# Derivation of confidence interval of incidence rate ratio

I am trying to understand the confidence interval equation for a Incidence Rate Ratio (IRR) given several places: $$95\text{% CL(IRR)} = \exp(\log(\text{IRR}) \pm 1.96\times \text{SE(log(IRR))})$$, with $$\text{SE(log(IRR))} = \sqrt{\frac{1}{e_1} + \frac{1}{e_2}}$$, where $$e_{1,2}$$ is the number of events in each arm. (See here for example)

Though the equation seems to be written many places I have a hard time finding a derivation or good discussion besides lofty sayings such as "because it is a ratio we take the log" which doesn't really explain anything (to me at least).

The setup for each Incidence Rate ($$\text{IR}$$) is given by:

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 $$e=\sum_i Y_i$$ and the total exposure $$T=\sum_i T_i$$ and the Incidence $$\text{IR} = \frac{e}{T}$$.

I understand why the standard deviation becomes $$\sigma\text{(IR)} = \frac{\sqrt{e}}{T}$$ as discussed here but how I get to the IRR equation is not clear to me.

I tried approximating both IR's as Gaussian and taking the ratio assuming that to be Gaussian as well (which is discussed more thoroughly here) which leads to:

$$95\text{% CL(IRR)} = \text{IRR} \pm 1.96\times \text{IRR} \times \sqrt{\frac{1}{e_1} + \frac{1}{e_2}}$$ which seems to be just the lowest order expansion of the top equation.

So why are we taking the logarithm and why is that alright given that it is a ratio of 2 scaled Poisson variables? Why does $$\log(\text{IRR})$$ end up being Gaussian and does that imply that $$\log(\text{IR}_{1,2})$$ is Gaussian as well?

• Thank you for your answer! I now understand what is being transformed is the expectation value. Using the propagation for error approximation I now have a clear path to the equation for $\sigma(\log(IRR))$. I still do not get the 1.96 from a chi-squared distribution. Also the approximation I have in mind does not seem to be dependent on log being the canonical transformation but "just" on ordinary log properties( meaning I expect the method for deriving $\sigma(log(some ratio))$ will work for any ratio). Can you elaborate a bit on the 1.96 and why it has to be the canonical transformation? Commented Jul 8, 2021 at 9:22
• @JensT, reading more on the Wald test may help: en.wikipedia.org/wiki/Wald_test, One way of constructing a confidence interval is to invert the test, i.e. find all the $theta_0$ values that would not cause the test to reject (this can be found with some algebra on the test statistic). Since the Wald statistic is asymptotically Chi-squared we plug in the x value that gives us the corresponding area under the curve for W (this gives us the 1.96 for 95% confidence). Commented Jul 8, 2021 at 15:13