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You're almost correct. As a reminder, you have to assume that $p, q \neq 0$.

Assume two independent sequences of random variables, $A_n$ and $B_n$, converge in distribution to $A$ and $B$: $A_n \rightarrow A$ and $B_n \rightarrow B$. Then for any real numbers, $c$ and $d$, we have $c \cdot A_n + d \cdot B_n \rightarrow c \cdot A + d \cdot B$.

In your example, we have $$A_n := \sqrt{n}\{\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p})\}$$ $$B_n := \sqrt{n}\{\log(\frac{\bar{Y}_n}{1-\bar{Y}_n})-\log (\frac{q}{1-q})\},$$ $$A_n - B_n \rightarrow N\{0, \frac{1}{p(1-p)}) - N(0, \frac{1}{q(1-q)}\}$$.

You made a mistake in calculating your variance. For independent random variables $A$ and $B$, $var(A - B) = var(A) + var(B)$. So your two sequences converge to $$N\{0, \frac{1}{p(1-p)} + \frac{1}{q(1-q)}\}.$$

I see two ways to get a consistent estimate of the variance, $\frac{1}{p(1-p)} + \frac{1}{q(1-q)}$.

1. Multivariate Delta Method

Since we're assuming $\bar{X}_n$ and $\bar{Y}_n$ are $\sqrt{n}$ consistent estimators of $p$ and $q$, we know that

$$\sqrt{n} \begin{bmatrix} \bar{X}_n - p \\ \bar{Y}_n - q\ \end{bmatrix} \rightarrow N(0, \Sigma), \text{where}$$ $$\Sigma = \begin{bmatrix} p(1-p) & 0 \\ 0 & q(1-q) \\ \end{bmatrix}. $$

You can definitely estimate $\Sigma$ since it's the variance of two independent Binomial random variables.

Let $h(p ,q) = log(\frac{p}{1-p}) - log(\frac{p}{1-p})$ and let $\nabla h(p, q)$ be the gradient of $h$.

The Delta Method gives the asymptotic distribution and covariance matrix: $$\sqrt{n} \{h(\bar{X}_n, \bar{Y}_n) - h(p, q)\} \rightarrow N \{0, \nabla h(p, q)^T \cdot \Sigma \cdot \nabla h(p, q) \}$$

2. Invariance Property of the Maximum Likelihood Estimate (MLE)

The Invariance Property of the MLE states that $\hat{\theta}$ is a MLE of $\theta$, then $f(\hat{\theta})$ is a MLE of $f(\theta)$. So plug $\hat{p} = \frac{1}{n} \sum_i X_i$ and $\hat{q} = \frac{1}{n} \sum_i Y_i$ into $f(p, q) = \frac{1}{p(1-p)} + \frac{1}{q(1-q)}$

You're almost correct. As a reminder, you have to assume that $p, q \neq 0$.

Assume two independent sequences of random variables, $A_n$ and $B_n$, converge in distribution to $A$ and $B$: $A_n \rightarrow A$ and $B_n \rightarrow B$. Then for any real numbers, $c$ and $d$, we have $c \cdot A_n + d \cdot B_n \rightarrow c \cdot A + d \cdot B$.

In your example, we have $$A_n := \sqrt{n}\{\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p})\}$$ $$B_n := \sqrt{n}\{\log(\frac{\bar{Y}_n}{1-\bar{Y}_n})-\log (\frac{q}{1-q})\},$$ $$A_n - B_n \rightarrow N\{0, \frac{1}{p(1-p)}\} - N\{0, \frac{1}{q(1-q)}\}$$.

You made a mistake in calculating your variance. For independent random variables $A$ and $B$, $var(A - B) = var(A) + var(B)$. So your two sequences converge to $$N\{0, \frac{1}{p(1-p)} + \frac{1}{q(1-q)}\}.$$

I see two ways to get a consistent estimate of the variance, $\frac{1}{p(1-p)} + \frac{1}{q(1-q)}$.

1. Multivariate Delta Method

Since we're assuming $\bar{X}_n$ and $\bar{Y}_n$ are $\sqrt{n}$ consistent estimators of $p$ and $q$, we know that

$$\sqrt{n} \begin{bmatrix} \bar{X}_n - p \\ \bar{Y}_n - q\ \end{bmatrix} \rightarrow N(0, \Sigma), \text{where}$$ $$\Sigma = \begin{bmatrix} p(1-p) & 0 \\ 0 & q(1-q) \\ \end{bmatrix}. $$

You can definitely estimate $\Sigma$ since it's the variance of two independent Binomial random variables. Plug $\hat{p}$ and $\hat{q}$ into $\Sigma$ to get $\hat{\Sigma}$.

Let $h(p ,q) = log(\frac{p}{1-p}) - log(\frac{p}{1-p})$ and let $\nabla h(p, q)$ be the gradient of $h$.

The Delta Method gives the asymptotic distribution and covariance matrix: $$\sqrt{n} \{h(\bar{X}_n, \bar{Y}_n) - h(p, q)\} \rightarrow N \{0, \nabla h(p, q)^T \cdot \Sigma \cdot \nabla h(p, q) \}$$

A consistent estimate of the variance is given by plugging $\hat{\Sigma}$ into the expression for the variance above.

2. Invariance Property of the Maximum Likelihood Estimate (MLE)

The Invariance Property of the MLE states that $\hat{\theta}$ is a MLE of $\theta$, then $f(\hat{\theta})$ is a MLE of $f(\theta)$. So plug $\hat{p} = \frac{1}{n} \sum_i X_i$ and $\hat{q} = \frac{1}{n} \sum_i Y_i$ into $f(p, q) = \frac{1}{p(1-p)} + \frac{1}{q(1-q)}$

You're almost correct.

Assume two independent sequences of random variables, $A_n$ and $B_n$, converge in distribution to $A$ and $B$: $A_n \rightarrow A$ and $B_n \rightarrow B$. Then for any real numbers, $c$ and $d$, we have $c \cdot A_n + d \cdot B_n \rightarrow c \cdot A + d \cdot B$.

In your example, we have $$A_n := \sqrt{n}\{\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p})\}$$ $$B_n := \sqrt{n}\{\log(\frac{\bar{Y}_n}{1-\bar{Y}_n})-\log (\frac{q}{1-q})\},$$ $$A_n - B_n \rightarrow N\{0, \frac{1}{p(1-p)}) - N(0, \frac{1}{q(1-q)}\}$$.

You made a mistake in calculating your variance. For independent random variables $A$ and $B$, $var(A - B) = var(A) + var(B)$. So your two sequences converge to $$N\{0, \frac{1}{p(1-p)} + \frac{1}{q(1-q)}\}.$$

I see two ways to get a consistent estimate of the variance, $\frac{1}{p(1-p)} + \frac{1}{q(1-q)}$.

1. Multivariate Delta Method

Since we're assuming $\bar{X}_n$ and $\bar{Y}_n$ are $\sqrt{n}$ consistent estimators of $p$ and $q$, we know that

$$\sqrt{n} \begin{bmatrix} \bar{X}_n - p \\ \bar{Y}_n - q\ \end{bmatrix} \rightarrow N(0, \Sigma), \text{where}$$ $$\Sigma = \begin{bmatrix} p(1-p) & 0 \\ 0 & q(1-q) \\ \end{bmatrix}. $$

You can definitely estimate $\Sigma$ since it's the variance of two independent Binomial random variables.

Let $h(p ,q) = log(\frac{p}{1-p}) - log(\frac{p}{1-p})$ and let $\nabla h(p, q)$ be the gradient of $h$.

The Delta Method gives the asymptotic distribution and covariance matrix: $$\sqrt{n} \{h(\bar{X}_n, \bar{Y}_n) - h(p, q)\} \rightarrow N \{0, \nabla h(p, q)^T \cdot \Sigma \cdot \nabla h(p, q) \}$$

2. Invariance Property of the Maximum Likelihood Estimate (MLE)

The Invariance Property of the MLE states that $\hat{\theta}$ is a MLE of $\theta$, then $f(\hat{\theta})$ is a MLE of $f(\theta)$. So plug $\hat{p} = \frac{1}{n} \sum_i X_i$ and $\hat{q} = \frac{1}{n} \sum_i Y_i$ into $f(p, q) = \frac{1}{p(1-p)} + \frac{1}{q(1-q)}$

You're almost correct. As a reminder, you have to assume that $p, q \neq 0$.

Assume two independent sequences of random variables, $A_n$ and $B_n$, converge in distribution to $A$ and $B$: $A_n \rightarrow A$ and $B_n \rightarrow B$. Then for any real numbers, $c$ and $d$, we have $c \cdot A_n + d \cdot B_n \rightarrow c \cdot A + d \cdot B$.

In your example, we have $$A_n := \sqrt{n}\{\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p})\}$$ $$B_n := \sqrt{n}\{\log(\frac{\bar{Y}_n}{1-\bar{Y}_n})-\log (\frac{q}{1-q})\},$$ $$A_n - B_n \rightarrow N\{0, \frac{1}{p(1-p)}) - N(0, \frac{1}{q(1-q)}\}$$.

You made a mistake in calculating your variance. For independent random variables $A$ and $B$, $var(A - B) = var(A) + var(B)$. So your two sequences converge to $$N\{0, \frac{1}{p(1-p)} + \frac{1}{q(1-q)}\}.$$

I see two ways to get a consistent estimate of the variance, $\frac{1}{p(1-p)} + \frac{1}{q(1-q)}$.

1. Multivariate Delta Method

Since we're assuming $\bar{X}_n$ and $\bar{Y}_n$ are $\sqrt{n}$ consistent estimators of $p$ and $q$, we know that

$$\sqrt{n} \begin{bmatrix} \bar{X}_n - p \\ \bar{Y}_n - q\ \end{bmatrix} \rightarrow N(0, \Sigma), \text{where}$$ $$\Sigma = \begin{bmatrix} p(1-p) & 0 \\ 0 & q(1-q) \\ \end{bmatrix}. $$

You can definitely estimate $\Sigma$ since it's the variance of two independent Binomial random variables.

Let $h(p ,q) = log(\frac{p}{1-p}) - log(\frac{p}{1-p})$ and let $\nabla h(p, q)$ be the gradient of $h$.

The Delta Method gives the asymptotic distribution and covariance matrix: $$\sqrt{n} \{h(\bar{X}_n, \bar{Y}_n) - h(p, q)\} \rightarrow N \{0, \nabla h(p, q)^T \cdot \Sigma \cdot \nabla h(p, q) \}$$

2. Invariance Property of the Maximum Likelihood Estimate (MLE)

The Invariance Property of the MLE states that $\hat{\theta}$ is a MLE of $\theta$, then $f(\hat{\theta})$ is a MLE of $f(\theta)$. So plug $\hat{p} = \frac{1}{n} \sum_i X_i$ and $\hat{q} = \frac{1}{n} \sum_i Y_i$ into $f(p, q) = \frac{1}{p(1-p)} + \frac{1}{q(1-q)}$

You're almost correct.

Assume two independent sequences of random variables, $A_n$ and $B_n$, converge in distribution to $A$ and $B$: $A_n \rightarrow A$ and $B_n \rightarrow B$. Then for any real numbers, $c$ and $d$, we have $c \cdot A_n + d \cdot B_n \rightarrow c \cdot A + d \cdot B$.

In your example, we have $$A_n := \sqrt{n}\{\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p})\}$$ $$B_n := \sqrt{n}\{\log(\frac{\bar{Y}_n}{1-\bar{Y}_n})-\log (\frac{q}{1-q})\},$$ so $$A_n - B_n \rightarrow N\{0, \frac{1}{p(1-p)}) - N(0, \frac{1}{q(1-q)}\}$$.

You made a mistake in calculating your variance. For independent random variables $A$ and $B$, $var(A - B) = var(A) + var(B)$. So your two sequences converge to $$N\{0, \frac{1}{p(1-p)} + \frac{1}{q(1-q)}\}.$$

I see two ways to get a consistent estimate of the variance, $\frac{1}{p(1-p)} + \frac{1}{q(1-q)}$.

1. Multivariate Delta Method

$\sqrt{n} \begin{bmatrix} \bar{X}_n - p \\ \bar{Y}_n - q\ \end{bmatrix} \rightarrow N(0, \Sigma)$ where $\Sigma = \begin{bmatrix} p(1-p) & 0 \\ 0 & q(1-q) \\ \end{bmatrix} $ . You can definitely estimate $\Sigma$ since it's the variance of two independent Binomial random variables. Let $h(p ,q) = log(\frac{p}{1-p}) - log(\frac{p}{1-p})$ and let $\nabla h(p, q)$ be the gradient of $h$. The Delta Method gives the asymptotic distribution and covariance matrix  : $\sqrt{n} \{h(\bar{X}_n, \bar{Y}_n) - h(p, q)\} \rightarrow N(0, \nabla h(p, q)^T \cdot \Sigma \cdot \nabla h(p, q))$

2. Invariance Property of the Maximum Likelihood Estimate (MLE)

The Invariance Property of the MLE states that $\hat{\theta}$ is a MLE of $\theta$, then $f(\hat{\theta})$ is a MLE of $f(\theta)$. So plug $\hat{p} = \frac{1}{n} \sum_i X_i$ and $\hat{q} = \frac{1}{n} \sum_i Y_i$ into $f(p, q) = \frac{1}{p(1-p)} + \frac{1}{q(1-q)}$

You're almost correct.

Assume two independent sequences of random variables, $A_n$ and $B_n$, converge in distribution to $A$ and $B$: $A_n \rightarrow A$ and $B_n \rightarrow B$. Then for any real numbers, $c$ and $d$, we have $c \cdot A_n + d \cdot B_n \rightarrow c \cdot A + d \cdot B$.

In your example, we have $$A_n := \sqrt{n}\{\log(\frac{\bar{X}_n}{1-\bar{X}_n})-\log (\frac{p}{1-p})\}$$ $$B_n := \sqrt{n}\{\log(\frac{\bar{Y}_n}{1-\bar{Y}_n})-\log (\frac{q}{1-q})\},$$ $$A_n - B_n \rightarrow N\{0, \frac{1}{p(1-p)}) - N(0, \frac{1}{q(1-q)}\}$$.

You made a mistake in calculating your variance. For independent random variables $A$ and $B$, $var(A - B) = var(A) + var(B)$. So your two sequences converge to $$N\{0, \frac{1}{p(1-p)} + \frac{1}{q(1-q)}\}.$$

I see two ways to get a consistent estimate of the variance, $\frac{1}{p(1-p)} + \frac{1}{q(1-q)}$.

1. Multivariate Delta Method

Since we're assuming $\bar{X}_n$ and $\bar{Y}_n$ are $\sqrt{n}$ consistent estimators of $p$ and $q$, we know that

$$\sqrt{n} \begin{bmatrix} \bar{X}_n - p \\ \bar{Y}_n - q\ \end{bmatrix} \rightarrow N(0, \Sigma), \text{where}$$ $$\Sigma = \begin{bmatrix} p(1-p) & 0 \\ 0 & q(1-q) \\ \end{bmatrix}. $$

You can definitely estimate $\Sigma$ since it's the variance of two independent Binomial random variables.

Let $h(p ,q) = log(\frac{p}{1-p}) - log(\frac{p}{1-p})$ and let $\nabla h(p, q)$ be the gradient of $h$.

The Delta Method gives the asymptotic distribution and covariance matrix: $$\sqrt{n} \{h(\bar{X}_n, \bar{Y}_n) - h(p, q)\} \rightarrow N \{0, \nabla h(p, q)^T \cdot \Sigma \cdot \nabla h(p, q) \}$$

2. Invariance Property of the Maximum Likelihood Estimate (MLE)

The Invariance Property of the MLE states that $\hat{\theta}$ is a MLE of $\theta$, then $f(\hat{\theta})$ is a MLE of $f(\theta)$. So plug $\hat{p} = \frac{1}{n} \sum_i X_i$ and $\hat{q} = \frac{1}{n} \sum_i Y_i$ into $f(p, q) = \frac{1}{p(1-p)} + \frac{1}{q(1-q)}$