For my study I am doubting about the following:

  1. First I want to present the incidence rate for patients who received an CT-scan.

    I thought I would just count all the CT-scans that took place and divide this by the time all patients are at risk. So, e.g.:

    Patient 1: 1 CT-scan in 365 days;
    Patient 2: 3 CT-scans in 250 days; 
    Patient 3: 1 CT-scan in 10 days;
    Total --> 5/625 = 0.008
  2. But then I thought about what would happen if I calculated the incidence rate for each patient separately and add this (instead of just the total as showed above).

    We would get something like this for the patients above:

    1/365 + 3/250 + 1/10 = 0.002739 + 0.0120 + 0.1 = 0.114739

What I understand is that the first way gives more weight to patients who are a long time at risk for getting a CT-scan, while the second way gives more weight to patients who are a short time at risk for getting a CT-scan.

What I don't know is if it is even justified to use the second way of calculating? And where I can't get my head around is what it means for my results if I use the second way instead of the first way - what is the difference between the first answer (0.008) and the second answer (0.114739)?


It's a simple rule that the sum of ratios is not the same as the ratio of sums. This is a result of Jensen's inequality. To really compare apples and oranges, you need to "average" out the second expression which is an average incidence of 0.038, still much higher than the first expression 0.008, but both can be described as the average incidence of the sample.

These are two different ways of summarizing incidence. The first approach is called a geometric mean. You would get 0.008 from $\exp(\log(1/365) + \log(3/250) + \log(1/10))$. The geometric mean tends toward the median more than the arithmetic mean. You can also refer to the geometric mean in this case as a Poisson intensity. It's desirable that the subject with the highest incidence: 1 scan every 10 days, is heavily downweighted by the fact they're only observed for 10 days. That's because this very high incidence has very high variability.

  • $\begingroup$ Thank you for your answer. So do I correctly conclude if I say that 0.008 is a geometric mean and tends toward the median more than the arithmetic mean (which is 0.038)? Thereby I am still wondering which way I do the most right (downweight the most) to the patient who receives 1 scan in only 10 days? $\endgroup$ – Isa Feb 28 at 10:49
  • $\begingroup$ @isa my answer claims that geometric mean downweights the observation with incidence 1/10. $\endgroup$ – AdamO Feb 28 at 16:02

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