# How can I compute the standard error of the Wald estimator?

According to Cameron and Trivedi Microeconometrics 2006, page 98-99, the Wald estimator can be written :

$$\widehat{\beta}_{Wald} = \frac{(\bar{y_1} - \bar{y_0})}{(\bar{x_1} - \bar{x_0})}$$

with :

• $\bar{y_1}$ : the mean of y for individuals with Z = 1
• $\bar{y_0}$ : the mean of y for individuals with Z = 0
• $\bar{x_1}$ : the mean of x for individuals with Z = 1
• $\bar{x_0}$ : the mean of x for individuals with Z = 0

I'm in a special case where $y_i$ is a dichotomic variable. I would like to know if there is a formula to compute the standard error in this simple case.

PS : I know there is a similar question (questions60893), but my question is really about the algebra and how to compute it knowing only $\bar{y_1}$, $\bar{y_0}$, $\bar{x_1}$, $\bar{x_0}$, $N_1$ the number of people with $Z=1$ and $N_0$ the number of people with $Z=0$, not about the stata command.

• – Arne Jonas Warnke Jun 17 '16 at 12:52
• I don't think so. See my PS. – PAC Jun 17 '16 at 13:48
• Ok, thanks for the clarification, I think soakely's comment referring to Fieller could be a helpful but the reference to the original Wald paper is certainly interesting. – Arne Jonas Warnke Jun 17 '16 at 18:54
• The Wald estimator is equivalent to the 2SLS (with binary endogeneous and binary instrument). You could just use a 2SLS and its standard errors? – Matifou Jun 18 '16 at 7:45
• I know but I don't have access to the raw data. I only have aggregate quantities. – PAC Jun 25 '16 at 10:26

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 !

Consider pages 287 - 290 in the original Wald(1940) paper. It walks you through the derivation of the variance.

• I think you meant Wald (1940). :-) – tchakravarty Jul 3 '16 at 5:12
• @tchakravarty, excellent catch. Who knew Wald could write papers 40 years after his death – Gene Burinsky Jul 3 '16 at 15:17