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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.

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2 Answers 2

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

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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)),
   from=min(ratio), to=max(ratio), col="red", add=TRUE) 

enter image description here

As a couple of extra comments:

  • 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 $e^{-(\lambda+\theta)}$
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  • $\begingroup$ 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. $\endgroup$ Sep 8, 2021 at 1:21
  • $\begingroup$ why do you cube the denominator? $\endgroup$ Aug 19, 2023 at 22:38
  • $\begingroup$ @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 $\endgroup$
    – Henry
    Aug 19, 2023 at 22:49
  • $\begingroup$ Got it, ty so much @Henry. $\endgroup$ Aug 19, 2023 at 23:23

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