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I'm looking for a detailed formulation of the Poisson Regression with time offset.

In the Wikipedia, it is written:

$$\log(\mathbb{E}[Y\mid x]) = \log(exposure) + \beta'x$$

with $Y$ following a Poisson distribution:

$$p(Y = y\mid x;\beta) = \frac{\lambda^y}{y!}e^{-\lambda} $$


This formulation is not clear for me:

  1. What exactly is the response variable?
  2. The Poisson distribution with a parameter $\lambda$ expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate and independently of the time since the last event.

So incorporating time in our regression model, the Poisson distribution should be expressed as below:

$$p(Y=y\mid x,t,\beta) = \frac{(\lambda t)^y}{y!}e^{-\lambda t} $$

with:

  • mean: $\mathbb{E}[Y \mid x,t,\beta] = \lambda t$;
  • variance: $\text{Var}[Y \mid x,t,\beta] = \lambda t$.

And the Poisson Regression model (with the $\log$ link function) is expressed as:

$$\log(\mathbb{E}[Y_i \mid T_i,X_i) = \log(\lambda_i T_i)=\log(T_i) + \beta'X_i$$

In which $\lambda_1 t_1 = y_1$, $\lambda_2 t_2 = y_2$, $\dotsc$ where $y_1,y_2,\dotsc$ are the count events. Is this true?

And here comes another confusion: in $\lambda t$ we already incorporate time element, why do we add an offset term $\log(T_i)$?

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1 Answer 1

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  1. The response variable is $Y$ which is a count variable. As it is in all GLMs, we do not write an "error term". In OLS models note that $Y = \beta'X + \epsilon$ is an equivalent formulation to $E[Y|x] = \beta'X$. GLMs expand this idea by incorporating a link function so that you can model a more general $f(E[Y|x]) = \beta'X$, $f$ is called a link function. For Poisson regression, the link function is a log transform, so the natural parameter scale is the log of a count variable. It is appreciably more complicated by the offset (here, a denominator of time or space); it is considered fixed and can be swapped from the LHS and the RHS of the equation. If you manipulate the equation, the response can be rewritten: $\log(E[Y|x]) - \log(exposure) = \log (E[Y|x] /exposure)$ which, exponentiated, has the appropriate ratio form.

  2. It is confusing and inadvisable to use the notation $\lambda_i t_i$ in my opinion, but essentially this is the more general expression of the mean value for the $i$-th individual. If you assume linearity, which is necessary for the GLM to be an appropriate model: $\log(\lambda_i t_i) = \log(T_i) + \beta'X_i$. Basically, the risk/rate for a single unit of time / space can be written as a linear combination of the risk/rate factors $X_i$ AND it will be PROPORTIONAL for arbitrarily longer durations of time. This enables comparisons between or within subjects, as there is no time/rate interactions.

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    $\begingroup$ If $Y$ is the count variable, then $\frac{Y}{exposure}$ can be very large for an individual presenting in the data base during a very small time interval with one count, making it an outlier. In fact it should not be because of very small time presenting in the data. Another point is that, the definition of Poisson distribution in the wikipedia doesn't make much sense because of the presence of the time factor here. $\endgroup$
    – Metariat
    Commented Jul 19, 2016 at 12:56
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    $\begingroup$ If you have an event within short time, then - under the stationarity assumption made - it actually is an outlier. $\endgroup$
    – Michael M
    Commented Jul 19, 2016 at 13:29
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    $\begingroup$ @matemattica I see no reason why that would be concerning. Not only does that appropriately count as evidence that the event rate is very high in that observation, but also poisson regression accounts for this variability using the mean variance relation. Since the variance is proportional to the mean, large observations are also highly variable ones. $\endgroup$
    – AdamO
    Commented Jul 19, 2016 at 14:17

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