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 $p=\lambda/(\lambda+\theta)$
