# Asymptotic conditional distribution of $\bar{Y}\mid\bar{X}=x$

I'm reviewing for my qualifying exam and I'm stuck on part of a problem.

Setup

Suppose that $(X,Y)$ are two random variables with joint distribution $$$f(x,y\mid\alpha,\beta)=c(\alpha,\beta)\exp(-\alpha x-\beta y)\sum_{j=0}^\infty\frac{x^j y^j}{j!^2}$$$ for $x,y>0$ and $c(\alpha,\beta)=\alpha\beta-1$.

Let $(X_1,Y_1)...(X_n,Y_n)$ be a random sample from $(X,Y)$.

Problem

Derive the asymptotic distribution of $\bar{Y}\mid\bar{X}=\bar{x}$.

Attempt/Thoughts

The joint distribution of $(X_i,Y_i)_{i\leq n}$ is given by: $$f((x_i,y_i)_n\mid\alpha,\beta)=c(\alpha,\beta)^n \exp\left(-\alpha \sum_i x_i-\beta \sum_i y_i\right) \prod^n_{i=1} \sum_{j=0}^{\infty}\frac{x_i^j y_i^j}{j!^2}$$

The joint distribution above is a 2-parameter exponential family (likely not full rank since we must have $\alpha\beta>1$, so there is no open rectangle on $\mathbb{R}^2$). Because it's an exponential family, the conditional distribution $(\sum_i y_i\mid \sum_i x_i=s)\iff (\bar{Y}\mid\bar{X}=\bar{x})$ will be a one-parameter exponential family, free of $\alpha$. However, I'm not sure how to derive this explicitly, let alone obtain its asymptotic distribution. If I follow Bayes' rule, then I have to compute somehow a marginal distribution of $\bar{X}$, in the presence of the pesky $\sum x^j y^j$ term.

Does anyone have any hints on how to get started? Thank you!

Edit

I believe I have a solution. Here is a rough outline for now. I can post the complete solution later.

Using multivariate CLT, we can compute the asymptotic distribution of $\bar{X},\bar{Y}$. The variances and covariances can be computed directly from the density. Then we use Bayes rule to compute the conditional density. This website also helped for calculating the marginal density of $\bar{X}$.