# Delta method for Poisson ratio

Let $$X_1,...,X_n$$ be drawn from $$Pois(\lambda)$$ and $$Y_1,...,Y_n$$ from $$Pois(\theta)$$. I would like to find the asymptotic distribution of $$\frac{\overline X}{\overline X + \overline Y }$$ using delta method.

My difficulty with the problem is the following: while I understand that $$\frac{\sqrt{n} (\overline X - \lambda)}{\sqrt{\lambda}} \sim N(0, 1)$$ asymptotically, there seems to be no proper transformation function $$g$$ so that satisfies $$\frac{\overline X}{\overline X + \overline Y } = g(\overline X)$$, since the statistics is a function of both $$\overline Y$$ and $$\overline X$$. I would appreciate thoughts on how delta method can be applied in this scenario.

Use a vector version of the delta method. You have convergence of $$\sqrt{n}(\bar X-\lambda,\, \bar Y-\theta)$$ to a bivariate Normal, and the function $$f(\bar X, \bar Y)=\frac{\bar X}{\bar X+\bar Y}$$ is differentiable (away from $$\lambda=\theta=0$$), so the delta method applies.

That isn't how I'd actually work out the answer, though. I would argue that conditional on $$N=\sum_i (X_i + Y_i)$$, the sum $$\sum_i X_i$$ is Binomial$$(N, \lambda/(\lambda+\theta))$$, so the ratio you're interested in is (conditionally) a binomial proportion, which is asymptotically Normal. Then I would note that $$N/n\stackrel{a.s.}{\to}\lambda+\theta$$, so that $$\sqrt{n}\left(\frac{\bar X}{\bar X+\bar Y}-p\right)\stackrel{d}{\to} N\left(0, \frac{p(1-p)}{\lambda+\theta}\right)$$ where $$p=\lambda/(\lambda+\theta)$$

This is just a visual comment on Thomas Lumley's answer (+1), illustrating it by simulation (blue) against his approximating normal distribution (red) using R

set.seed(2021)
lambda <- 2
theta  <- 5
n      <- 1000
cases  <- 10^5
Xbar   <- rpois(cases, n * lambda) / n
Ybar   <- rpois(cases, n * theta ) / n
ratio  <- Xbar / (Xbar + Ybar)
plot(density(ratio), col="blue")
curve(dnorm(x, lambda/(lambda+theta), sqrt(lambda*theta/(lambda+theta)^3/n)),

• There is a slight issue that there is always a positive probability that you get $$\frac00$$. So the actual ratio distribution is not well defined, though if $$n\lambda$$ and $$n\mu$$ are both large the probability of this is extremely small
• You do not have to take means, as the ratio of the sums $$\frac{\sum X_i}{\sum X_i +\sum Y_i}$$ has the same distribution
• Since the sums are themselves Poisson distributed, you might then, in a handwaving way, say that for large $$\lambda$$ and $$\theta$$ the distribution of the ratio $$\frac{X}{X+Y}$$ is approximately $$N\left(\frac{\lambda}{\lambda+\theta}, \frac{\lambda\theta}{(\lambda+\theta)^3}\right)$$ and that the probability of seeing $$\frac00$$ is only $$e^{-(\lambda+\theta)}$$
• The probability of seeing 0/0 is $e^{-n(\lambda+\theta)}$, which is much smaller. You can define the ratio variable to have some other value when it's 0/0 and the asymptotic distribution won't be changed: 0 or 1 or 69/420 or anything you like. Sep 8, 2021 at 1:21
• @Estimatetheestimators - see Thomas Lumley's answer with a variance of $\frac{p(1-p)}{\lambda+\theta}$ which needs to be scaled by $\frac1n$. Then let $p = \frac{\lambda}{\lambda+\theta}$ and tidy up Aug 19, 2023 at 22:49