How to derive the asymptotic distribution of the test statistic of a large sample test for equality of two binomial populations?

Question

Let $X_1, \ldots, X_{n_X}$ and $Y_1, \ldots, Y_{n_Y}$ be $n_x$ and $n_Y$ iid observations from two independent Bernoulli populations with probabilities of success $p_X$ and $p_Y$. Define the statistics $T_X = \sum_{i=1}^{n_X}X_i$ and $T_Y = \sum_{i=1}^{n_Y}Y_i$. I am testing the hypothesis:

$$ H_0 : p_X = p_Y \ \ \ \text{and} \ \ \ H_1 : p_X \neq p_Y $$

Consider the test statistic:

$$ T = \dfrac{\hat{p}_X-\hat{p}_Y}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_X}+\frac{1}{n_Y}\right)}} $$

where $\hat{p}_X = \frac{T_X}{n_X}$ and $\hat{p}_Y = \frac{T_Y}{n_Y}$, and $\hat{p} = \frac{T_X+T_Y}{n_X+n_Y}$.

I would like to derive the asymptotic distribution of $T$ under $H_0$ as both $n_X$ and $n_Y$ go to infinity. My works is as follows: By the asymptotic properties of the MLE:

$$ \sqrt{n_X}(\hat{p}_X-p_X) \to_{D}N(0,p_X(1-p_X)) $$

and

$$ \sqrt{n_Y}(\hat{p}_Y-p_Y) \to_{D}N(0,p_Y(1-p_Y)) $$

I would like to find the distribution of $\hat{p}_X-\hat{p}_Y$, but am not sure how. In general two "in distribution" results are not additive. That is, it is genreally NOT the case that $X_n+Y_n \to_{D} X+Y$. Does anyone have any ideas?

I know in general that under the null:

$$ T = \dfrac{\hat{p}_X-\hat{p}_Y}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_X}+\frac{1}{n_Y}\right)}} \to_D N(0,1) $$

Answer 1

The OP is correct that you just cannot "add" convergence in distribution, for this you have to be aware of the covariance structure, which is fortunately enough trivial in this case.

update: Thanks for pointing out the earlier, crucial (and pretty bad) mistake. Hopefully the following answer is more or less correct...

Suppose, for the moment, that $n_Y = \lceil c n_X \rceil$, for some $c > 0$.

Define $$ \theta_X = \frac {\hat p_X - p} { \sqrt{ \left(\frac 1 {n_X} + \frac 1 {n_Y}\right) p(1-p)}} \quad \mbox{and} \quad \theta_Y = \frac {\hat p_Y - p} { \sqrt{ \left(\frac 1 {n_X} + \frac 1 {n_Y}\right) p(1-p)}},$$ so that $$T = \sqrt{\frac{p(1-p)}{\hat p(1 - \hat p)}} (\theta_X - \theta_Y).$$

Now $\theta_X$ converges in distribution to $N\left(0,\frac c {1+c} \right)$, whereas $\theta_Y$ converges in distribution to $N\left(0, \frac 1 {1 + c} \right)$, and jointly they converge to the independent product of these two distributions.

It follows that $\theta_X - \theta_Y$ converges (using the continuous mapping theorem, hopefully correctly this time) to a $N(0,1)$ distribution. Since $\hat p(1-\hat p) \rightarrow p(1-p)$ almost surely, it follows that $T \stackrel{d}{\rightarrow} N(0,1)$.

In a similar way, when e.g. $n_Y = n_X^2$ you can find that $\theta_X \stackrel{d}{\rightarrow} N(0,1)$ and $\theta_Y \stackrel{a.s.}{\rightarrow} 0$, so that again $T \stackrel{d}{\rightarrow} N(0,1)$.

Unfortunately I don't see how to avoid making some assumption on the relative growth of $n_X$ and $n_Y$, but perhaps a more general argument is possible.