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What's the right way to compute the Standard Error of the Mean of the ratio of two random variables that follow a binomial distribution?

I asked a similar question here using Weibull distributions and now I'm interested on the relationship between binomial distributions.

I'm running an experiment that tracks success/failure rate of two distinct groups over 7 days in a row. These two two groups follow a binomial distribution:

$$X \sim \mbox{B}({ n }_{ 1 },{ p }_{ 1 })$$ $$Y \sim \mbox{B}({ n }_{ 2 },{ p }_{ 2 })$$

The experiment starts with a sample of ~ 10k observations in each group but the sample size differs during the following days (eg: I can have [10k 5k 7k 3k 2k 1k 0.5k 0.8k] observations for a group. Each number equals the number of observations of that group in a specific day, no day will have more than the initial sample size observations, an observation on day 3 can also be observed on day 5 and all observations will always be included on day 0).

I want to calculate the $SEM$ of the following ratio:

$$Z=\frac {X-Y}{Y}$$

The method I'm following is obtaining the Expected Value and Variation of the ratio by using Taylor Expansions.

Since both distributions have $N*\hat{p} > 5$ and $N*(1-\hat{p})> 5$ we can assume it follows a normal distribution and compute the $SEM$ by using:

$${SEM_i}= \frac{{\sigma_i}}{\sqrt{N_i}}$$

where ${N_i}$ is the sample size in day $i$.

It turns out that this yields

  • a very small $SEM$ (about 0.2~1% of the mean) when I use ${N_i}={Nx_i}+{Ny_i}$ and
  • a really huge $SEM$ (about 2~3 times the mean) when I use ${N_i} = 7-i$ (the number of days I run the experiment), so it makes me question the right $N$ to use.

Which one of these points is the correct one to get the $ {N_i} $?

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    $\begingroup$ $Y$ has a positive probability of being $0$ (so too does $X$) which is going to cause issues with $\frac {X-Y}{Y}$ $\endgroup$
    – Henry
    Commented May 25, 2015 at 21:37
  • $\begingroup$ Yea, but that probability is way too small, that will rarely occur - if not never. $\endgroup$
    – Thiago
    Commented May 25, 2015 at 22:00
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    $\begingroup$ Even if the probability is $1/(10^{10^{100}})$, the population variance of the ratio will not be finite. You'd have to explicitly exclude zero (making it truncated binomial) to get something it makes sense to try to calculate. $\endgroup$
    – Glen_b
    Commented May 26, 2015 at 5:46
  • $\begingroup$ Thanks for the answers, it's a really good point to consider and clarifies the question a lot. Just to get a little bit back to the point (that I'm still struggling a bit) of the sample size ${N_i}$: for simplicity, suppose that I change my $Z$ to $Z=X-Y$. In that case (now excluding the issues with division by 0), which ${N_i}$ should I use to obtain the $SEM$ of $Z$? (here I'm more trying to understand the theory behind choosing the right ${N_i}$) $\endgroup$
    – Thiago
    Commented May 27, 2015 at 21:21
  • $\begingroup$ What is the applied problem which leads you to consider the ratio of binomials? Knowing that, maybe we can propose something better. $\endgroup$
    – kjetil b halvorsen ♦
    Commented Aug 1, 2017 at 19:07

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