# How to calculate confidence interval of incidence rate under the Poisson 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: http://epid.blogspot.com/2012/08/how-to-calculate-confidence-interval-of.html) 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?

• 1. Can you show your setup (what's IR, what's N? Are they the same as at the link?) and the derivation, including assumptions? 2. Take a look at the discussion in comments under your link, particularly with what's going on with the N and T, since some of that may be relevant to your issue Commented Sep 7, 2017 at 1:03
• I'm guessing you're asking about the situation where 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 that the $Y_i$ variates are independent. Then we define the random variable $N=\sum_i Y_i$ and $T=\sum_i T_i$, in which case $\hat{\lambda}=N/T$ is the MLE of $\lambda$ (this sort of thing is what I meant by "setup and derivation including assumptions"); you seek a CI for $\hat{\lambda}$. Do I have that correct? Commented Sep 7, 2017 at 1:16

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".