Solved – How to calculate confidence interval of incidence rate under the Poisson distribution

confidence intervalnormal-approximationpoisson distribution

I'd like to be able to calculate the confidence interval for an incidence rate under the Poisson distribution. I am inclined to use a normal approximation formula (the Wald interval) which is $\text{IR} \mp 1.96\,\sqrt{\text{IR}/N}$

However, most sources on line (for example here: use $\text{IR} \mp 1.96\,\text{IR}/\sqrt{N}$, which is not the same thing and the results are quite different.

Does anyone happen to know what is this formula and which one should I use?

Best Answer

Assuming I understand the situation correctly:

the number of incidents for an individual is $Y_i \sim \text{Pois}(\lambda \, T_i)$ where the incidence rate per unit time is $\lambda$, $T_i$ is the exposure time for individual $i$ and the $Y_i$ variates are independent. Then we define the random variable $N=\sum_i Y_i$ and the total exposure $T=\sum_i T_i$, in which case $\hat{\lambda}=N/T$ is the MLE of $\lambda$

you seek a CI for $\hat{\lambda}$

$\text{Var}(\hat{\lambda}) = \text{Var}(\frac{N}{T}) = \frac{1}{T^2}\text{Var}(N)=\frac{1}{T^2}\text{Var}(\sum_iY_i)=\frac{1}{T^2}(\sum_i\lambda T_i)=\lambda/T$

Hence the estimated standard error is $\sqrt{\hat{\lambda}/T} = \frac{\sqrt{N}}{T} = \frac{N\times \frac{1}{\sqrt{N}}}{T}=\frac{N}{T}\times \frac{1}{\sqrt{N}}=\frac{\text{IR}}{\sqrt{N}}$

So (if $N>0$), either of these two formulas is okay:

$\text{s.e.}(\text{IR}) = \sqrt{\frac{IR}{T}} = \frac{IR}{\sqrt{N}}$

and so a large sample (large-$N$) interval for the $\text{IR}$ would multiply that by a suitable $Z$ value to get the interval half-width either side of the estimate.

You seem to have perhaps muddled up the roles of $N$ and $T$ in your calculation

Actually, it looks like the link you gave explained how their formula came about pretty clearly in the couple of lines under "If $N$ is large".

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