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?


1 Answer 1


Reading up on the background of generalized linear models (glm) may help. The Wikipedia article may help: https://en.wikipedia.org/wiki/Generalized_linear_model

Generalized linear models work with distributions that can be factored into a certain pattern (Exponential family), part of that factoring gives what is called the "canonical link" function between the mean of the distribution and the linear combination of the predictors. In the case of the Poisson distribution the canonical link is the natural log. You can use other link functions, but the canonical link has some theoretical benefits and is the default for most software. You do not have to use the log, but without good reason to use another link function, it makes the most sense.

There are multiple ways to create confidence intervals, the intervals where you take the maximum likelihood estimate and add and subtract a constant times the standard error is called a Wald Style interval. The constant to multiply actually comes from a chi-squared distribution, but since the chi-squared with 1 degree of freedom is just a standard Gaussian squared, the constant is the same for a single estimated interval. These intervals work well when the estimate of the standard error is stable at the ML estimate and the likelihood function is close to symmetric around the ML estimate. This is exact in the case of a Gaussian likelihood, but is a pretty good approximation most of the time for the Poisson and can be reasonable for the Binomial and other distributions, and it is fairly quick and simple, so a good starting place. But there are also cases where the Wald style intervals do not work well (the main example I am aware of is with the binomial/logistic regression, google for Hauk-Donner effect) and the likelihood ratio or other intervals become much preferred.

  • $\begingroup$ 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? $\endgroup$
    – JensT
    Commented Jul 8, 2021 at 9:22
  • $\begingroup$ @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). $\endgroup$
    – Greg Snow
    Commented Jul 8, 2021 at 15:13
  • $\begingroup$ Well this have been a journey. I think I finally got most of your pointers (still don't quite get why the transformation has to be canonical. Seems to be just a coincidence in this case). Thank you very much for your patience! I really appreciate it. $\endgroup$
    – JensT
    Commented Jul 13, 2021 at 6:33

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