# Incorporating denominator uncertainty into a proportion

I am calculating an incidence risk (r): number of cases of a disease in a population over one year (c) divided by the total mid-year population (N).

$$r = \frac{c}{N}$$

Let's assume that c is a precise measure, with no uncertainty around it.

N, however, is a point estimate. I also have a 95% CI around it:

$$N=125$$

$$(u(N),v(N))=(100,150)$$

Can I incorporate this uncertainty into the risk, in order to calculate a 95% CI around r?

This is not something I've ever done. After some online searching, I've come across error propagation, that I think would apply here - error propagation for a quotient.

$$\sigma_r = |r| \sqrt{(\frac{\sigma_c}{c})^2 + (\frac{\sigma_N}{N})^2 + 2\frac{\sigma_{cN}}{cN}}$$

However, I'm not sure:

1. If this is appropriate to use in this circumstance
2. How to go from a confidence interval to sigma_{N} and sigma_{cN}.

1. In your case, $$\sigma_c$$ and $$\sigma_{cN}$$ equal zero, as $$c$$ is measured without error. But actually, there's a simpler approach...
2. Since the function $$1/N$$ is monotonic, a valid 95% confidence interval of $$1/N$$ is just equal to a 95% confidence interval of $$N$$ run through the function $$1/N$$. So your 95% CI of $$N$$ would be $$(1/150, 1/100)$$, and the associated 95% CI of $$c/N$$ is $$(c/150, c/100)$$.