Convergence of random variables with hypergeometric distribution

This is a very interesting conjecture of large scale property of hypergeometric distribution.

Let $a>1$ be a integer constant, $N\in\mathbb{N_+}$, for any $x<N-1$, consider $N+(a-1)x$ balls in a bag, in which $ax$ of them are colored white and the other $N-x$ are colored black. Now randomly select $N$ balls without replacement from the bag. Let random variable $X$ denote the number of white balls selected. For another $y=x+1$, we also select $N$ balls from a bag with $N+(a-1)y$ balls, in which $ay$ are colored white, then we get a similar random variable $Y$.

In short, $X \sim H(N+(a-1)x,ax,N)$ and $Y \sim H(N+(a-1)y,ay,N)$, with $y=x+1$.

We want to prove that $\mathbb{E}(\frac{xy}{X}-\frac{xy}{Y})\rightarrow \frac{1}{a}$, when $N\rightarrow\infty$. The convergence is uniform w.r.t to $x$, namely, the convergence rate is independent of $x$. (For large enough $N$, the expectation can be arbitrarily close to $1/a$, regardless of which $x<N-1$ we choose. See the comment of the first answer.)

It is likely that the convergence must use a suitable coupling of $X$ and $Y$.

In your setting, when $N\to\infty$ we have $\mathbb{P}(X=ax)\to 1$ and so $\mathbb{E}(\frac{xy}{X})\to \frac{y}{a}$. Similarly, $\mathbb{E}(\frac{xy}{Y})\to \frac{x}{a}$, which together with $y=x+1$ yield the desired result.
The convergence rate cannot be independent of $x$ as the model is only defined when $N>x$.
• Thank you for your answer. Here independence means, $\forall \epsilon>0$, there is an number $N(\epsilon)$, such that $\forall N>N(\epsilon)$, we have $|\mathbb{E}(xy/X-xy/Y)-1/a|<\epsilon$. Note that $\mathbb{E}(\frac{xy}{X})$ may tend to infinity, because $x$ can tend to infinity when $N$ grows, so this answer is not correct :-) Feb 3 '15 at 13:09