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.
 A: 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 !
A: Consider pages 287 - 290 in the original Wald(1940) paper. It walks you through the derivation of the variance. 
A: For future readers: the $\beta_1$ in @MichaelChirico's post is $\beta_\mathrm{wald}$ (the average treatment effect, in usual use), and the covariance formula follows from $E[Y|X] = \beta_0 + \beta_1 X$ without loss of generality (since X is binary).
(Apologies for the extra answer; I have insufficient reputation to comment)
