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Suppose my observed data $y$ and $x$ is generated by the following relationship for each observation $i$:

$$ y_i = h(y_i,\theta) + x_i + \varepsilon_i$$

where $x_i$ is a strictly exogenous variable , $\varepsilon_i$ is an i.i.d. error term and $\theta$ the vector or parameters of interest. Assume that the solution to this equation is unique.

When $h$ is sufficiently simple (e.g. linear), I can manually solve for $y_i$ and consistently estimate $\theta$ with OLS.

However, when $h$ is more involved, e.g., $h(y_i, \theta) = (1+\exp(\theta_1 - \theta_2 y_i)) / \theta_2$, then there is no closed-form solution. It feels tempting to estimate $\theta$ using non-linear least squares: $$ \arg \min_\theta \sum_i(y_i - h(y_i, \theta) - x_i)^2$$

but there are obvious endogeneity concerns with respect to the right-hand side of this expression.

Is there any instrument that allows dealing with these endogeneity concerns? Or how else is such a non-linear relationship without closed-form solution approached?

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  • $\begingroup$ $y_i$ is generated by a function of $y_i$ itself? Are you sure that this is correct? $h(y_i, \theta) = y_i - x_i - \varepsilon_i $. $\endgroup$
    – Tim
    Commented Jan 7, 2020 at 21:49
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    $\begingroup$ Why not write the relation as $x_i=f(y_i,\theta)+\epsilon_i$ where $\epsilon_i=-\varepsilon_i$ and $f(y_i,\theta)=y_i-h(y_i,\theta)$? That's a standard regression setting, presenting no special problems. $\endgroup$
    – whuber
    Commented Jan 7, 2020 at 22:07
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    $\begingroup$ How is that supposed to help when $x$ is assumed exogenous but $y_i$ is assumed endogenous (if that is indeed what @bonifaz means by "obvious endogeneity concerns") $\endgroup$ Commented Jan 7, 2020 at 22:56
  • $\begingroup$ @Stop "Endogenous" and "exogenous" do not add anything to the model. If they do, then that needs to be incorporated in the model statement! $\endgroup$
    – whuber
    Commented Jan 8, 2020 at 2:26
  • $\begingroup$ @Tim, yes, that's what I meant with 'implicitly defined'. $\endgroup$
    – bonifaz
    Commented Jan 8, 2020 at 9:03

2 Answers 2

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Here is some rough code in R implementing the non-linear two stage least squares estimator under the assumption that $\mathbb E[\epsilon\lvert x] = 0$. Just to see if it works ... it seems to be working up to the point of standard over-underflow issues there might be. But the non-linear two stage least squares estimator should be implemented in most statistical software.

N <- 10000
x <- rnorm(N)
e <- rnorm(N)
x <- c(0,x)
e <- c(0,e)
theta_1 <- 1
theta_2 <- 0.5

# function to solve for y for simulation of data given x and error e
makefunction <- function(x,e)
    {
        g <- function(y)
         {
         out <- y - (1 + exp(theta_1 - theta_2*y)/theta_2) - x - e
           return(out)
         }
      return(g)
    }

y <- rep(NA,N+1)
for (i in 1:(N+1))
    {
        g <- makefunction(x[i],e[i])
        y[i] <- uniroot(g,lower=-100,upper=100)$root 
    }

# y is endogenous ... non-zero covariance
cov(y,e)
# x is exogenous
cov(x,e)

# The error-function
r <- function(theta_1,theta_2)
    {
        out <- y - (1 + exp(theta_1 - theta_2*y)/theta_2) - x 
        return(out)
    }

# Instruments are function of x
X <- cbind(x,x^2,x^3) 
objective <- function(theta)
    {
        a1 <- theta[1]
        a2 <- theta[2]
        out <- rbind(r(a1,a2))%*%X%*%solve(t(X)%*%X)%*%t(X)%*%cbind(r(a1,a2))
        return(out)
    }

optim(c(2,3),objective)

based on Amemiya (1983) in the Handbook of Econometrics formula (5.10).

I divide with $\theta_2$ within a prentheses instead of outside, but the code still illustrates how the estimator works so I'm not gonna alter it.

The residual function is

$$r(y,x,\theta) = y -h(y,\theta) - x$$

and

$$\mathbb E[r(y,x,\theta)\lvert x]=0$$ implying that

$$\mathbb E[t(x)r(y,x,\theta)]=0$$

for any function $t(x)$. Basically the two stage non linear least squares estimator is as such a general method of moments estimator.

A more recent text is Wooldridge, J.M. (1996) "Estimating systems of equations with different instruments for different equations" in Journal of Econometrics an article the gist of which is recapitulated in his (2010) Econometric Analysis of Cross Section and Panel Data in the chapter on GMM estimation page 530 - ...

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Or how else is such a non-linear relationship without closed-form solution approached?

For maximum likelihood we want the density of $y$ conditional on $x$ (because this is our sample), $f_{y|x}(y|x)$.

We start with making an assumption on the density of $\varepsilon$ conditional on $x$, $f_{\varepsilon|x}(\varepsilon|x)$.

When the relation is not implicit, say $y = ax + \varepsilon$ the change-of-variables method has a Jacobian determinant equal to unity so we simply have

$$f_{y|x}(y|x) = f_{\varepsilon|x}(y-ax|x)$$

and we can proceed as usual. But when the relation is implicit we have

$$\varepsilon = y-h(y,\theta)-x \implies \frac{\partial \varepsilon}{\partial y} = 1- \frac{\partial h(y,\theta)}{\partial y}$$

so here, the observation density will be

$$f_{y|x}(y|x) = \left|1- \frac{\partial h(y,\theta)}{\partial y}\right| \cdot f_{\varepsilon|x}(y-h(y,\theta)-x|x)$$

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