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edit approved May 4, 2017 at 22:17
MichaelChirico
edit approved May 4, 2017 at 22:17
MichaelChirico
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Here is my answer to my question. I hope there is no mistake in calculus.

We have :

  • $y_{1,i}$ a dichotomicdichotomous random variable following a Bernouilli distribution with parameter $\mu_{y_1}$
  • $y_{0,i}$ a dichotomicdichotomous random variable following a Bernouilli distribution with parameter $\mu_{y_0}$
  • $x_{1,i}$ a dichotomicdichotomous random variable following a Bernouilli distribution with parameter $\mu_{x_1}$
  • $x_{0,i}$ a dichotomicdichotomous random variable following a Bernouilli distribution with parameter $\mu_{x_0}$

The Wald estimator is defined as :

$$ \beta_{Wald} = \frac{\mu_{y_1} - \mu_{y_0}}{\mu_{x_1} - \mu_{x_0}} $$

This can be estimated using the plug-in estimator :

$$ \widehat{\beta_{Wald}} = \frac{\bar{y_1} - \bar{y_0}}{\bar{x_1} - \bar{x_0}} $$

I want to know the distribution of $\widehat{\beta_{Wald}}$. Since $\bar{y_1} - \bar{y_0}$ and $\bar{x_1} - \bar{x_0}$ converge to a normal distribution, I know that I can derive the distribution of $\widehat{\beta_{Wald}}$ using the Delta method (See Larry Wasserman, All of Statistics : A Concise Course in Statistical Inference, Springer, coll. « Springer Texts in Statistics », 2004, page 79).

I define two new variables :

  • $U = \bar{y_1} - \bar{y_0}$
  • $V = \bar{X_1} - \bar{X_0}$

I know that :

  • $U \xrightarrow{L}\, \mathcal{N}(\mu_U, \sigma^2_U)$
  • $V \xrightarrow{L}\, \mathcal{N}(\mu_V, \sigma^2_V)$

I define the function $g(U,V) = U/V$. According to the Delta method, I know that :

$$ g(U,V) \xrightarrow{L}\, \mathcal{N}\left(g(\mu_U, \mu_V), Dg(\mu_U, \mu_V)^T\Sigma Dg(\mu_U, \mu_V)\right) $$

with $Dg(\mu_U, \mu_V)$ the Jacobian matrix of function g and $Sigma$$\Sigma$ the variance-covariance matrix of vector $(U,V)$.

So I compute the Jacobian :

$$Dg \left( \begin{array}{c} \mu_U \\ \mu_V \end{array} \right) = \left(\begin{array}{rcl} \frac{1}{\mu_V} \\ \frac{-\mu_U}{\mu_V^2} \end{array} \right) $$

and I have the variance-covariance matrix :

$$ \Sigma = \left( \begin{array}{cc} \sigma^2_U & \sigma_{U,V} \\ \sigma_{U,V} & \sigma^2_V \end{array} \right) $$

So the variance of $g(U/V)$ is :

$$ Dg(\mu_U, \mu_V)^T \Sigma Dg(\mu_U, \mu_V) = \frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V $$

In this case, since $y_i$ follow a Bernouilli distribution, its variance is just $\mu_{y_i} (1-\mu_{y_i})$ and can be estimated using the plug-in estimator as $\bar{y_i} (1-\bar{y_i})$. Therefore I can estimate the following quantities :

$$ \begin{eqnarray} \sigma^2_U & = & V(\bar{y_1} - \bar{y_0})\\ & = & V(\bar{y_1}) + V(\bar{y_0}) & = & \frac{1}{N_1} \bar{y_1} (1 - \bar{y_1}) + \frac{1}{N_0} \bar{y_0} (1 - \bar{y_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma^2_V & = & V(\bar{x_1} - \bar{x_0})\\ & = & V(\bar{x_1}) + V(\bar{x_0}) & = & \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} \bar{x_0} (1 - \bar{x_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma_{U,V} & = & cov(\bar{y_1} - \bar{y_0}, \bar{x_1} - \bar{x_0})\\ & = & \beta_1 V(\bar{x_1}) + \beta_1 V(\bar{x_0})\\ & = & \beta_1 \left( \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} (\bar{x_0} (1-\bar{x_0}))\right) \end{eqnarray} $$

So I can have sample estimates of all quantities in the equation $\frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V$. Therefore I can get the variance of my Wald estimator and compute my standard error !

Here is my answer to my question. I hope there is no mistake in calculus.

We have :

  • $y_{1,i}$ a dichotomic random variable following a Bernouilli distribution with parameter $\mu_{y_1}$
  • $y_{0,i}$ a dichotomic random variable following a Bernouilli distribution with parameter $\mu_{y_0}$
  • $x_{1,i}$ a dichotomic random variable following a Bernouilli distribution with parameter $\mu_{x_1}$
  • $x_{0,i}$ a dichotomic random variable following a Bernouilli distribution with parameter $\mu_{x_0}$

The Wald estimator is defined as :

$$ \beta_{Wald} = \frac{\mu_{y_1} - \mu_{y_0}}{\mu_{x_1} - \mu_{x_0}} $$

This can be estimated using the plug-in estimator :

$$ \widehat{\beta_{Wald}} = \frac{\bar{y_1} - \bar{y_0}}{\bar{x_1} - \bar{x_0}} $$

I want to know the distribution of $\widehat{\beta_{Wald}}$. Since $\bar{y_1} - \bar{y_0}$ and $\bar{x_1} - \bar{x_0}$ converge to a normal distribution, I know that I can derive the distribution of $\widehat{\beta_{Wald}}$ using the Delta method (See Larry Wasserman, All of Statistics : A Concise Course in Statistical Inference, Springer, coll. « Springer Texts in Statistics », 2004, page 79).

I define two new variables :

  • $U = \bar{y_1} - \bar{y_0}$
  • $V = \bar{X_1} - \bar{X_0}$

I know that :

  • $U \xrightarrow{L}\, \mathcal{N}(\mu_U, \sigma^2_U)$
  • $V \xrightarrow{L}\, \mathcal{N}(\mu_V, \sigma^2_V)$

I define the function $g(U,V) = U/V$. According to the Delta method, I know that :

$$ g(U,V) \xrightarrow{L}\, \mathcal{N}\left(g(\mu_U, \mu_V), Dg(\mu_U, \mu_V)^T\Sigma Dg(\mu_U, \mu_V)\right) $$

with $Dg(\mu_U, \mu_V)$ the Jacobian matrix of function g and $Sigma$ the variance-covariance matrix of vector $(U,V)$.

So I compute the Jacobian :

$$Dg \left( \begin{array}{c} \mu_U \\ \mu_V \end{array} \right) = \left(\begin{array}{rcl} \frac{1}{\mu_V} \\ \frac{-\mu_U}{\mu_V^2} \end{array} \right) $$

and I have the variance-covariance matrix :

$$ \Sigma = \left( \begin{array}{cc} \sigma^2_U & \sigma_{U,V} \\ \sigma_{U,V} & \sigma^2_V \end{array} \right) $$

So the variance of $g(U/V)$ is :

$$ Dg(\mu_U, \mu_V)^T \Sigma Dg(\mu_U, \mu_V) = \frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V $$

In this case, since $y_i$ follow a Bernouilli distribution, its variance is just $\mu_{y_i} (1-\mu_{y_i})$ and can be estimated using the plug-in estimator as $\bar{y_i} (1-\bar{y_i})$. Therefore I can estimate the following quantities :

$$ \begin{eqnarray} \sigma^2_U & = & V(\bar{y_1} - \bar{y_0})\\ & = & V(\bar{y_1}) + V(\bar{y_0}) & = & \frac{1}{N_1} \bar{y_1} (1 - \bar{y_1}) + \frac{1}{N_0} \bar{y_0} (1 - \bar{y_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma^2_V & = & V(\bar{x_1} - \bar{x_0})\\ & = & V(\bar{x_1}) + V(\bar{x_0}) & = & \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} \bar{x_0} (1 - \bar{x_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma_{U,V} & = & cov(\bar{y_1} - \bar{y_0}, \bar{x_1} - \bar{x_0})\\ & = & \beta_1 V(\bar{x_1}) + \beta_1 V(\bar{x_0})\\ & = & \beta_1 \left( \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} (\bar{x_0} (1-\bar{x_0}))\right) \end{eqnarray} $$

So I can have sample estimates of all quantities in the equation $\frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V$. Therefore I can get the variance of my Wald estimator and compute my standard error !

Here is my answer to my question. I hope there is no mistake in calculus.

We have :

  • $y_{1,i}$ a dichotomous random variable following a Bernouilli distribution with parameter $\mu_{y_1}$
  • $y_{0,i}$ a dichotomous random variable following a Bernouilli distribution with parameter $\mu_{y_0}$
  • $x_{1,i}$ a dichotomous random variable following a Bernouilli distribution with parameter $\mu_{x_1}$
  • $x_{0,i}$ a dichotomous random variable following a Bernouilli distribution with parameter $\mu_{x_0}$

The Wald estimator is defined as :

$$ \beta_{Wald} = \frac{\mu_{y_1} - \mu_{y_0}}{\mu_{x_1} - \mu_{x_0}} $$

This can be estimated using the plug-in estimator :

$$ \widehat{\beta_{Wald}} = \frac{\bar{y_1} - \bar{y_0}}{\bar{x_1} - \bar{x_0}} $$

I want to know the distribution of $\widehat{\beta_{Wald}}$. Since $\bar{y_1} - \bar{y_0}$ and $\bar{x_1} - \bar{x_0}$ converge to a normal distribution, I know that I can derive the distribution of $\widehat{\beta_{Wald}}$ using the Delta method (See Larry Wasserman, All of Statistics : A Concise Course in Statistical Inference, Springer, coll. « Springer Texts in Statistics », 2004, page 79).

I define two new variables :

  • $U = \bar{y_1} - \bar{y_0}$
  • $V = \bar{X_1} - \bar{X_0}$

I know that :

  • $U \xrightarrow{L}\, \mathcal{N}(\mu_U, \sigma^2_U)$
  • $V \xrightarrow{L}\, \mathcal{N}(\mu_V, \sigma^2_V)$

I define the function $g(U,V) = U/V$. According to the Delta method, I know that :

$$ g(U,V) \xrightarrow{L}\, \mathcal{N}\left(g(\mu_U, \mu_V), Dg(\mu_U, \mu_V)^T\Sigma Dg(\mu_U, \mu_V)\right) $$

with $Dg(\mu_U, \mu_V)$ the Jacobian matrix of function g and $\Sigma$ the variance-covariance matrix of vector $(U,V)$.

So I compute the Jacobian :

$$Dg \left( \begin{array}{c} \mu_U \\ \mu_V \end{array} \right) = \left(\begin{array}{rcl} \frac{1}{\mu_V} \\ \frac{-\mu_U}{\mu_V^2} \end{array} \right) $$

and I have the variance-covariance matrix :

$$ \Sigma = \left( \begin{array}{cc} \sigma^2_U & \sigma_{U,V} \\ \sigma_{U,V} & \sigma^2_V \end{array} \right) $$

So the variance of $g(U/V)$ is :

$$ Dg(\mu_U, \mu_V)^T \Sigma Dg(\mu_U, \mu_V) = \frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V $$

In this case, since $y_i$ follow a Bernouilli distribution, its variance is just $\mu_{y_i} (1-\mu_{y_i})$ and can be estimated using the plug-in estimator as $\bar{y_i} (1-\bar{y_i})$. Therefore I can estimate the following quantities :

$$ \begin{eqnarray} \sigma^2_U & = & V(\bar{y_1} - \bar{y_0})\\ & = & V(\bar{y_1}) + V(\bar{y_0}) & = & \frac{1}{N_1} \bar{y_1} (1 - \bar{y_1}) + \frac{1}{N_0} \bar{y_0} (1 - \bar{y_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma^2_V & = & V(\bar{x_1} - \bar{x_0})\\ & = & V(\bar{x_1}) + V(\bar{x_0}) & = & \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} \bar{x_0} (1 - \bar{x_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma_{U,V} & = & cov(\bar{y_1} - \bar{y_0}, \bar{x_1} - \bar{x_0})\\ & = & \beta_1 V(\bar{x_1}) + \beta_1 V(\bar{x_0})\\ & = & \beta_1 \left( \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} (\bar{x_0} (1-\bar{x_0}))\right) \end{eqnarray} $$

So I can have sample estimates of all quantities in the equation $\frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V$. Therefore I can get the variance of my Wald estimator and compute my standard error !

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PAC
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Here is my answer to my question. I hope there is no mistake in calculus.

We have :

  • $y_{1,i}$ a dichotomic random variable following a Bernouilli distribution with parameter $\mu_{y_1}$
  • $y_{0,i}$ a dichotomic random variable following a Bernouilli distribution with parameter $\mu_{y_0}$
  • $x_{1,i}$ a dichotomic random variable following a Bernouilli distribution with parameter $\mu_{x_1}$
  • $x_{0,i}$ a dichotomic random variable following a Bernouilli distribution with parameter $\mu_{x_0}$

$$ \beta_{Wald} = \frac{\bar{y_1} - \bar{y_0}}{\bar{x_1} - \bar{x_0}} $$ The Wald estimator is defined as :

$$ \beta_{Wald} = \frac{\mu_{y_1} - \mu_{y_0}}{\mu_{x_1} - \mu_{x_0}} $$

This can be estimated using the plug-in estimator :

$$ \widehat{\beta_{Wald}} = \frac{\bar{y_1} - \bar{y_0}}{\bar{x_1} - \bar{x_0}} $$

I want to know the distribution of $\beta_{Wald}$$\widehat{\beta_{Wald}}$. Since $\bar{y_1} - \bar{y_0}$ and $\bar{x_1} - \bar{x_0}$ converge to a normal distribution, I know that I can derive the distribution of $\beta_{Wald}$$\widehat{\beta_{Wald}}$ using the Delta method (See Larry Wasserman, All of Statistics : A Concise Course in Statistical Inference, Springer, coll. « Springer Texts in Statistics », 2004, page 79).

I define two new variables :

  • $U = \bar{y_1} - \bar{y_0}$
  • $V = \bar{X_1} - \bar{X_0}$

I know that :

  • $U \xrightarrow{L}\, \mathcal{N}(\mu_U, \sigma^2_U)$
  • $V \xrightarrow{L}\, \mathcal{N}(\mu_V, \sigma^2_V)$

I define the function $g(U,V) = U/V$. According to the Delta method, I know that :

$$ g(U,V) \xrightarrow{L}\, \mathcal{N}\left(g(\mu_U, \mu_V), Dg(\mu_U, \mu_V)^T\Sigma Dg(\mu_U, \mu_V)\right) $$

with $Dg(\mu_U, \mu_V)$ the Jacobian matrix of function g and $Sigma$ the variance-covariance matrix of vector $(U,V)$.

So I compute the Jacobian :

$$Dg \left( \begin{array}{c} \mu_U \\ \mu_V \end{array} \right) = \left(\begin{array}{rcl} \frac{1}{\mu_V} \\ \frac{-\mu_U}{\mu_V^2} \end{array} \right) $$

and I have the variance-covariance matrix :

$$ \Sigma = \left( \begin{array}{cc} \sigma^2_U & \sigma_{U,V} \\ \sigma_{U,V} & \sigma^2_V \end{array} \right) $$

So the variance of $g(U/V)$ is :

$$ Dg(\mu_U, \mu_V)^T \Sigma Dg(\mu_U, \mu_V) = \frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V $$

In this case, since $y_i$ follow a Bernouilli distribution, its variance is just $\mu_{y_i} (1-\mu_{y_i})$ and can be estimated using the plug-in estimator as $\bar{y_i} (1-\bar{y_i})$. Therefore I can estimate the following quantities :

$$ \begin{eqnarray} \sigma^2_U & = & V(\bar{y_1} - \bar{y_0})\\ & = & V(\bar{y_1}) + V(\bar{y_0}) & = & \frac{1}{N_1} \bar{y_1} (1 - \bar{y_1}) + \frac{1}{N_0} \bar{y_0} (1 - \bar{y_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma^2_V & = & V(\bar{x_1} - \bar{x_0})\\ & = & V(\bar{x_1}) + V(\bar{x_0}) & = & \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} \bar{x_0} (1 - \bar{x_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma_{U,V} & = & cov(\bar{y_1} - \bar{y_0}, \bar{x_1} - \bar{x_0})\\ & = & \beta_1 V(\bar{x_1}) + \beta_1 V(\bar{x_0})\\ & = & \beta_1 \left( \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} (\bar{x_0} (1-\bar{x_0}))\right) \end{eqnarray} $$

So I can have sample estimates of all quantities in the equation $\frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V$. Therefore I can get the variance of my Wald estimator and compute my standard error !

Here is my answer to my question. I hope there is no mistake in calculus.

We have :

$$ \beta_{Wald} = \frac{\bar{y_1} - \bar{y_0}}{\bar{x_1} - \bar{x_0}} $$

I want to know the distribution of $\beta_{Wald}$. Since $\bar{y_1} - \bar{y_0}$ and $\bar{x_1} - \bar{x_0}$ converge to a normal distribution, I know that I can derive the distribution of $\beta_{Wald}$ using the Delta method (See Larry Wasserman, All of Statistics : A Concise Course in Statistical Inference, Springer, coll. « Springer Texts in Statistics », 2004, page 79).

I define two new variables :

  • $U = \bar{y_1} - \bar{y_0}$
  • $V = \bar{X_1} - \bar{X_0}$

I know that :

  • $U \xrightarrow{L}\, \mathcal{N}(\mu_U, \sigma^2_U)$
  • $V \xrightarrow{L}\, \mathcal{N}(\mu_V, \sigma^2_V)$

I define the function $g(U,V) = U/V$. According to the Delta method, I know that :

$$ g(U,V) \xrightarrow{L}\, \mathcal{N}\left(g(\mu_U, \mu_V), Dg(\mu_U, \mu_V)^T\Sigma Dg(\mu_U, \mu_V)\right) $$

with $Dg(\mu_U, \mu_V)$ the Jacobian matrix of function g and $Sigma$ the variance-covariance matrix of vector $(U,V)$.

So I compute the Jacobian :

$$Dg \left( \begin{array}{c} \mu_U \\ \mu_V \end{array} \right) = \left(\begin{array}{rcl} \frac{1}{\mu_V} \\ \frac{-\mu_U}{\mu_V^2} \end{array} \right) $$

and I have the variance-covariance matrix :

$$ \Sigma = \left( \begin{array}{cc} \sigma^2_U & \sigma_{U,V} \\ \sigma_{U,V} & \sigma^2_V \end{array} \right) $$

So the variance of $g(U/V)$ is :

$$ Dg(\mu_U, \mu_V)^T \Sigma Dg(\mu_U, \mu_V) = \frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V $$

In this case, since $y_i$ follow a Bernouilli distribution, its variance is just $\mu_{y_i} (1-\mu_{y_i})$ and can be estimated using the plug-in estimator as $\bar{y_i} (1-\bar{y_i})$. Therefore I can estimate the following quantities :

$$ \begin{eqnarray} \sigma^2_U & = & V(\bar{y_1} - \bar{y_0})\\ & = & V(\bar{y_1}) + V(\bar{y_0}) & = & \frac{1}{N_1} \bar{y_1} (1 - \bar{y_1}) + \frac{1}{N_0} \bar{y_0} (1 - \bar{y_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma^2_V & = & V(\bar{x_1} - \bar{x_0})\\ & = & V(\bar{x_1}) + V(\bar{x_0}) & = & \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} \bar{x_0} (1 - \bar{x_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma_{U,V} & = & cov(\bar{y_1} - \bar{y_0}, \bar{x_1} - \bar{x_0})\\ & = & \beta_1 V(\bar{x_1}) + \beta_1 V(\bar{x_0})\\ & = & \beta_1 \left( \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} (\bar{x_0} (1-\bar{x_0}))\right) \end{eqnarray} $$

So I can have sample estimates of all quantities in the equation $\frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V$. Therefore I can get the variance of my Wald estimator and compute my standard error !

Here is my answer to my question. I hope there is no mistake in calculus.

We have :

  • $y_{1,i}$ a dichotomic random variable following a Bernouilli distribution with parameter $\mu_{y_1}$
  • $y_{0,i}$ a dichotomic random variable following a Bernouilli distribution with parameter $\mu_{y_0}$
  • $x_{1,i}$ a dichotomic random variable following a Bernouilli distribution with parameter $\mu_{x_1}$
  • $x_{0,i}$ a dichotomic random variable following a Bernouilli distribution with parameter $\mu_{x_0}$

The Wald estimator is defined as :

$$ \beta_{Wald} = \frac{\mu_{y_1} - \mu_{y_0}}{\mu_{x_1} - \mu_{x_0}} $$

This can be estimated using the plug-in estimator :

$$ \widehat{\beta_{Wald}} = \frac{\bar{y_1} - \bar{y_0}}{\bar{x_1} - \bar{x_0}} $$

I want to know the distribution of $\widehat{\beta_{Wald}}$. Since $\bar{y_1} - \bar{y_0}$ and $\bar{x_1} - \bar{x_0}$ converge to a normal distribution, I know that I can derive the distribution of $\widehat{\beta_{Wald}}$ using the Delta method (See Larry Wasserman, All of Statistics : A Concise Course in Statistical Inference, Springer, coll. « Springer Texts in Statistics », 2004, page 79).

I define two new variables :

  • $U = \bar{y_1} - \bar{y_0}$
  • $V = \bar{X_1} - \bar{X_0}$

I know that :

  • $U \xrightarrow{L}\, \mathcal{N}(\mu_U, \sigma^2_U)$
  • $V \xrightarrow{L}\, \mathcal{N}(\mu_V, \sigma^2_V)$

I define the function $g(U,V) = U/V$. According to the Delta method, I know that :

$$ g(U,V) \xrightarrow{L}\, \mathcal{N}\left(g(\mu_U, \mu_V), Dg(\mu_U, \mu_V)^T\Sigma Dg(\mu_U, \mu_V)\right) $$

with $Dg(\mu_U, \mu_V)$ the Jacobian matrix of function g and $Sigma$ the variance-covariance matrix of vector $(U,V)$.

So I compute the Jacobian :

$$Dg \left( \begin{array}{c} \mu_U \\ \mu_V \end{array} \right) = \left(\begin{array}{rcl} \frac{1}{\mu_V} \\ \frac{-\mu_U}{\mu_V^2} \end{array} \right) $$

and I have the variance-covariance matrix :

$$ \Sigma = \left( \begin{array}{cc} \sigma^2_U & \sigma_{U,V} \\ \sigma_{U,V} & \sigma^2_V \end{array} \right) $$

So the variance of $g(U/V)$ is :

$$ Dg(\mu_U, \mu_V)^T \Sigma Dg(\mu_U, \mu_V) = \frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V $$

In this case, since $y_i$ follow a Bernouilli distribution, its variance is just $\mu_{y_i} (1-\mu_{y_i})$ and can be estimated using the plug-in estimator as $\bar{y_i} (1-\bar{y_i})$. Therefore I can estimate the following quantities :

$$ \begin{eqnarray} \sigma^2_U & = & V(\bar{y_1} - \bar{y_0})\\ & = & V(\bar{y_1}) + V(\bar{y_0}) & = & \frac{1}{N_1} \bar{y_1} (1 - \bar{y_1}) + \frac{1}{N_0} \bar{y_0} (1 - \bar{y_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma^2_V & = & V(\bar{x_1} - \bar{x_0})\\ & = & V(\bar{x_1}) + V(\bar{x_0}) & = & \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} \bar{x_0} (1 - \bar{x_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma_{U,V} & = & cov(\bar{y_1} - \bar{y_0}, \bar{x_1} - \bar{x_0})\\ & = & \beta_1 V(\bar{x_1}) + \beta_1 V(\bar{x_0})\\ & = & \beta_1 \left( \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} (\bar{x_0} (1-\bar{x_0}))\right) \end{eqnarray} $$

So I can have sample estimates of all quantities in the equation $\frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V$. Therefore I can get the variance of my Wald estimator and compute my standard error !

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Here is my answer to my question. I hope there is no mistake in calculus.

We have :

$$ \beta_{Wald} = \frac{\bar{y_1} - \bar{y_0}}{\bar{x_1} - \bar{x_0}} $$

I want to know the distribution of $\beta_{Wald}$. Since $\bar{y_1} - \bar{y_0}$ and $\bar{x_1} - \bar{x_0}$ converge to a normal distribution, I know that I can derive the distribution of $\beta_{Wald}$ using the Delta method (See Larry Wasserman, All of Statistics : A Concise Course in Statistical Inference, Springer, coll. « Springer Texts in Statistics », 2004, page 79).

I define two new variables :

  • $U = \bar{y_1} - \bar{y_0}$
  • $V = \bar{X_1} - \bar{X_0}$

I know that :

  • $U \xrightarrow{L}\, \mathcal{N}(\mu_U, \sigma^2_U)$
  • $V \xrightarrow{L}\, \mathcal{N}(\mu_V, \sigma^2_V)$

I define the function $g(U,V) = U/V$. According to the Delta method, I know that :

$$ g(U,V) \xrightarrow{L}\, \mathcal{N}\left(g(\mu_U, \mu_V), Dg(\mu_U, \mu_V)^T\Sigma Dg(\mu_U, \mu_V)\right) $$

with $Dg(\mu_U, \mu_V)$ the Jacobian matrix of function g and $Sigma$ the variance-covariance matrix of vector $(U,V)$.

So I compute the Jacobian :

$$Dg(\mu_U, \mu_V) = \left(\begin{array} \frac{1}{V} \\ \frac{-U}{V^2} \end{array} \right) $$$$Dg \left( \begin{array}{c} \mu_U \\ \mu_V \end{array} \right) = \left(\begin{array}{rcl} \frac{1}{\mu_V} \\ \frac{-\mu_U}{\mu_V^2} \end{array} \right) $$

and I have the variance-covariance matrix :

$$ \Sigma = \left( \begin{array}{cc} \sigma^2_U & \sigma_{U,V} \\ \sigma_{U,V} & \sigma^2_V \end{array} \right) $$

So the variance of $g(U/V)$ is :

$$ Dg(\mu_U, \mu_V)^T \Sigma Dg(\mu_U, \mu_V) = \frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V $$

In this case, since $y_i$ follow a Bernouilli distribution, its variance is just $\mu_{y_i} (1-\mu_{y_i})$ and can be estimated using the plug-in estimator as $\bar{y_i} (1-\bar{y_i})$. Therefore I can estimate the following quantities :

$$ \begin{eqnarray} \sigma^2_U & = & V(\bar{y_1} - \bar{y_0})\\ & = & V(\bar{y_1}) + V(\bar{y_0}) & = & \frac{1}{N_1} \bar{y_1} (1 - \bar{y_1}) + \frac{1}{N_0} \bar{y_0} (1 - \bar{y_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma^2_V & = & V(\bar{x_1} - \bar{x_0})\\ & = & V(\bar{x_1}) + V(\bar{x_0}) & = & \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} \bar{x_0} (1 - \bar{x_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma_{U,V} & = & cov(\bar{y_1} - \bar{y_0}, \bar{x_1} - \bar{x_0})\\ & = & \beta_1 V(\bar{x_1}) + \beta_1 V(\bar{x_0})\\ & = & \beta_1 \left( \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} (\bar{x_0} (1-\bar{x_0}))\right) \end{eqnarray} $$

So I can have sample estimates of all quantities in the equation $\frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V$. Therefore I can get the variance of my Wald estimator and compute my standard error !

Here is my answer to my question. I hope there is no mistake in calculus.

We have :

$$ \beta_{Wald} = \frac{\bar{y_1} - \bar{y_0}}{\bar{x_1} - \bar{x_0}} $$

I want to know the distribution of $\beta_{Wald}$. Since $\bar{y_1} - \bar{y_0}$ and $\bar{x_1} - \bar{x_0}$ converge to a normal distribution, I know that I can derive the distribution of $\beta_{Wald}$ using the Delta method (See Larry Wasserman, All of Statistics : A Concise Course in Statistical Inference, Springer, coll. « Springer Texts in Statistics », 2004, page 79).

I define two new variables :

  • $U = \bar{y_1} - \bar{y_0}$
  • $V = \bar{X_1} - \bar{X_0}$

I know that :

  • $U \xrightarrow{L}\, \mathcal{N}(\mu_U, \sigma^2_U)$
  • $V \xrightarrow{L}\, \mathcal{N}(\mu_V, \sigma^2_V)$

I define the function $g(U,V) = U/V$. According to the Delta method, I know that :

$$ g(U,V) \xrightarrow{L}\, \mathcal{N}\left(g(\mu_U, \mu_V), Dg(\mu_U, \mu_V)^T\Sigma Dg(\mu_U, \mu_V)\right) $$

with $Dg(\mu_U, \mu_V)$ the Jacobian matrix of function g and $Sigma$ the variance-covariance matrix of vector $(U,V)$.

So I compute the Jacobian :

$$Dg(\mu_U, \mu_V) = \left(\begin{array} \frac{1}{V} \\ \frac{-U}{V^2} \end{array} \right) $$

and I have the variance-covariance matrix :

$$ \Sigma = \left( \begin{array}{cc} \sigma^2_U & \sigma_{U,V} \\ \sigma_{U,V} & \sigma^2_V \end{array} \right) $$

So the variance of $g(U/V)$ is :

$$ Dg(\mu_U, \mu_V)^T \Sigma Dg(\mu_U, \mu_V) = \frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V $$

In this case, since $y_i$ follow a Bernouilli distribution, its variance is just $\mu_{y_i} (1-\mu_{y_i})$ and can be estimated using the plug-in estimator as $\bar{y_i} (1-\bar{y_i})$. Therefore I can estimate the following quantities :

$$ \begin{eqnarray} \sigma^2_U & = & V(\bar{y_1} - \bar{y_0})\\ & = & V(\bar{y_1}) + V(\bar{y_0}) & = & \frac{1}{N_1} \bar{y_1} (1 - \bar{y_1}) + \frac{1}{N_0} \bar{y_0} (1 - \bar{y_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma^2_V & = & V(\bar{x_1} - \bar{x_0})\\ & = & V(\bar{x_1}) + V(\bar{x_0}) & = & \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} \bar{x_0} (1 - \bar{x_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma_{U,V} & = & cov(\bar{y_1} - \bar{y_0}, \bar{x_1} - \bar{x_0})\\ & = & \beta_1 V(\bar{x_1}) + \beta_1 V(\bar{x_0})\\ & = & \beta_1 \left( \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} (\bar{x_0} (1-\bar{x_0}))\right) \end{eqnarray} $$

Here is my answer to my question. I hope there is no mistake in calculus.

We have :

$$ \beta_{Wald} = \frac{\bar{y_1} - \bar{y_0}}{\bar{x_1} - \bar{x_0}} $$

I want to know the distribution of $\beta_{Wald}$. Since $\bar{y_1} - \bar{y_0}$ and $\bar{x_1} - \bar{x_0}$ converge to a normal distribution, I know that I can derive the distribution of $\beta_{Wald}$ using the Delta method (See Larry Wasserman, All of Statistics : A Concise Course in Statistical Inference, Springer, coll. « Springer Texts in Statistics », 2004, page 79).

I define two new variables :

  • $U = \bar{y_1} - \bar{y_0}$
  • $V = \bar{X_1} - \bar{X_0}$

I know that :

  • $U \xrightarrow{L}\, \mathcal{N}(\mu_U, \sigma^2_U)$
  • $V \xrightarrow{L}\, \mathcal{N}(\mu_V, \sigma^2_V)$

I define the function $g(U,V) = U/V$. According to the Delta method, I know that :

$$ g(U,V) \xrightarrow{L}\, \mathcal{N}\left(g(\mu_U, \mu_V), Dg(\mu_U, \mu_V)^T\Sigma Dg(\mu_U, \mu_V)\right) $$

with $Dg(\mu_U, \mu_V)$ the Jacobian matrix of function g and $Sigma$ the variance-covariance matrix of vector $(U,V)$.

So I compute the Jacobian :

$$Dg \left( \begin{array}{c} \mu_U \\ \mu_V \end{array} \right) = \left(\begin{array}{rcl} \frac{1}{\mu_V} \\ \frac{-\mu_U}{\mu_V^2} \end{array} \right) $$

and I have the variance-covariance matrix :

$$ \Sigma = \left( \begin{array}{cc} \sigma^2_U & \sigma_{U,V} \\ \sigma_{U,V} & \sigma^2_V \end{array} \right) $$

So the variance of $g(U/V)$ is :

$$ Dg(\mu_U, \mu_V)^T \Sigma Dg(\mu_U, \mu_V) = \frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V $$

In this case, since $y_i$ follow a Bernouilli distribution, its variance is just $\mu_{y_i} (1-\mu_{y_i})$ and can be estimated using the plug-in estimator as $\bar{y_i} (1-\bar{y_i})$. Therefore I can estimate the following quantities :

$$ \begin{eqnarray} \sigma^2_U & = & V(\bar{y_1} - \bar{y_0})\\ & = & V(\bar{y_1}) + V(\bar{y_0}) & = & \frac{1}{N_1} \bar{y_1} (1 - \bar{y_1}) + \frac{1}{N_0} \bar{y_0} (1 - \bar{y_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma^2_V & = & V(\bar{x_1} - \bar{x_0})\\ & = & V(\bar{x_1}) + V(\bar{x_0}) & = & \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} \bar{x_0} (1 - \bar{x_0}) \end{eqnarray} $$

$$ \begin{eqnarray} \sigma_{U,V} & = & cov(\bar{y_1} - \bar{y_0}, \bar{x_1} - \bar{x_0})\\ & = & \beta_1 V(\bar{x_1}) + \beta_1 V(\bar{x_0})\\ & = & \beta_1 \left( \frac{1}{N_1} \bar{x_1} (1 - \bar{x_1}) + \frac{1}{N_0} (\bar{x_0} (1-\bar{x_0}))\right) \end{eqnarray} $$

So I can have sample estimates of all quantities in the equation $\frac{\sigma^2_U}{\mu_V^2} - 2 \frac{\mu_U}{\mu_V^3} \sigma_{U,V} + \frac{\mu_U^2}{\mu_V^4} \sigma^2_V$. Therefore I can get the variance of my Wald estimator and compute my standard error !

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