# Estimator for an incidence rate

While going through a statistics course for medicine students, I ran accross a problem related to incidence rates. The context of the problem is a chapter about the Poisson distribution. In the problem, 2300 smokers are followed over a 1 year span during which 24 of them develop lung cancer. They then want to compute the incidence rate of the process and proceed as follows:

$$\text{Incidence rate} = \frac{24}{2300-24/2}$$

At first, I didn't understand why they subtracted $24/2$, but I assumed it was some correction for the fact that since those 24 persons develop the cancer during the year, their time at risk is shorter than that of the ones not developing the disease. No further information was given in the textbook itself, at least not in the problem. A quick search confirmed that I'm thinking along the correct lines.

But I still don't understand the rationale for the formula. Can someone enlighten me? Also, if some references accessible to medical students could be given. I don't mind having more techical references as well.

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I don't completely understand your question - can you flesh it out a bit more? Incidence rates are just that, rates, and thus have person-time "exposed" in the denominator. You are correct about the 24/2, which reflects the assumption that the people who develop lung cancer did so at the mid-point of the interval, and thus censors them at 6 months. In contrast, you could estimate a prevalence ratio (24/2300), but a prevalence is a function of incidence and disease duration, thus less useful if you are interested in identifying causes of disease. – D L Dahly Dec 25 '12 at 21:07
But why is this assumption OK? – Raskolnikov Dec 25 '12 at 23:50
Only because it's a better assumption than censoring at 3 or 9 months, for example. Unless you had some kind of seasonal effect, or something similar, your best guess is the mid point of the interval. The only way to improve on this is to collect your data at a higher temporal resolution. – D L Dahly Dec 26 '12 at 14:53

I propose modeling cancer occurrence as a Poisson process. Multiple events (appearance of tumors) are possible within the same individual over the time period of observation. If $\lambda$ is the rate of tumor appearance by year, the probability of 0 events is $e^{-\lambda}$, and the probability of 1 event or more is $p=1-e^{-\lambda}$.

You follow $n$ individuals during a year. The number of individuals with 1 event or more is $X \sim \mathrm{Bin}(n,p)$. The expected number is $E(X) = np = n(1-e^{-\lambda})$.

Now you observe $x$ events and want to estimate $\lambda$. First estimate $\hat p = {x\over n}$, then $\hat \lambda = - \log\left(1 - {x \over n}\right) \approx {x\over n} + {x^2 \over 2 n^2}$. By invariance of maximum-likelihood estimators, $\hat \lambda$ is the MLE of $\lambda$.

Your estimator is ${ x/n \over 1 - x/2n} \approx {x\over n} + {x^2 \over 2 n^2}$. The difference between the two estimators is about $x^3/6n^3$, which is very small if $x/n$ is small. I guess this provides some justification, even if some other modeling could possibly lead directly to your estimator.

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This does look convincing. Thank you very much! – Raskolnikov Dec 26 '12 at 12:23
@Raskolnikov and Elvis (+1), note also that standard continued fractions for $-\log(1-z)$ truncated at the second convergent yield $-\log(1-z) \approx z/(1-z/2)$ and this approximation is always better than the second-order Taylor series expansion for the range of $z$ of interest. – cardinal Dec 26 '12 at 15:54
In fact, I made the same dertivation but made a mistake by equating $p$ with $e^{-\lambda}$. That's why I couldn't figure out the link with the other estimator. – Raskolnikov Dec 26 '12 at 16:34
@cardinal great comment, thanks. – Elvis Dec 26 '12 at 16:34
@Raskolnikov I thank you for this nice question which leads to a nice exercise for my students ;) nice pseudo, too – Elvis Dec 26 '12 at 16:36

Assuming diagnoses of cancer are uniformly spread across the year, the persons who are diagnosed are exposed to the risk of being diagnosed for (on average) half a year before that diagnosis.

Your link mentions the assumption of occurrence at the half-way point in the observation period but not where it comes from - which is just the assumption of uniformity. This assumption isn't always reasonable, and there are times when it can make a substantive difference. I'd recommend being aware of the assumption every time you use the formula, because you should consider its suitability and if it isn't suitable, whether it is likely to have a substantive impact on the estimate (in which case, a better assumption about the occurrence should be investigated)

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So, that's all there is to it? Assumption of uniformity? But why would the uniformity of diagnoses matter? Why not the probability of contracting the disease, which I assume would more likely be Poisson distributed (at least as a null model)? – Raskolnikov Dec 25 '12 at 23:53
Assumption of uniformity is where it comes from, so yes, that's all it is; in the absence of other information (and sometimes even in the presence of it) it's a common assumption in calculating exposure to risk. As for the bit about diagnosis, I assumed the data is on diagnosis, not incidence, because we don't observe undiagnosed incidence - whatever is analyzed is what the assumption would need to apply to. – Glen_b Dec 28 '12 at 2:15
To be more explicit, your link mentions the formula as coming from the 'actuarial method'. The relevant actuarial material is the elements of exposed-to-risk that is pretty much standard in every actuarial syllabus I am aware of. That specific assumption of uniformity is not something I just made up, it's absolutely explicit in the actuarial training. You asked where it comes from; the link mentions the actuarial method; that in turn arises from the standard actuarial approach to exposed to risk. – Glen_b Dec 28 '12 at 2:26