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01-12-2017 12:10 PM

Hi All,

I ran into a Poisson regression analysis issue. My data contain "counts" for an event and "days" for the event. I am fosusing on the "rate", which is counts/days. Usually people would use Poisson regression to fit the rate data by doing:

**Model Event = treat /offset=log(days).**

However, I am wondering if it makes sense to add a "rate" variable in the data set by doing: event_rate = events/days, and then fit the Poisson model:

**Model event_rate = treat.**

Thanks for your comments and help.

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Solution

01-13-2017
09:19 AM

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01-13-2017 12:10 AM - edited 01-13-2017 12:10 AM

The models look the same, but the likelihood equations aren't. The estimates will be the same (for offset and event_rate= event/days models) but the standard error of the estimates will differ. Look at this small simulation:

```
data a;
call streaminit(7576);
rate = 10;
do days = 10, 100;
logDays = log(days);
do i = 1 to 20;
event = rand("Poisson", rate*days);
event_rate1 = event / days;
event_rate2 = event / days * mean(10, 100);
event_rate3 = event / days * 1000;
output;
end;
end;
run;
ods select ParameterEstimates(persist);
title "Estimate of log number per day with Offset";
proc glimmix data=a;
model event = / dist=poisson offset=logDays solution;
run;
title "Estimate of log rate per day";
proc glimmix data=a;
model event_rate1 = / dist=poisson solution;
run;
title "Estimate of log rate per 55 days";
proc glimmix data=a;
model event_rate2 = / dist=poisson solution;
run;
title "Estimate of log rate per 1000 days";
proc glimmix data=a;
model event_rate3 = / dist=poisson solution;
run;
```

Estimate of log number per day with Offset The GLIMMIX Procedure Parameter Estimates Standard Effect Estimate Error DF t Value Pr > |t| Intercept 2.3091 0.006720 39 343.62 <.0001 Estimate of log rate per day Parameter Estimates Standard Effect Estimate Error DF t Value Pr > |t| Intercept 2.2992 0.05008 39 45.91 <.0001 Estimate of log rate per 55 days Parameter Estimates Standard Effect Estimate Error DF t Value Pr > |t| Intercept 6.3065 0.006753 39 933.83 <.0001 Estimate of log rate per 1000 days Parameter Estimates Standard Effect Estimate Error DF t Value Pr > |t| Intercept 9.2070 0.001584 39 5813.16 <.0001

PG

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01-12-2017 06:07 PM - edited 01-12-2017 06:10 PM

Unless your number of days varies very little, you should stick with the offset model to get valid inferences.

If you decide to go with event_rate, it should be defined as event/days*meanDays, where meanDays is the mean number of days in your data.

PG

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01-12-2017 06:56 PM

Hi PG,

Thanks for the comment. That is very helpful. Just for my personal education. Why the offset model gives better inference? Mathematically the offset model and the event_rate are the same, right?

log(u/t) = alpha+beta*x --> log(u) - log(t) = alpha+beta*x --> log(u) = log(t) + alpha+beta*x

Thanks,

Solution

01-13-2017
09:19 AM

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01-13-2017 12:10 AM - edited 01-13-2017 12:10 AM

The models look the same, but the likelihood equations aren't. The estimates will be the same (for offset and event_rate= event/days models) but the standard error of the estimates will differ. Look at this small simulation:

```
data a;
call streaminit(7576);
rate = 10;
do days = 10, 100;
logDays = log(days);
do i = 1 to 20;
event = rand("Poisson", rate*days);
event_rate1 = event / days;
event_rate2 = event / days * mean(10, 100);
event_rate3 = event / days * 1000;
output;
end;
end;
run;
ods select ParameterEstimates(persist);
title "Estimate of log number per day with Offset";
proc glimmix data=a;
model event = / dist=poisson offset=logDays solution;
run;
title "Estimate of log rate per day";
proc glimmix data=a;
model event_rate1 = / dist=poisson solution;
run;
title "Estimate of log rate per 55 days";
proc glimmix data=a;
model event_rate2 = / dist=poisson solution;
run;
title "Estimate of log rate per 1000 days";
proc glimmix data=a;
model event_rate3 = / dist=poisson solution;
run;
```

Estimate of log number per day with Offset The GLIMMIX Procedure Parameter Estimates Standard Effect Estimate Error DF t Value Pr > |t| Intercept 2.3091 0.006720 39 343.62 <.0001 Estimate of log rate per day Parameter Estimates Standard Effect Estimate Error DF t Value Pr > |t| Intercept 2.2992 0.05008 39 45.91 <.0001 Estimate of log rate per 55 days Parameter Estimates Standard Effect Estimate Error DF t Value Pr > |t| Intercept 6.3065 0.006753 39 933.83 <.0001 Estimate of log rate per 1000 days Parameter Estimates Standard Effect Estimate Error DF t Value Pr > |t| Intercept 9.2070 0.001584 39 5813.16 <.0001

PG

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01-13-2017 09:20 AM

Thanks @PG, this is very helpful!

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01-13-2017 10:07 AM

Again, thanks @PG. Sorry for having so many questions. Your simulation is very helpful. However, if one wants to pick an estimate, which one should he use?

It seems that using "Offset" gives pretty similar "estimate" as "Rate" does, although the SEs are different. But if using "Rate*Mean(days)", it gives completely different "estimate" but similar SEs.

I also tried my real data using PROC GENMOD on both Offset model and Rate model. It seems that the two give pretty close results in terms of "estimate" and "SE".

Here are my code snippets:

1. Poisson model with offset

proc genmod data=final; title "BV-003 Part 2 - Poisson Model"; class ptno treat; model numcr = treat /type3 dist=poi offset=logdayct; repeated subject=ptno; estimate 'D vs A' treat -1 1 0/exp; estimate 'E vs A' treat -1 0 1/exp; estimate 'D vs E' treat 0 1 -1/exp; run;

2. Poisson model with rate data:

proc genmod data=final; title "BV-003 Part 2 - Poisson Model on the Cramp/Day Ratio"; class ptno treat; model crp_day = treat /type3 dist=poi; repeated subject=ptno; estimate 'D vs A' treat -1 1 0/exp; estimate 'E vs A' treat -1 0 1/exp; estimate 'E vs D' treat 0 1 -1/exp; run;

And here are the outputs:

For 1,

For 2,

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01-18-2017 04:31 AM

I think that you can model the rate and get exactly same estimates and standard errors as if you model the count. But, the person-years should then be used in the weight statement.

The two proc genmod statements below seems to give exactly same output.

**data** test;

do i=**1** to **100**;

pyrs=rand('uniform',**0**,**5**);

a=rand('bernoulli',**0.5**);

y=rand('poisson',pyrs*exp(**2**+a***1**));

logpyrs=log(pyrs);

rate=y/pyrs;

output;

end;

**run**;

**proc** **genmod** data=test;

class a(ref="0");

model y=a/dist=poisson link=log offset=logpyrs;

**run**;

**proc** **genmod** data=test;

class a(ref="0");

model rate=a/dist=poisson link=log;

weight pyrs;

**run**;

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01-18-2017 07:45 AM

Thanks, @JacobSimonsen. This is very helpful.

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01-18-2017 08:13 AM

its because the expression for the likelihood function λ^{x} e^{-pyrs*λ} is the same as (λ^{x/pyrs} e^{-λ})^{pyrs}