# estimation of population ratio using delta method [closed]

am writing a seminar on estimation of population ratio using delta method and am having problem on the literature review. i have written the introduction but need help on on how to use delta method to estimate the population ratio and the methodology

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• What ratio is this? A ratio of means? A population attributable fraction? – AdamO Jun 12 '18 at 15:53

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.

• I believe that the evaluation of the quadratic form should give a slightly different result. $$\sigma^2_R=\frac{\sigma_y^2}{n\mu_x^2} - 2\frac{\sigma_{xy}\mu_y}{\mu_x^3}+\frac{\sigma^2_x\mu_y^2}{n\mu_x^4}$$ – Martijn Weterings Mar 28 at 19:19
• @MartijnWeterings you are correct that is a typo going from the quadratic form to the equation for $\sigma^2_R$. Will fix that now, thanks for catching my error. – Lucas Roberts Mar 29 at 0:52