My aim is to find the 95% confidence interval of the ratio of two variables for which I have summary statistics. More specifically, I have the prevalence of mothers drinking during their pregnancy (assumed to follow a normal distribution) and I have the prevalence of children born with foetal alcohol syndrome (FAS, supposed to follow a binomial distribution for which I know all parameters). By dividing the former by the latter I obtain the number of pregnant women that drink per 1 birth with FAS.

I thought the easiest way would be to sample both distributions N times, therefore generate N samples of this ratio and the 2.5% and 97.5% quantiles should give me the right CIs. The mean of this ratio is very far from the ratio of the means of the initial distribution though. This is a known thing and arises from Jensen's inequality (I assume this comes simply from the nonlinearity of the function f(x)=1/x). How do I interpret my Monte Carlo samples? Are they representative of the error?

The literature now proposes to use Taylor series approximations to estimate the variance of the ratio mean of 2 normally distributed variables. This would generate a variance around the ratio of the means and make it all nice and symmetrical. Given that these approaches will not give the same results at all, one giving you symmetric CIs, the other one giving you asymmetric results, which one is correct?

  • $\begingroup$ Shouldn't the prevalence of drinking mothers also be a binomial? $\endgroup$ – gung - Reinstate Monica Apr 10 '16 at 3:34
  • $\begingroup$ @gung Yes it should be, but it's approximated well enough by a normal distribution. $\endgroup$ – Moppentapper Apr 10 '16 at 3:35
  • $\begingroup$ So you are independently sampling from the two distributions? But wouldn't drinking and the chances of a child being born with FAS be dependent? $\endgroup$ – Wolfgang Apr 10 '16 at 10:48
  • $\begingroup$ @Wolfgang you're entirely right, the data comes from different sources and covariance is unknown. I'm open to suggestions on how to deal with this too though, although it might get a bit off topic! Any help and insight is appreciated! $\endgroup$ – Moppentapper Apr 10 '16 at 10:53
  • $\begingroup$ @Moppentapper Ok, fair enough. Just wanted to throw that out there. $\endgroup$ – Wolfgang Apr 10 '16 at 12:57

Here is an analytic solution:

Let $$\bar{x} \sim N(\mu, \sigma^2 / n_1)$$ and $$y \sim Binomial(n_2, \pi)$$ and assume that $\bar{x}$ and $y$ are independent. Let $p = y/n_2$, whose large-sample distribution is $N\left(\pi, \frac{\pi(1-\pi)}{n_2}\right)$. You want a 95% CI for $\mu / \pi$. Then based on the delta method, we can show that $log(\bar{x} / p)$ has a large-sample distribution that is normal with expected value $log(\mu / \pi)$ and variance $\frac{\sigma^2}{n_1 \mu^2} + \frac{(1-\pi)}{n_2 \pi}$. The variance can be estimated by substituting the sample estimates, so let $$Var[log(\bar{x} / p)] = \frac{s^2}{n_1 \bar{x}^2} + \frac{(1-p)}{n_2 p}.$$ Now compute an approximate 95% CI for $log(\mu / \pi)$ with $$log(\bar{x} / p) \pm 1.96 \sqrt{Var[log(\bar{x} / p)]}$$ and then exponentiate the bounds to obtain an approximate 95% CI for $\mu / \pi$.

How well this works in your case depends on how large $n_1$ and $n_2$ are, that is, whether you can actually appeal to the large-sample behavior of the distribution of $log(\bar{x} / p)$.

  • $\begingroup$ Wow! I'm going to have to go over this in detail to make sure I understand but wanted to thank you already for this! So what do you make of the Monte Carlo approach? Would the 2.5% and 97.5% quantiles of the samples still reflect the CI even if the average samples ratio is very much different from the ratio of the means of each distribution? $\endgroup$ – Moppentapper Apr 12 '16 at 13:56
  • $\begingroup$ Hard to say. You could do a simulation study (using parameter values that are realistic for your particular application) to sort out how well these approaches actually work. $\endgroup$ – Wolfgang Apr 12 '16 at 21:17

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