mean in log scale for all cells with at least 20 observation shows clearly that, the assumption of constant variance is not suitable and the relationship is close to proportional. Thus, a Poisson regression models is more suitable than a ordinary linear models. Table 6.1: Number of Children Ever Born to Women of Indian Race By Marital Duration, Type of Place of Residence and Educational Level (Each cell shows the mean, variance and sample size) Marr. Suva Urban Rural Dur. N LP UP S+ N LP UP S+ N LP UP S SydU MSH3 GLM (2015) First semester Dr. J. Chan 209

4 6.2 The Poisson Distribution A random variable Y is said to have a Poisson distribution with parameter µ if it takes integer values y = 0, 1, 2,... with probability Pr(Y = y) = e µ µ y for µ > 0. The mean and variance are y! E(Y ) = Var(Y ) = µ. Since the mean is equal to the variance, any factor that affects one will also affect the other and hence homoscedasticity fails for Poisson data. Poisson model captures the usual fact with count data that the variance tends to increase with the mean. The Poisson distribution can be derived as a limiting form of the binomial distribution. Specifically, if Y B(n, π), Y approx P (µ) with fixed µ = nπ as n and π 0. Thus, the Poisson distribution provides an approximation to the binomial for the analysis of rare events, where µ is small and n is large. Alternatively, the Poisson distribution is defined in terms of a stochastic process described as follows. Suppose events occur randomly in time with the following conditions: The probability of at least one occurrence of the event in a given time interval is proportional to the length of the interval. The probability of two or more occurrences of the event in a very small time interval is negligible. The numbers of occurrences of the event in disjoint time intervals are mutually independent. SydU MSH3 GLM (2015) First semester Dr. J. Chan 210

5 Then the probability distribution of the number of occurrences of the event in a fixed time interval is Poisson with mean µ = λt, where λ is the rate of occurrence of the event per unit time and t is the length of the time interval. The proof will be given in the Appendix. A process satisfying the three assumptions is called a Poisson process. A useful property of the Poisson distribution is that the sum of independent Poisson random variables is also Poisson. Specifically, if Y 1 and Y 2 are independent with Y i P (µ i ) for i = 1, 2 then Y 1 + Y 2 P (µ 1 + µ 2 ). This result generalizes in an obvious way to the sum of more than two Poisson observations. For a group of n i individuals with identical covariate values, let Y ij denote the number of events experienced by the j-th unit in the i-th group, and let Y i denote the total number of events in group i. Then, under the usual assumption of independence, if Y ij P (µ i ) for j = 1, 2,... n i, then Y i P (n i µ i ). In terms of estimation, we obtain exactly the same likelihood function if we work with the individual counts Y ij or the group counts Y i. 6.3 Log-Linear Models Let Y i P (µ i ) independently and y 1, y 2,... y n are n observed values. Let the mean µ i (variance as well) depend on a vector of explanatory variables x i in a simple linear model: µ i = x iβ. However, while x i β can assume any real value, the Poisson mean µ i has to be non-negative. One solution is to model the logarithm of the µ i, a canonical link for Poisson distribution, using a linear model: log(µ i ) = x iβ SydU MSH3 GLM (2015) First semester Dr. J. Chan 211

6 where β j in β represents the expected change in the log of the mean per unit change in the predictor x ij. Exponentiating the equation, we obtain a multiplicative model for the mean: µ i = exp(x iβ) = exp( p p p x ij β j ) = exp(x ij β j ) = [exp(β j )] x ij j=1 which agrees with what we observe that the size of effects is proportional to the count itself. Increasing x ij by one unit, the mean µ i is multiplied by exp(β j ). 6.4 ML estimation The log-likelihood function for n independent Poisson observations is n l(β) = {y i log(µ i ) µ i log(y i!)}. i=1 Taking derivatives of l(β) with respect to the elements of β, and setting them zero l n y i n = µ i x ij µ i x ij = x ij (y i µ i ), β j µ i i=1 the maximum likelihood estimates in log-linear Poisson models satisfy the estimating equations X y = X µ where X is the design matrix, with row x i for each observation and one column for each predictor, including the constant (if any). This estimating equation arises in any generalized linear model with canonical link. Thus, in models with a constant, one of the estimating equations matches the sum of observed and fitted values. In models with a discrete factor, SydU MSH3 GLM (2015) First semester Dr. J. Chan 212 j=1 i=1 j=1

7 such as the Children Ever Born example, the observed and fitted total number of children ever born will be matched in each category of marital duration since first marriage. This result generalizes to higher order terms. The interaction model A+B+AB would match the totals in each combination of categories of A and B, or the AB margin. To estimate the parameters, we will use the iteratively-reweighted least squares (IRLS) algorithm the working dependent variable z z i = η i + y i ˆµ i ˆµ i since η i µ i = ln µ i µ i = 1 µ i and the diagonal matrix W of iterative weights has elements w ii = [V ar(z i )] 1 = [V ar(y i ) ( η i µ i ) 2 ] 1 = [ˆµ i ( 1ˆµ i ) 2 ] 1 = ˆµi where ˆµ i is the fitted values based on the current parameter estimates. Initial values can be obtained by applying the link to the data, that is taking the log of the response, and regressing it on the predictors using OLS. The procedure usually converges in a few iterations. 6.5 Goodness of fit and hypothesis test 1. Deviance: It measures the discrepancy between observed and fitted values for Poisson responses: n [ ( ) ] yi n D = 2 y i log (y i ˆµ i ) χ 2 ˆµ n p i i=1 where p the number of parameters. The first term is identical to the binomial deviance and the second term, a sum of differences between observed and fitted values, is usually zero for MLE which march marginal totals. SydU MSH3 GLM (2015) First semester Dr. J. Chan 213

8 2. Pearson s chi-squared statistic: It is the squared difference between observed and fitted values to the variance of the observed value: n X 2 (y i µ i ) 2 n = χ 2 ˆµ n p i i=1 which the same for Poisson and binomial data. 3. Likelihood ratio test: It can be constructed in terms the difference in deviances between two nested model: 2 ln λ = D(ω 1) D(ω 2 ) φ n χ 2 p2 p 1 under the assumption that the smaller model ω 1 is correct in H Wald test: To test H 0 : β = β 0, the test statistics is W 0 = ( β β 0 ) [var( β)] 1 ( β β 0 ) n based on the fact that the MLE β N p (β, X W X) where W is the diagonal matrix of estimation weights w ii. 6.6 Children ever born example Grouped Data and the Offset Let Y ijkl denote the number of children from the l-th woman in the (i, j, k)-th group, where i denotes marital duration, j residence and k education and with mean µ ijk. Let Y ijk = l Y ijkl denote the group total. The group total is a realization of a Poisson variate with mean n ijk µ ijk, where n ijk is the number of observations in the (i, j, k)-th cell. Suppose that a log-linear model is postulated for the individual means: log E(Y ijkl ) = log(µ ijk ) = x ijkβ SydU MSH3 GLM (2015) First semester Dr. J. Chan 214 χ 2 p

9 where x ijk is a vector of covariates. Then the log of the expected value of the group total is log E(Y ijk ) = log(n ijk µ ijk ) = log(n ijk ) + x ijkβ. Thus the group totals have the same coefficients β as the individual means, except that the linear predictor includes the term log(n ijk ) called an offset which is the log of exposure measures, the number of women here. Thus, we can fit log-linear models to the individual counts or group totals with the log of the number of women in each cell as an offset. The deviances is different because they measure goodness of fit to different sets of counts but differences of deviances between nested models are exactly the same The Deviance Table Table 7.2 reports the results of fitting a variety of Poisson models. The null model assuming the same number of children for all the groups is rejected based (D = 3732 on 69 df). Introducing marital duration reduces the deviance to on 64 df. The drop of 3566 on 5 df shows that the number of CEB depends on the duration since the first marriage. To test the effect of education controlling for duration, the additive model D + E relative to the one-factor D model gives a chi-squared statistic of 65.8 ( ) on 3 (64-61) df and is highly significant. We can also test the net effect of education controlling for residence & duration, by comparing the 3-factor additive model D + R + E with the 2-factor model D + R. The difference in deviances of 50.1 ( ) on 3 (62-59) df is highly significant. This smaller Chi-square statistic indicates that part of the education effect may due to the fact that more educated women live in Suva or in other urban areas. SydU MSH3 GLM (2015) First semester Dr. J. Chan 215

10 Table 6.2: Deviances for Poisson log-linear models fitted to CEB data Type Model Deviance d.f. Null factor Models Duration Residence Education factor Models D + R D + E DR DE factor Models D + R + E D + RE E + DR R + DE DR + RE DE + RE DR + DE DR + DE + RE Does education make more of a difference in rural areas than in urban areas? To answer this question we compare models D+RE to D+R+E. The reduction in deviance is 10.8 ( ) on 6 (59-53) df and is not significant, with a P-value of Does the effect of education increase with marital duration? Adding DE (model R+DE) to model D+R+E reduces the deviance by 15.7 ( ) at the cost of 15 df (59-44), hardly a bargain. Thus, we conclude that the additive model which has a deviance of on 59 d.f. (an average of 1.2 per d.f.) is adequate for these data. SydU MSH3 GLM (2015) First semester Dr. J. Chan 216

11 6.6.3 The Additive Model The associated P-value of the D+R+E model is 0.14 (Pr(χ 2 59 > 70.65) = 0.14), so the model passes the goodness-of-fit test. Table 6.3 shows its parameter estimates and standard errors. Duration. The constant represents the log of the mean number of CEB for the reference cell, which is Suvanese women with no education and 0-4 years marriage duration. Since exp(0.1173) = 0.89, these women have 0.89 children on average at this time in their lives. The duration parameters trace the increase in CEB with duration for any residence and education group. The model is multiplicative in the original scale. From duration 0-4 to 5-9 years, the number of CEB gets multiplied by exp(0.9977) = By duration years, the number of CEB is exp(1.977) = 7.22 times those at duration 0-4 for any residence type and education level. Residence. The effects of residence show that Suvanese women have the lowest fertility. At any given marriage duration, women living in other urban areas have 12% larger families (exp(0.1123) = 1.12) than Suvanese women with the same level of education. Similarly, at any fixed duration, women who live in rural areas have 16% more children (exp(0.1512) = 1.16), than Suvanese women with the same level of education. Education. Higher education is associated with smaller family sizes net of duration and residence. At any given marriage duration, women with upper primary education have 10% fewer kids, and women with secondary or higher education have 27% (1 exp( ) = 0.27) fewer kids, than women with no education in the same type of residence. SydU MSH3 GLM (2015) First semester Dr. J. Chan 217

12 Table 6.3: Estimates for additive log-linear model of CEB data Parameter Estimate Std. Error z-ratio Constant Duration Residence Suva - Urban Rural Education None - Lower Upper Sec The educational differential of 27% between these two groups translates into a quarter of a child at durations 0-4, increases to about one child around duration 15, and reaches almost one and three quarter children by duration 25+. Thus, the effect of education increases with marital duration. (5-9, Sec+: exp( ) = 1.77) Table 6.4: Fitted values for Suvanese women with no education and with secondary or higher Education Marital Duration No Education Secondary Difference SydU MSH3 GLM (2015) First semester Dr. J. Chan 218

13 > rm(list=ls()) > data=read.table("ceb.txt",skip=1) > mu=data[,1] > ni=data[,3] > ln=log(ni) > dur=as.factor(data[,4]) > resid=as.factor(data[,5]) > edu=as.factor(data[,6]) > y=round(mu*ni,0) > d0=glm(y~offset(ln),family=poisson)$dev > d1=glm(y~offset(ln)+dur,family=poisson)$dev > d2=glm(y~offset(ln)+resid,family=poisson)$dev > d3=glm(y~offset(ln)+edu,family=poisson)$dev > d4=glm(y~offset(ln)+dur+resid,family=poisson)$dev > d5=glm(y~offset(ln)+dur+edu,family=poisson)$dev > d6=glm(y~offset(ln)+dur*resid,family=poisson)$dev > d7=glm(y~offset(ln)+dur*edu,family=poisson)$dev > d8=glm(y~offset(ln)+dur+edu+resid,family=poisson)$dev > d9=glm(y~offset(ln)+dur+edu*resid,family=poisson)$dev > d10=glm(y~offset(ln)+dur*resid+edu,family=poisson)$dev > d11=glm(y~offset(ln)+dur*edu+resid,family=poisson)$dev > d12=glm(y~offset(ln)+dur*resid+edu*resid,family=poisson)$dev > d13=glm(y~offset(ln)+dur*edu+resid*edu,family=poisson)$dev > d14=glm(y~offset(ln)+dur*resid+dur*edu,family=poisson)$dev > d15=glm(y~offset(ln)+dur*resid+dur*edu+resid*edu,family=poisson)$dev > c(d0,d1,d2,d3,d4,d5,d6) [1] > c(d7,d8,d9,d10,d11,d12,d13) [1] > summary(glm(y~offset(ln)+dur+resid+edu,family=poisson)) Call: glm(formula = y ~ offset(ln) + dur + resid + edu, family = poisson) Deviance Residuals: Min 1Q Median 3Q Max SydU MSH3 GLM (2015) First semester Dr. J. Chan 219

14 Coefficients: Estimate Std. Error z value Pr(> z ) (Intercept) * dur < 2e-16 *** dur < 2e-16 *** dur < 2e-16 *** dur < 2e-16 *** dur < 2e-16 *** resid *** resid e-08 *** edu edu ** edu e-08 *** --- Signif. codes: 0 *** ** 0.01 * (Dispersion parameter for poisson family taken to be 1) Null deviance: on 69 degrees of freedom Residual deviance: on 59 degrees of freedom AIC: Number of Fisher Scoring iterations: Negative binomial distribution Negative binomial (NB) distribution is a discrete probability distribution for the number of successes in a sequence of iid Bernoulli trials before a specified number of failures, r, occurs. It s pmf, mean and variance are ( ) y + r 1 f(y) = (1 p) r p y, x E(Y ) = pr 1 p, and V ar(y ) = pr (1 p) 2. SydU MSH3 GLM (2015) First semester Dr. J. Chan 220

15 As V ar(y ) > E(Y ), it is overdispersed. Being a Poisson-Gamma mixture, p Y NB(r, p) Y P (λ), λ G(r, 1 p ) it has higher variability than Poisson. The proof is: f(y; r, p) = 0 λ y e λ y! = (1 p)r p r y!γ(r) = (1 p)r p r = λ r 1 e λ(1 p)/p dλ ( p 1 p )r Γ(r) y!γ(r) Γ(r + y) y!γ(r) py (1 p) r = 0 λ r+y 1 e λ/p dλ p r+y Γ(r + y) ( y + r 1 Poisson and NB distributions are related as: ( r, P (λ) = lim r NB y λ λ + r since taking expectation on both sides, λ = rp 1 p p = E(Y ) = V ar(y ) = lim r lim r pr 1 p = lim r pr (1 p) 2 = lim r λ λ+r r 1 λ λ+r = λ λ λ+r r (1 λ λ+r ) ) (1 p) r p y. = lim λλ + r )2 r r In fitting a negative binomial distribution to the CEB data, > library(mass) > glm.nb(y ~ offset(ln)+dur+resid+edu) λ λ+r. Note that = λ. Call: glm.nb(formula = y ~ offset(ln) + dur + resid + edu, init.theta = , link = log) SydU MSH3 GLM (2015) First semester Dr. J. Chan 221

16 Coefficients: (Intercept) dur2 dur3 dur4 dur5 dur resid2 resid3 edu2 edu3 edu Degrees of Freedom: 69 Total (i.e. Null); 59 Residual Null Deviance: 3730 Residual Deviance: AIC: There were 27 warnings (use warnings() to see them) > data=read.table("cebmeanvar.txt") > mug=data[,1] #mean by group > var=data[,2] #var by group > mean(mug) [1] > mean(var) [1] There is no improvement in model fit (Residual Deviance & AIC) as the data is equi-dispersed across groups. 6.8 Model for excess zero The model concerns a random event containing excess zero counts Zero-inflated Poisson model Proposed by Lambert (1992), the model contains two components for two zero generating processes: a binary distribution that generates structural zeros and a Poisson or Negative Binomial distributions that generates counts, some of which may be zero. Essentially, it is a mixture model is SydU MSH3 GLM (2015) First semester Dr. J. Chan 222

17 given by: MSH3 Generalized linear model Ch. 6 Count data models Pr(Y = 0) = π + (1 π)e λ Pr(Y = y) = (1 π) λy e λ, y 1 y! where λ is the expected Poisson count here and π is the probability of extra zeros. An example is the number of certain insurance claims within a population that is zero-inflated by those people who have not taken out insurance against the risk and thus are unable to claim. The mean and variance are since E(Y ) = (1 π)λ and V ar(y ) = λ(1 π)(1 + λπ) V ar(y ) = E(Y 2 ) [E(Y )] 2 = (1 π)(λ 2 + λ) (1 π) 2 λ 2 = (1 π)[λ 2 + λ (1 π)λ 2 ] = λ(1 π)(1 + λπ). The method of moments estimators are given by λ = s2 + ȳ 2 ȳ ȳ and π = s2 ȳ s 2 + ȳ 2 ȳ where ȳ is the sample mean and s 2 is the sample variance. The maximum likelihood estimator can be found by solving the following equation ( ȳ(1 e λ ) = λ 1 n ) 0 (1) n for ˆλ where n 0 n is the observed proportion of zeros. This can be solved by iterations or the R module uniroot, and the maximum likelihood estimator for π is given below (proofs in assignment): ˆπ = 1 ȳ ˆλ. SydU MSH3 GLM (2015) First semester Dr. J. Chan 223

18 6.8.2 Hurdle model MSH3 Generalized linear model Ch. 6 Count data models Proposed by Mullahy (1986), the model contains two components: a binary model for measuring whether the outcome overcomes the hurdle (say zero for zero-inflated data) and a truncated model like Poisson or negative binomial (NB) distributions explaining those outcomes that pass the hurdle. The model is given by: Pr(Y = 0) = π λ y e λ Pr(Y = y) = (1 π) y!(1 e λ ), y 1 The loglikelihood function is l(π, λ) = n 0 ln π+n 1 ln(1 π)+ln λ n 1 i=1 n 1 y i n 1 λ y i! n 1 ln(1 e λ ) where n 0 is the count of 0 and n 1 is the count of y i > 0. Hence the MLE are ˆπ = n 0 n and ˆλ is the solution to (1). 6.9 Model selection example Maxell (1961) discusses a 5 4 table giving the number of boys with 4 different ratings for disturbed dreams in 5 different age groups. The higher the rating the more the boys suffer from disturbed dreams. Rating Age i=1 SydU MSH3 GLM (2015) First semester Dr. J. Chan 224

19 One way of modeling the data is to let the number in the (i, j)-th cell be Y ij and assume that Y ij Poisson(µ ij ). Given 5 i=1 4 j=1 Y ij = 223, the joint conditional distribution of the Y ij s is multinomial. We look for a pattern in the µ ij relating to the factors age and rating. We use log link. The saturated model Ω : ln µ ij = µ 0 + α i + β j + (αβ) ij where α i is the age group effect, α 1 = β 1 = 0, β j is the dream rating effect, (αβ) ij is the interaction effects and (αβ) 1j = 0 = (αβ) i1. The null model φ : ln µ ij = µ 0 where ˆµ ij = ȳ = = 11.15, µ 0 = ln µ ij = ln(11.15) = > y=c(7,3,4,7,13,11,15,10,7,11,9,23,10,12,9,28,3,4,5,32) > age=factor(c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5)) > rate=factor(c(1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4)) > glm1=glm(y~age+rate,family=poisson) > summary(glm1) Call: glm(formula = y ~ age + rate, family = poisson) Deviance Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error z value Pr(> z ) (Intercept) e-07 *** age ** age *** age e-05 *** SydU MSH3 GLM (2015) First semester Dr. J. Chan 225

20 age ** rate rate rate e-07 *** --- Signif. codes: 0 *** ** 0.01 * (Dispersion parameter for poisson family taken to be 1) Null deviance: on 19 degrees of freedom #n-p=20-1 Residual deviance: on 12 degrees of freedom #n-p= AIC: Number of Fisher Scoring iterations: 5 > ey=glm1$fitted.values > ey > n=length(y) > mu=mean(y) > mu [1] > D.null=2*sum(y*log(y/mu)-(y-mu)) > D.null [1] > D.add=2*sum(y*log(y/ey)-(y-ey)) #additive age_rate model > D.add [1] > AIC.add=-2*sum(y*log(ey)-ey-log(factorial(y)))+2*(1+3+4) > AIC.add [1] To test if β 1 = β 2 = β 3, i.e., rating can be collapsed to 2-levels, the SydU MSH3 GLM (2015) First semester Dr. J. Chan 226

21 change in deviance, as compared to the saturated model with 0 deviance and 0 d.f., is which distributed as χ Since the p-value > 0.05, accept H 0. > r1=factor(c(1,1,1,2,1,1,1,2,1,1,1,2,1,1,1,2,1,1,1,2)) > glm3=glm(y~age*r1,family=poisson) > summary(glm3) Call: glm(formula = y ~ age * r1, family = poisson) Deviance Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error z value Pr(> z ) (Intercept) e-09 *** age ** age * age * age r age2:r age3:r age4:r age5:r ** --- Signif. codes: 0 *** ** 0.01 * (Dispersion parameter for poisson family taken to be 1) Null deviance: on 19 degrees of freedom Residual deviance: on 10 degrees of freedom #n-p= AIC: SydU MSH3 GLM (2015) First semester Dr. J. Chan 227

22 Number of Fisher Scoring iterations: 4 For residual checks in GLM, the mean and variance are functionally related in interpreting the residuals plot. To help in this problem, we generally use the scaled residuals of some form: 1. Ordinary residuals: R i = Y i ˆµ i but it is useful only in normal case. 2. Pearson (or standardized) residuals: R pi = Y i µ i a(φ)b (ˆθ i ) where Var(Y i ) = a(φ)b (θ i ). For the Poisson case, as Var(Y i ) = E(Y i ). R pi = Y i ˆµ i ˆµi > r=residuals.glm(glm3,type="pearson") > r e e e e e e e e e e e e e e e e e e e e-15 > par=glm3$coeff > r1=(y[1]-exp(par[1]))/sqrt(exp(par[1])) > r1 #agree with the first pearson residual (Intercept) Results of the four models are summarized below: SydU MSH3 GLM (2015) First semester Dr. J. Chan 228

23 Model Dev df Diff dev Diff df p-value Conclusion 1. Saturated age*rate Reduced sat. age*rate Accept M2 3. Additive age+rate Reject M3, favor M1 4. Null Reject M4, favor M3 Note that M2 and M3 are not nested. Reference: 1. Little, J.A. (1979) The General Linear Model and Direct Standardization A Comparison. Sociological methods and research, 7, Lambert, D. (1992) Zero-Inflated Poisson Regression, with an Application to Defects in Manufacturing. Technometrics, 34: Mullahy, J. (1986) Specification and testing of some modified count data models. Journal of Econometrics, 33, SydU MSH3 GLM (2015) First semester Dr. J. Chan 229

24 Appendix MSH3 Generalized linear model Ch. 6 Count data models Definition: A stochastic process {N(t), t 0} is a counting process if N(t) represents the total number of events that have occured by time t. Note: N(t) 0. If s < t, N(s) N(t). Definition: A stochastic process is independent increment the number of events in disjoint intervals are independent. Definition: A stochastic process is stationary increment if the distributions of N(t 1 + s) N(t 2 + s) and N(t 1 ) N(t 2 ) are equal. Lemma: The counting process {N(t), t 0} is a Poisson process with rate λ if 1. N(0) = 0 2. The process has stationary and independent increment. 3. Pr(N(h) = 1) = λh + o(h) 4. Pr(N(h) 2) = o(h) where f(x) = o(x) if f(x) x Proof: Let p n (t) = Pr(N(t) = n). We have 0 as x 0 and o(h) ± o(h) = o(h). p 0 (t + h) = Pr(N(t) = 0, N(t + h) N(t) = 0) Hence = Pr(N(t) = 0) Pr(N(h) = 0) = p 0 (t)(1 λh + o(h)) by (2-4) p p 0 (t + h) p 0 (t) 0(t) = lim = lim λp 0 (t) + o(h) h 0 h h 0 h p 0(t) = λp 0 (t) p 0(t) p 0 (t) = λ ln p 0(t) = λt + c p 0 (t) = ke λt SydU MSH3 GLM (2015) First semester Dr. J. Chan 230

25 Since p 0 (0) = Pr(N(0) = 0) = 1 k = 1, we have p 0 (t) = e λt. Then assuming p n 1 (t) = (λt)n 1 e λt using mathematical induction, we have (n 1)! p n (t + h) = Pr(N(t) = n, N(t + h) N(t) = 0) + Pr(N(t) = n 1, N(t + h) N(t) = 1) n + Pr(N(t) = n k, N(t + h) N(t) = k) k=2 = Pr(N(t) = n) Pr(N(h) = 0) + Pr(N(t) = n 1) Pr(N(h) = 1) + o(h) = p n (t)(1 λh + o(h)) + p n 1 (t)(λh + o(h)) + o(h) p n (t + h) p n (t) = p n (t)λ + p n (t) o(h) h h p n(t) = λp n (t) + λp n 1 (t) + p n 1(t)λ + p n 1 (t) o(h) h + o(h) h Implies p n(t) + λp n (t) = λ (λt)n 1 e λt (n 1)! = λn t n 1 e λt (n 1)! The solution to the differential equation is dy + h(t)y = g(t) dt e h(t)dt g(t)dt + c y =. e h(t)dt Now h(t) = λ and g(t) = λn t n 1 e λt. We have (n 1)! (e λdt )[ λn t n 1 e λt (n 1)! ]dt + c (e λt )[ λn t n 1 e λt (n 1)! ]dt + c p n (t) = = e λdt tn 1 λn (n 1)! = dt + c = ( e λt ) (λt) = e λt n + c. n! λn tn n(n 1)! + c e λt SydU MSH3 GLM (2015) First semester Dr. J. Chan 231 e λt

26 Since p n (0) = Pr(N(0) = n) = 0 c = 0, n 1, we have p n (t) = e λt (λt) n n! which is the pmf of Poisson distribution. SydU MSH3 GLM (2015) First semester Dr. J. Chan 232