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The multivariate delta method has a heuristic justification here: https://en.wikipedia.org/wiki/Delta_method#Multivariate_delta_method. For the multivariate delta method you have a specific function $f$ that takes a vector argument which is $p$ dimensional and maps this to a $k$ dimensional space. In the case of a ratio estimator $p=2$ and $k=1$. The function $f$ is

$$f\left(\begin{bmatrix} \bar{y} \\ \bar{x} \\ \end{bmatrix}\right) = \bar{y}/\bar{x}$$ Now what are needed are a few more quantities, the first is:

$$f(\vec{\mu})=f\left(\begin{bmatrix} \mu_{y} \\ \mu_{x} \\ \end{bmatrix}\right) = \mu_{y}/\mu_{x}$$

These are the $h(B)$ and $h(\beta)$ respectively in notation in the Wikipedia link.
Next you need the vector of partial derivatives of $f(\vec{\mu})$, this is:

$$\nabla f(\vec{\mu})=\begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}$$ Also we need the variance covariance matrix of the vector

$$\begin{bmatrix} \bar{y} \\ \bar{x} \\ \end{bmatrix}$$ which is

$$\begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix}.$$ Note this variance-covariance matrix is the $\Sigma/n$ in the Wikipedia notation. For a proof that $\mathbb{C}ov(\bar{y},\bar{x}) =\mathbb{C}ov(x,y)$ see Estimating the covariance of the means from two samples? Now the only thing left is to calculate the quadratic form:

$$\nabla f(\vec{\mu})^T\begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix}\nabla f(\vec{\mu}) = \begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}^T \begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix} \begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}.$$

Which when I worked this out gives you the equation:

$$\sigma^2_R=\frac{\sigma_y^2}{n\mu_x^2} - 2\frac{\mu_y\sigma_{xy}}{\mu_x^3}+\frac{\sigma^2_x\mu_y^2}{n\mu_x^4},$$

where this quantity is the variance of the delta method normal distribution. Putting this altogether gives us that

$$\sqrt{n}\left(\frac{\bar{y}}{\bar{x}}-\frac{\mu_y}{\mu_x}\right) \sim N(0,\sigma^2_R)$$

So you can estimate the ratio of the population means by the ratio of the sample means provided you can estimate variances and the covariance, or equivalently, the correlation $\rho = \frac{\sigma_{xy}}{\sigma_x\sigma_y}$, by substitution, $\sigma_x\sigma_y\rho = \sigma_{xy}$. This is how the delta method is most commonly used in the derivation of the ratio estimator distribution.

The multivariate delta method has a heuristic justification here: https://en.wikipedia.org/wiki/Delta_method#Multivariate_delta_method. For the multivariate delta method you have a specific function $f$ that takes a vector argument which is $p$ dimensional and maps this to a $k$ dimensional space. In the case of a ratio estimator $p=2$ and $k=1$. The function $f$ is

$$f\left(\begin{bmatrix} \bar{y} \\ \bar{x} \\ \end{bmatrix}\right) = \bar{y}/\bar{x}$$ Now what are needed are a few more quantities, the first is:

$$f(\vec{\mu})=f\left(\begin{bmatrix} \mu_{y} \\ \mu_{x} \\ \end{bmatrix}\right) = \mu_{y}/\mu_{x}$$

These are the $h(B)$ and $h(\beta)$ respectively in notation in the Wikipedia link.
Next you need the vector of partial derivatives of $f(\vec{\mu})$, this is:

$$\nabla f(\vec{\mu})=\begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}$$ Also we need the variance covariance matrix of the vector

$$\begin{bmatrix} \bar{y} \\ \bar{x} \\ \end{bmatrix}$$ which is

$$\begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix}.$$ Note this variance-covariance matrix is the $\Sigma/n$ in the Wikipedia notation. For a proof that $\mathbb{C}ov(\bar{y},\bar{x}) =\mathbb{C}ov(x,y)$ see Estimating the covariance of the means from two samples? Now the only thing left is to calculate the quadratic form:

$$\nabla f(\vec{\mu})^T\begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix}\nabla f(\vec{\mu}) = \begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}^T \begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix} \begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}.$$

Which when I worked this out gives you the equation:

$$\sigma^2_R=\frac{\sigma_y^2}{n\mu_x^2} - 2\frac{\mu_y\sigma_{xy}}{\mu_x^3}+\frac{\sigma^2_x\mu_y^2}{n\mu_x^4},$$

where this quantity is the variance of the delta method normal distribution. Putting this altogether gives us that

$$\left(\frac{\bar{y}}{\bar{x}}-\frac{\mu_y}{\mu_x}\right) \sim N(0,\sigma^2_R)$$

So you can estimate the ratio of the population means by the ratio of the sample means provided you can estimate variances and the covariance, or equivalently, the correlation $\rho = \frac{\sigma_{xy}}{\sigma_x\sigma_y}$, by substitution, $\sigma_x\sigma_y\rho = \sigma_{xy}$. This is how the delta method is most commonly used in the derivation of the ratio estimator distribution.

The multivariate delta method has a heuristic justification here: https://en.wikipedia.org/wiki/Delta_method#Multivariate_delta_method. For the multivariate delta method you have a specific function $f$ that takes a vector argument which is $p$ dimensional and maps this to a $k$ dimensional space. In the case of a ratio estimator $p=2$ and $k=1$. The function $f$ is

$$f\left(\begin{bmatrix} \bar{y} \\ \bar{x} \\ \end{bmatrix}\right) = \bar{y}/\bar{x}$$ Now what are needed are a few more quantities, the first is:

$$f(\vec{\mu})=f\left(\begin{bmatrix} \mu_{y} \\ \mu_{x} \\ \end{bmatrix}\right) = \mu_{y}/\mu_{x}$$

These are the $h(B)$ and $h(\beta)$ respectively in notation in the Wikipedia link.
Next you need the vector of partial derivatives of $f(\vec{\mu})$, this is:

$$\nabla f(\vec{\mu})=\begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}$$ Also we need the variance covariance matrix of the vector

$$\begin{bmatrix} \bar{y} \\ \bar{x} \\ \end{bmatrix}$$ which is

$$\begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix}.$$ Note this variance-covariance matrix is the $\Sigma/n$ in the Wikipedia notation. For a proof that $\mathbb{C}ov(\bar{y},\bar{x}) =\mathbb{C}ov(x,y)$ see Estimating the covariance of the means from two samples? Now the only thing left is to calculate the quadratic form:

$$\nabla f(\vec{\mu})^T\begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix}\nabla f(\vec{\mu}) = \begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}^T \begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix} \begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}.$$

Which when I worked this out gives you the equation:

$$\sigma^2_R=\frac{\sigma_x^2}{n\mu_x^2} - 2\frac{\sigma_{xy}}{\mu_x^2}+\frac{\sigma^2_y\mu_x^2}{n\mu_x^4},$$

where this quantity is the variance of the delta method normal distribution. Putting this altogether gives us that

$$\sqrt{n}\left(\frac{\bar{y}}{\bar{x}}-\frac{\mu_y}{\mu_x}\right) \sim N(0,\sigma^2_R)$$

So you can estimate the ratio of the population means by the ratio of the sample means provided you can estimate variances and the covariance, or equivalently, the correlation $\rho = \frac{\sigma_{xy}}{\sigma_x\sigma_y}$, by substitution, $\sigma_x\sigma_y\rho = \sigma_{xy}$. This is how the delta method is most commonly used in the derivation of the ratio estimator distribution.

The multivariate delta method has a heuristic justification here: https://en.wikipedia.org/wiki/Delta_method#Multivariate_delta_method. For the multivariate delta method you have a specific function $f$ that takes a vector argument which is $p$ dimensional and maps this to a $k$ dimensional space. In the case of a ratio estimator $p=2$ and $k=1$. The function $f$ is

$$f\left(\begin{bmatrix} \bar{y} \\ \bar{x} \\ \end{bmatrix}\right) = \bar{y}/\bar{x}$$ Now what are needed are a few more quantities, the first is:

$$f(\vec{\mu})=f\left(\begin{bmatrix} \mu_{y} \\ \mu_{x} \\ \end{bmatrix}\right) = \mu_{y}/\mu_{x}$$

These are the $h(B)$ and $h(\beta)$ respectively in notation in the Wikipedia link.
Next you need the vector of partial derivatives of $f(\vec{\mu})$, this is:

$$\nabla f(\vec{\mu})=\begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}$$ Also we need the variance covariance matrix of the vector

$$\begin{bmatrix} \bar{y} \\ \bar{x} \\ \end{bmatrix}$$ which is

$$\begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix}.$$ Note this variance-covariance matrix is the $\Sigma/n$ in the Wikipedia notation. For a proof that $\mathbb{C}ov(\bar{y},\bar{x}) =\mathbb{C}ov(x,y)$ see Estimating the covariance of the means from two samples? Now the only thing left is to calculate the quadratic form:

$$\nabla f(\vec{\mu})^T\begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix}\nabla f(\vec{\mu}) = \begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}^T \begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix} \begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}.$$

Which when I worked this out gives you the equation:

$$\sigma^2_R=\frac{\sigma_y^2}{n\mu_x^2} - 2\frac{\mu_y\sigma_{xy}}{\mu_x^3}+\frac{\sigma^2_x\mu_y^2}{n\mu_x^4},$$

where this quantity is the variance of the delta method normal distribution. Putting this altogether gives us that

$$\sqrt{n}\left(\frac{\bar{y}}{\bar{x}}-\frac{\mu_y}{\mu_x}\right) \sim N(0,\sigma^2_R)$$

So you can estimate the ratio of the population means by the ratio of the sample means provided you can estimate variances and the covariance, or equivalently, the correlation $\rho = \frac{\sigma_{xy}}{\sigma_x\sigma_y}$, by substitution, $\sigma_x\sigma_y\rho = \sigma_{xy}$. This is how the delta method is most commonly used in the derivation of the ratio estimator distribution.

The multivariate delta method has a heuristic justification here: https://en.wikipedia.org/wiki/Delta_method#Multivariate_delta_method. For the multivariate delta method you have a specific function $f$ that takes a vector argument which is $p$ dimensional and maps this to a $k$ dimensional space. In the case of a ratio estimator $p=2$ and $k=1$. The function $f$ is

$$f\left(\begin{bmatrix} \bar{y} \\ \bar{x} \\ \end{bmatrix}\right) = \bar{y}/\bar{x}$$ Now what are needed are a few more quantities, the first is:

$$f(\vec{\mu})=f\left(\begin{bmatrix} \mu_{y} \\ \mu_{x} \\ \end{bmatrix}\right) = \mu_{y}/\mu_{x}$$

These are the $h(B)$ and $h(\beta)$ respectively in notation in the Wikipedia link.
Next you need the vector of partial derivatives of $f(\vec{\mu})$, this is:

$$\nabla f(\vec{\mu})=\begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}$$ Also we need the variance covariance matrix of the vector

$$\begin{bmatrix} \bar{y} \\ \bar{x} \\ \end{bmatrix}$$ which is

$$\begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix}.$$ Note this variance-covariance matrix is the $\Sigma/n$ in the Wikipedia notation. For a proof that $\mathbb{C}ov(\bar{y},\bar{x}) =\mathbb{C}ov(x,y)$ see Estimating the covariance of the means from two samples? Now the only thing left is to calculate the quadratic form:

$$\nabla f(\vec{\mu})^T\begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix}\nabla f(\vec{\mu}) = \begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}^T \begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix} \begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}.$$

Which when I worked this out gives you the equation:

$$\sigma^2_R=\frac{\sigma_x^2}{n\mu_y^2} - 2\frac{\sigma_{xy}}{\mu_x^2}+\frac{\sigma^2_y\mu_y^2}{n\mu_x^4},$$

where this quantity is the variance of the delta method normal distribution. Putting this altogether gives us that

$$\sqrt{n}\left(\frac{\bar{y}}{\bar{x}}-\frac{\mu_y}{\mu_x}\right) \sim N(0,\sigma^2_R)$$

So you can estimate the ratio of the population means by the ratio of the sample means provided you can estimate variances and the covariance, or equivalently, the correlation $\rho = \frac{\sigma_{xy}}{\sigma_x\sigma_y}$, by substitution, $\sigma_x\sigma_y\rho = \sigma_{xy}$. This is how the delta method is most commonly used in the derivation of the ratio estimator distribution.

The multivariate delta method has a heuristic justification here: https://en.wikipedia.org/wiki/Delta_method#Multivariate_delta_method. For the multivariate delta method you have a specific function $f$ that takes a vector argument which is $p$ dimensional and maps this to a $k$ dimensional space. In the case of a ratio estimator $p=2$ and $k=1$. The function $f$ is

$$f\left(\begin{bmatrix} \bar{y} \\ \bar{x} \\ \end{bmatrix}\right) = \bar{y}/\bar{x}$$ Now what are needed are a few more quantities, the first is:

$$f(\vec{\mu})=f\left(\begin{bmatrix} \mu_{y} \\ \mu_{x} \\ \end{bmatrix}\right) = \mu_{y}/\mu_{x}$$

These are the $h(B)$ and $h(\beta)$ respectively in notation in the Wikipedia link.
Next you need the vector of partial derivatives of $f(\vec{\mu})$, this is:

$$\nabla f(\vec{\mu})=\begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}$$ Also we need the variance covariance matrix of the vector

$$\begin{bmatrix} \bar{y} \\ \bar{x} \\ \end{bmatrix}$$ which is

$$\begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix}.$$ Note this variance-covariance matrix is the $\Sigma/n$ in the Wikipedia notation. For a proof that $\mathbb{C}ov(\bar{y},\bar{x}) =\mathbb{C}ov(x,y)$ see Estimating the covariance of the means from two samples? Now the only thing left is to calculate the quadratic form:

$$\nabla f(\vec{\mu})^T\begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix}\nabla f(\vec{\mu}) = \begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}^T \begin{bmatrix} \sigma^2_{y}/n & \sigma_{yx} \\ \sigma_{yx} & \sigma^2_{x}/n\\ \end{bmatrix} \begin{bmatrix} \frac{1}{\mu_{x}} \\ \frac{-\mu_{y}}{\mu_{x}^2} \\ \end{bmatrix}.$$

Which when I worked this out gives you the equation:

$$\sigma^2_R=\frac{\sigma_x^2}{n\mu_x^2} - 2\frac{\sigma_{xy}}{\mu_x^2}+\frac{\sigma^2_y\mu_x^2}{n\mu_x^4},$$

where this quantity is the variance of the delta method normal distribution. Putting this altogether gives us that

$$\sqrt{n}\left(\frac{\bar{y}}{\bar{x}}-\frac{\mu_y}{\mu_x}\right) \sim N(0,\sigma^2_R)$$

So you can estimate the ratio of the population means by the ratio of the sample means provided you can estimate variances and the covariance, or equivalently, the correlation $\rho = \frac{\sigma_{xy}}{\sigma_x\sigma_y}$, by substitution, $\sigma_x\sigma_y\rho = \sigma_{xy}$. This is how the delta method is most commonly used in the derivation of the ratio estimator distribution.