Solved – Use of glm() and graph of regression line

generalized linear modelr

I am new to glms and have picked up the following text, I am trying to do the exercises and I'm a little stuck on exercises 4.5 question 4.1. The question states that a possible model for the data is a poisson distribution with parameter $\lambda_i = i^\theta$ where $i= (1,2,\dots,20)$ is the time index. I want to fit a poisson regression in R using the log link function, such that:
$$
g(\lambda_i)=\log(\lambda_i) = \beta_1 + \beta_2 \log i
$$
In R, I've done the following:

y<-c(1,6,16,23,27,39,31,30,43,51,63,70,88,97,91,104,110,113,149,159)
x<-1:20

I'm confused about the glm function, I'm pretty sure I should be fitting:

n1<-glm( y~log(x), family = poisson (link = log) )
plot(log(x),y)

What I'm finding hard to understand is when plotting the regression line, we should be plotting:

$$
\lambda_i =\exp ( \beta_1 + \beta_2 x_i)
$$
So we should have:

plot(log(x),y)
lines(log(x), exp(n1$fit))

which doesn't give a decent looking result, although

lines(log(x),n1$fit)

seems to be the right way to go, but doesn't make intuitive sense to me, aren't the fit values giving the linear part of the model??

Best Answer

The documentation clears this up. About halfway through it describes the object glm returns, which has the attribute

fitted.values   
  the fitted mean values, obtained by transforming the linear predictors by 
the inverse of the link function.

When you do n1$fit, R does partial name matching, and gives you n1$fitted.values.

Related Solutions

Solved – Fitting a Generalized Linear Model (GLM) in R

There are three components to the GLM: an outcome variable, a linear predictor and a link function. The link function in the GLM relates the expected value of the outcome variable to the linear predictor. In other words, not the expected value itself, but a function of it is modeled by the linear predictor. An example with the logarithm as the link function and the linear predictor $\beta_0 + \beta_1*x$ is:

$$\log(E(y)) = \beta_0 + \beta_1*x$$

In your case, the linear predictor is $\log(\beta_0) + \beta_1*\log({\rm exp}_1) + \beta_2*\log({\rm exp}_2)$. So the equation for your model becomes:

$$\log(E(y)) = \log(\beta_0) + \beta_1*\log({\rm exp}_1) + \beta_2*\log({\rm exp}_2)$$

I think this is a bit weird and I would argue that possibly that's not the model you are supposed to fit. Anyway, to fit this model with R, the code should look like this:

model <- glm(formula = Y ~ log(exp1) + log(exp2), family = poisson(link="log"), 
             data = CSV_table)

The only thing you have to take care of after running the model is to take the exponential function of the intercept, if you want to write the intercept as a log. A good book if you want to learn about the GLM and categorical data analysis in general is the one by Agresti (2007).

References:

_{Agresti, A. (1996). An introduction to categorical data analysis (Vol. 135). New York: Wiley.}

GLM Analysis – Understanding OLS vs. Poisson GLM with Identity Link Function

Yes, they're the same thing. MLE for a Gaussian is least squares, so when you do a Gaussian GLM with identity link, you're doing OLS.
a) "I thought both methods would try to estimate E(Y|X)"

Indeed, they do, but the way that conditional expectation is estimated as a function of the data is not the same. Even if we ignore the distribution (and hence how the data enter the likelihood) and think about the GLM just in terms of mean and variance (as if it were just a weighted regression), the variance of a Poisson increases with the mean, so the relative weights on observations would be different.

b) "What does the likelihood function look like when I use the identity link for Poisson?"

$\mathcal{L}(\beta_0,\beta_1) = \prod_i e^{-\lambda_i}\lambda_i^{y_i}/y_i!$

$\qquad\qquad\,=\exp(\sum_i -\lambda_i+{y_i}\log(\lambda_i)-\log{(y_i!)}\,)\quad$ where $\lambda_i=\beta_0+\beta_1 x_i$

$\qquad\qquad\,=\exp(\sum_i -(\beta_0+\beta_1 x_i)+{y_i}\log(\beta_0+\beta_1 x_i)-\log{(y_i!)}\,)$

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