Limiting distribution of the sum of two scaled dependent random variables.

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I have an interesting question that I've somehow partially figured out, but can't get to a full conclusion.

$\textbf{Question:}$

Suppose that $X_1, ..., X_n \overset{iid}{\thicksim}$ with a cumulative distribution function $F$.

Let $V = \sum_{i=1}^n I(X_i \le 0)$ and $W = \sum_{i=1}^n I(X_i > 0)$ and take that $\theta_1 = P(X_1 \le 0)$ and $\theta_2 = 1 - \theta_1$, where $0< \theta_1 < 1$.

Assume $Z_n = \frac{\left(V - n\theta_1\right)^2}{n\theta_1} + \frac{\left(W - n\theta_2\right)^2}{n\theta_2}$

Show that $Z_n$ converges to a limiting distribution and identify this distribution.

$\\$

Here is my attempt:

Let $V_i = I(X_i \le 0)$. Then the $V_i's$ are $iid$ with mean $\theta_1$ and variance $\theta_1(1- \theta_1)$.

Similarly, $W_i = I(X_i > 0)$, $i=1, ... , n$ will also be $iid$ with mean $\theta_2$ and variance $\theta_2(1- \theta_2)$.

I can apply CLT to obtain that:

$$\frac{\left(V - n\theta_1 \right) }{\sqrt{n\theta_1}} \overset{d}{\rightarrow} N(0, 1- \theta_1)$$

$$\frac{\left( W - n\theta_2 \right) }{\sqrt{n\theta_2}} \overset{d}{\rightarrow} N(0, 1- \theta_2)$$

Now if we let

$$Z_{n_1} = \frac{\left(V - n\theta_1 \right) }{\sqrt{n\theta_1}} \hspace{1cm} and \hspace{1cm} Z_{n_2} = \frac{\left( W - n\theta_2 \right) }{\sqrt{n\theta_2}}$$

we observe that

$$Z_n = Z_{n_1}^2 + Z_{n_2}^2$$

Applying Slutsky's theorem, we obtain that:

$$Z^2_{n_1} = (1-\theta_1)\frac{Z^2_{n_1}}{1-\theta_1} \overset{d}{\rightarrow} \theta_2 \chi^2_1$$

and

$$Z^2_{n_2} = (1-\theta_2)\frac{Z^2_{n_2}}{1-\theta_2} \overset{d}{\rightarrow} \theta_1 \chi^2_1$$

This is where I got stuck. If $Z_{n_1}^2$ and $Z_{n_2}^2$ were independent, it'll be easy to derive the limiting distribution of their sum, but there are not.

Any pointers on how to proceed will be appreciated.

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Just write $W=n-V$ and $\theta_2 = 1-\theta_1$ to get $$Z_{n_2} = \frac{W-n\theta_2}{\sqrt{n\theta_2}} = - \frac{V-n\theta_1}{\sqrt{n(1-\theta_1)}}$$ so $$Z_{n_1}^2 + Z_{n_2}^2 = (V-n\theta_1)^2\left(\frac{1}{n\theta_1} + \frac{1}{n(1-\theta_1)}\right) = \frac{(V-n\theta_1)^2}{n\theta_1(1-\theta_1)} \overset{d}{\to} \chi^2(1).$$