In Poisson regression I need to compute the deviance, in order to do that I need to compute the log-likelihood function. It seems not difficult because I have the estimated model and my data set I need only to apply the next formulas given in Wikipedia.

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But I need to compare this model, with the saturated model i.e a regression with 61 parameters that is the number of observations, and the null model that is the model only with the intercept.

In both cases, how to compute the log-likelihood function?

Also, I am confused since I am trying to replicate the computations of the log-likelihood function in excel, In R I have used the function $logLik(model)$ in excel the formula given above, but when I sum I got completely different values. How to replicate the R estimator?


In Poisson regression there are two Deviances.

The Null Deviance shows how well the response variable is predicted by a model that includes only the intercept (grand mean).

And the Residual Deviance is −2 times the difference between the log-likelihood evaluated at the maximum likelihood estimate (MLE) and the log-likelihood for a "saturated model" (a theoretical model with a separate parameter for each observation and thus a perfect fit).

Now let us write down those likelihood functions.

Suppose $Y$ has a Poisson distribution whose mean depends on vector $\bf{x}$, for simplicity, we will suppose $\bf{x}$ only has one predictor variable. We write $$ E(Y|x)=\lambda(x) $$ For Poisson regression we can choose a log or an identity link function, we choose a log link here.

$$Log(\lambda (x))=\beta_0+\beta_1x$$

$\beta_0$ is the intercept.

The Likelihood function with the parameter $\beta_0$ and $\beta_1$ is $$ L(\beta_0,\beta_1;y_i)=\prod_{i=1}^{n}\frac{e^{-\lambda{(x_i)}}[\lambda(x_i)]^{y_i}}{y_i!}=\prod_{i=1}^{n}\frac{e^{-e^{(\beta_0+\beta_1x_i)}}\left [e^{(\beta_0+\beta_1x_i)}\right ]^{y_i}}{y_i!} $$ The log likelihood function is:

$$ l(\beta_0,\beta_1;y_i)=-\sum_{i=1}^n e^{(\beta_0+\beta_1x_i)}+\sum_{i=1}^ny_i (\beta_0+\beta_1x_i)-\sum_{i=1}^n\log(y_i!) \tag{1} $$ When we calculate null deviance we will plug in $\beta_0$ into $(1)$. $\beta_0$ will be calculated by a intercept only regression, $\beta_1$ will be set to zero. We write $$ l(\beta_0;y_i)=-\sum_{i=1}^ne^{\beta_0}+\sum_{i=1}^ny_i\beta_0-\sum_{i=1}^n \log(y_i!) \tag{2} $$ Next we need to calculate the log likelihood for "saturated model" (a theoretical model with a separate parameter for each observation), therefore, we have $\mu_1,\mu_2,...,\mu_n$ parameters here.

(Note, in $(1)$, we only have two parameters, i.e. as long as the subject have same value for the predictor variables we think they are the same).

The log likelihood function for "saturated model" is $$ l(\mu)=\sum_{i=1}^n y_i \log\mu_i-\sum_{i=1}^n\mu_i-\sum_{i=1}^n \log(y_i!) $$ Then it can be written as: $$ l(\mu)=\sum y_i I_{(y_i>0)} \log\mu_i-\sum\mu_iI_{(y_i>0)}-\sum \log(y_i!)I_{(y_i>0)}-\sum\mu_iI_{(y_i=0)} \tag{3} $$ (Note, $y_i\ge 0$, when $y_i=0,y_i\log\mu_i=0$ and $\log(y_i!)=0$, this will be useful later, not now) $$ \frac{\partial}{\partial \mu_i}l(\mu)=\frac{y_i}{\mu_i}-1 $$ set to zero, we get

$$ \hat{\mu_i}=y_i $$ Now put $\hat{\mu_i}$ into $(3)$, since when $y_i=0$ we can see that $\hat{\mu_i}$ will be zero.

Now for the likelihood function $(3)$of the "saturated model" we can only care $y_i>0$, we write $$ l(\hat{\mu})=\sum y_i \log{y_i}-\sum y_i-\sum \log(y_i!) \tag{4} $$

From $(4)$ you can see why we need $(3)$ since $\log y_i$ will be undefied when $y_i=0$

Now the let us calcualte the deviances.

The Residual Deviance=$$-2[(1)-(4)]=-2*[l(\beta_0,\beta_1;y_i)-l(\hat{\mu})]\tag{5}$$

The Null Deviance= $$-2[(2)-(4)]=-2*[l(\beta_0;y_i)-l(\hat{\mu})]\tag{6} $$

Ok, next let us calculate the two Deviances by R then by "hand" or excel.

x<- c(2,15,19,14,16,15,9,17,10,23,14,14,9,5,17,16,13,6,16,19,24,9,12,7,9,7,15,21,20,20)
p_glm<-glm(y~x, family=poisson, data=p_data)

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You can see $\beta_0=0.30787,\beta_1=0.07636$, Null Deviance=48.31, Residual Deviance=27.84.

Here is the intercept only model

p_glm2<-glm(y~1,family=poisson, data=p_data)

enter image description here

You can see $\beta_0=1.44299$

Now let us calculate these two Deviances by hand (or by excel)


for(i in 1:30){
l_regression[i]<--exp( 0.30787 +0.07636 *x[i])+y[i]*(0.30787+0.07636 *x[i])-

# -60.25116 ###log likelihood for regression model

for(i in 1:30){
}     #there is one y_i=0 need to take care
#-46.33012 ###log likelihood for saturated model

for(i in 1:30){


#-70.48501 ##log likelihood for intercept model only


#27.84209 ##Residual Deviance


##48.30979 ##Null Deviance

You can see use these formulas and calculate by hand you can get exactly the same numbers as calculated by GLM function of R.

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  • $\begingroup$ Is it valid to compare two models performance by the log likelihoods with anova(glm1, glm2, test 0 "Chisq") in R then? This being for a model with one dropped parameter in an attempt to see what contribution that parameter has on the model. $\endgroup$ – Lamma Jul 10 at 11:30

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