# Regressions with a Mis-measured, Binary Outcome

Many outcomes of interest in economics are binary. For example, we may want to learn how employment status $$Y^*$$ varies with demographics $$X$$, where $$Y^*=1$$ means “employed” and $$Y^*=0$$ means unemployed or not in the labor force. But how do we know if someone is employed? Typically we ask them, perhaps as part of a large, nationally representative survey such as the CPS. Researchers who study labor market dynamics have long known, however, that observed data on labor market status are often inaccurate (Poterba & Summers, 1986). Administrative errors creep into even the most carefully-administered surveys. But more importantly, survey respondents do not always tell the truth, whether by mistake or deliberately, and this problem seems to have gotten worse in recent years (Meyer et al. 2015) Instead of true employment status $$Y^*$$ researchers only observe a noisy measure $$Y\in \{0, 1\}$$.

In my previous post I showed that classical measurement error an outcome variable is basically innocuous. But I also showed that measurement error in a binary random variable cannot be classical. In this post, I’ll explore the consequences of this fact when we want to learn $$\mathbb{P}(Y^*=1|X)$$ but only observe $$Y$$ and $$X$$, not the true outcome variable $$Y^*$$. To keep things concrete, I will assume throughout that $$\mathbb{P}(Y^*=1|X) = F(X'\beta)$$ where $$F$$ is a strictly increasing, differentiable function. This covers all the usual suspects: logit, probit, and the linear probability model.1 The parameter $$\beta$$ may have a causal interpretation or may simply have a predictive one. Either way, the question I’ll focus on here is whether, and if so how, $$\beta$$ can be identified in the presence of measurement error. For simplicity I will assume throughout that that the covariates $$X$$ are measured without error.

# What’s the problem?

Why does observing $$Y$$ rather than $$Y^*$$ present a problem? To answer this question, we need to derive the relationship between $$\mathbb{P}(Y=1|X)$$ and $$\mathbb{P}(Y^*=1|X)$$. Since $$Y$$ and $$Y^*$$ are both binary, $$\mathbb{E}(Y|X) = \mathbb{P}(Y=1|X)$$ and similarly $\mathbb{E}(Y^*|X) = \mathbb{P}(Y^*=1|X) = F(X'\beta).$ Now define the measurement error $$W$$ as $$W = Y - Y^*$$ so we can write $$Y = Y^* + W$$. By the linearity of expectation, \begin{aligned} \mathbb{P}(Y=1|X) &= \mathbb{E}(Y|X)\\ &= \mathbb{E}(Y^* + W|X) = \mathbb{E}(Y^*|X) + \mathbb{E}(W|X)\\ &= \mathbb{P}(Y^*=1|X) + \mathbb{E}(W|X) \end{aligned} so we see that if $$\mathbb{E}(W|X)=0$$ then $$\mathbb{P}(Y=1|X)$$ and $$\mathbb{P}(Y^*=1|X)$$ will coincide. Unfortunately $$\mathbb{E}(W|X)$$ in general cannot be zero. This means that learning $$\mathbb{P}(Y=1|X)$$ will not tell us what we want to know: $$\mathbb{P}(Y^*=1|X)$$.

To see why $$\mathbb{E}(W|X) \neq 0$$, first define the mis-classification probabilities $$\alpha_0(\cdot)$$ and $$\alpha_1(\cdot)$$ \begin{aligned} \alpha_0(X) &\equiv P(Y=1|Y^*=0,X)\\ \alpha_1(X) &\equiv P(Y=0|Y^*=1,X). \end{aligned} The subscripts on $$\alpha$$ refer to the value of $$Y^*$$ on which we condition: $$\alpha_0(\cdot)$$ conditions on $$Y^*=0$$ while $$\alpha_1(\cdot)$$ conditions on $$Y^*=1$$. You can interpret the mis-classification probabilities by analogy to null hypothesis testing: $$\alpha_0(X)$$ is effectively the type I error rate as a function of $$X$$ and $$\alpha_1(X)$$ is the type II error rate as a function of $$X$$.

To keep things fully general for the moment, we allow the mis-classification probabilities to depend on $$X$$. Perhaps a young male worker with five years of experience is more likely to make an erroneous self-report in the CPS than and older female worker with more experience, for example.2 Since $$Y$$ and $$Y^*$$ are both binary, $$W \in \{-1, 0, 1\}$$ and we calculate $$\mathbb{E}(W|X)$$ as follows: \begin{aligned} \mathbb{E}(W|X) &= -1 \times \mathbb{P}(W=-1|X) + 0 \times \mathbb{P}(W=0|X) + 1 \times \mathbb{P}(W=1|X) \\ &= \mathbb{P}(W=1|X) - \mathbb{P}(W=-1|X). \end{aligned} Now consider the event $$\{W = -1\}$$. The only way that this could occur is if $$Y = 0$$ and $$Y^* = 1$$. Accordingly, \begin{aligned} \mathbb{P}(W = -1|X) &= \mathbb{P}(Y = 0, Y^* = 1|X)\\ &= \mathbb{P}(Y=0|Y^*=1,X)\mathbb{P}(Y^*=1|X) \\ &= \alpha_1(X) F(X'\beta). \end{aligned} Similarly, the only way that $$\{W=1\}$$ can occur is if $$Y=1$$ and $$Y^*=0$$ so that \begin{aligned} \mathbb{P}(W = 1|X) &= \mathbb{P}(Y = 1, Y^* = 0|X)\\ &= \mathbb{P}(Y=1|Y^*=0,X) \mathbb{P}(Y^*=0|X)\\ &= \alpha_0(X) \left[1 - F(X'\beta)\right]. \end{aligned} Therefore, \begin{aligned} \mathbb{E}(W|X) &= \mathbb{P}(W=1|X) - \mathbb{P}(W=-1|X) \\ &= \alpha_0(X)\left[1 - F(X'\beta) \right] -\alpha_1(X) F(X'\beta). \end{aligned} So how could $$\mathbb{E}(W|X) = 0$$? Re-arranging the preceding to solve for $$F(X'\beta)$$, $\mathbb{E}(W|X) = 0 \iff F(X'\beta) = \frac{\alpha_0(X)}{\alpha_0(X) + \alpha_1(X)}.$ This shows that that $$\mathbb{E}(W|X)$$ can only be zero in an extremely peculiar case where $$\alpha_0(\cdot)$$ and $$\alpha_1(\cdot)$$ depend on $$X$$ in just the right way. If the mis-classification probabilities are constants that do not depend on $$X$$, we would require $$F(X'\beta) = \alpha_0/(\alpha_0 + \alpha_1)$$. This is only possible if all the elements of $$\beta$$ besides the intercept are zero. Since $$\mathbb{E}(W|X)$$ will not in general equal zero, $$\mathbb{P}(Y=1|X)$$ will not in general equal $$\mathbb{P}(Y^*=1|X)$$.

# What happens if we ignore the problem?

Substituting our expression for $$\mathbb{E}(W|X)$$ and factoring the result, \begin{aligned} \mathbb{P}(Y=1|X) &= \mathbb{P}(Y^*=1|X) + E(W|X) \\ &= F(X'\beta) + \alpha_0(X)\left[1 - F(X'\beta) \right] -\alpha_1(X) F(X'\beta)\\ &= \alpha_0(X) + F(X'\beta) [1 - \alpha_0(X) - \alpha_1(X)]. \end{aligned}

Because we observe $$(Y, X)$$, $$\mathbb{P}(Y=1|X)$$ is identified. With enough data, we can learn this conditional probability as a function of $$X$$ as accurately as we wish. The problem is that $$\alpha_0(X)$$ and $$\alpha_1(X)$$ drive a wedge between what we can observe, $$\mathbb{P}(Y=1|X)$$, and what we’re trying to learn $$\mathbb{P}(Y^*=1|X) = F(X'\beta)$$. Without knowing more about the functions $$\alpha_0(\cdot)$$ and $$\alpha_1(\cdot)$$ we can’t say much about how $$\mathbb{P}(Y=1|X)$$ and $$\mathbb{P}(Y^*=1|X)$$ will differ. Because they are probabilities, $0\leq \alpha_0(X) \leq 1, \quad 0\leq \alpha_1(X) \leq 1.$ But because they are conditional probabilities that condition on different events, $$\{Y^*=0, X=x\}$$ versus $$\{Y^*=1, X=x\}$$, the sum $$\alpha_0(x) + \alpha_1(x)$$ could be greater than one. This means that $$1 - \alpha_0(X) - \alpha_1(X)$$ could be negative, at least for certain values of $$X$$. It’s common in practice, however, to assume that $$\alpha_0(X) + \alpha_1(X) < 1$$ for all possible values that the covariates $$X$$ could take on. To understand this assumption, and the problem more generally, it’s helpful to consider a simple special case in which the mis-classification probabilities do not depend on $$X$$. In this case we can say precisely how measurement error in the outcome affects what we can learn about the relationship between $$X$$ and $$Y^*$$ and make clear why $$\alpha_0(X) + \alpha_1(X) < 1$$ is usually a reasonable assumption.

## A Special Case: Fixed Mis-classification

Suppose that the mis-classification probabilities are fixed, i.e. that \begin{aligned} \alpha_0(X) &\equiv \mathbb{P}(Y=1|Y^*=0|X) = \mathbb{P}(Y=1|Y^*=0)\equiv \alpha_0 \\ \alpha_1(X) &\equiv \mathbb{P}(Y=0|Y^*=1|X) = \mathbb{P}(Y=0|Y^*=1)\equiv \alpha_1. \end{aligned} This is a fairly strong assumption. It says that both self-reporting and administrative errors occur at the same rate for everyone, regardless of their observed characteristics. In this case, our expression for $$\mathbb{P}(Y=1|X)$$ from above becomes \begin{aligned} \mathbb{P}(Y=1|X) &= \alpha_0 + F(X'\beta) (1 - \alpha_0 - \alpha_1). \end{aligned} Defining $$f$$ as the derivative of $$F$$, this means that the observed partial effect of a continuous covariate $$X_j$$ with respect to $$Y$$ is \begin{align*} \frac{\partial}{\partial X_j} \mathbb{P}(Y=1|X) &= \frac{\partial}{\partial X_j} \left[ \alpha_0 + F(X'\beta)(1 - \alpha_0 - \alpha_1)\right]\\ &= f(X'\beta)\beta_j (1 - \alpha_0 - \alpha_1) \end{align*} whereas the true partial effect, with respect to $$Y^*$$, is $\frac{\partial}{\partial X_j} \mathbb{P}(Y^*=1|X) = \frac{\partial}{\partial X_j} F(X'\beta) = f(X'\beta) \beta_j.$ If $$(\alpha_0 + \alpha_1) > 1$$ then $$(1 - \alpha_0 - \alpha_1)$$ will be negative. This means that the measurement error problem is so severe that all of the observed partial effects have the wrong sign. A bit of tedious algebra shows that $$Y$$ and $$Y^*$$ must be negatively correlated in this case: $$Y$$ is such a noisy measure of $$Y^*$$ that when $$Y=1$$ we’re better off predicting that $$Y^*=0$$. For this reason, it’s traditional to assume that $$\alpha_0 + \alpha_1 < 1$$. In this case $$0 < (1 - \alpha_0 - \alpha_1) \leq 1$$ so the observed partial effects are attenuated versions of the true partial effects, in that $0 < \frac{\partial}{\partial X_j} \mathbb{P}(Y=1|X) \leq \frac{\partial}{\partial X_j} \mathbb{P}(Y^*=1|X).$ So in this special case, non-classical measurement error in a binary outcome variable has the same effect as classical measurement in a continuous regressor: attenuation bias.

## Observational Equivalence and $$(\alpha_0 + \alpha_1)$$

You may be wondering: do we really need the assumption $$\alpha_0 + \alpha_1<1$$ or is it merely convenient? Couldn’t the observed data tell us whether $$\alpha_0 + \alpha_1$$ is less than one or greater than one? The answer turns out to be no, and it’s easy to show why if $$F(t) = 1 - F(-t)$$ as in the probit, logit, and the linear probability models. Suppose that this condition on $$F$$ holds. Then we can write \begin{aligned} \mathbb{P}(Y=1|X) &= \alpha_0 + (1 - \alpha_0 - \alpha_1) F(X'\beta)\\ &= \alpha_0 + (1 - \alpha_0 - \alpha_1) \left[ 1 - F\big(X'(-\beta)\big)\right]\\ &= \left(\alpha_0 + 1 - \alpha_0 - \alpha_1\right) - (1 - \alpha_0 - \alpha_1) F\big(X'(-\beta)\big)\\ &= (1 - \alpha_1) + \left[ \alpha_0 - (1 - \alpha_1) \right] F\big(X' (-\beta)\big)\\ &= (1 - \alpha_1) + \left[1 - (1 - \alpha_1) - (1 - \alpha_0) \right] F\big(X' (-\beta)\big). \end{aligned} Defining $$\widetilde{\alpha}_0 \equiv (1 - \alpha_1)$$, $$\widetilde{\alpha}_1 \equiv (1 - \alpha_0)$$, and $$\widetilde{\beta} \equiv -\beta$$, we have established that \begin{aligned} \mathbb{P}(Y=1|X) &= \alpha_0 + (1 - \alpha_0 - \alpha_1) F(X'\beta)\\ &= \widetilde{\alpha}_0 + \left( 1 - \widetilde{\alpha}_0 - \widetilde{\alpha}_1 \right) F\big(X'\widetilde{\beta}\big). \end{aligned} Since $$Y$$ is binary, $$\mathbb{P}(Y=1|X)$$ tells us everything that there is to know about the distribution of $$Y$$ given $$X$$. The preceding pair of equalities shows that the observed conditional distribution $$\mathbb{P}(Y=1|X)$$ could just as well have arisen from $$\mathbb{P}(Y^*=1|X) = F(X'\widetilde{\beta})$$ with mis-classification probabilities $$(\widetilde{\alpha}_0, \widetilde{\alpha}_1)$$ as it could from $$\mathbb{P}(Y^*=1|X) = F(X'\beta)$$ with mis-classification probabilities $$(\alpha_0, \alpha_1)$$. From observations of $$(Y, X)$$ alone there is no way to tell these possibilities apart: we say that they are observationally equivalent. Notice that if $$\alpha_0 + \alpha_1 < 1$$ then $$\widetilde{\alpha}_0 + \widetilde{\alpha}_1 > 1$$ and vice-versa. This shows that the only way to point identify $$\beta$$ is to assume either that $$\alpha_0 + \alpha_1 < 1$$ or the reverse inequality.3 For the reasons discussed above, it usually makes sense to choose $$\alpha_0 + \alpha_1 < 1$$.

# Some Solutions

So what is an applied researcher to do? If we could somehow learn the mis-classification probabilities, we could use them to “adjust” $$\mathbb{P}(Y=1|X)$$ and identify $$\mathbb{P}(Y^*=1|X) = F(X'\beta)$$ as follows: $F(X'\beta) = \frac{\mathbb{P}(Y=1|X) - \alpha_0(X)}{1 - \alpha_0(X) - \alpha_1(X)}.$ Broadly speaking there are two ways learn the mis-classification probabilities. The first approach estimates $$\alpha_0(X)$$ and $$\alpha_1(X)$$ using a second dataset. The second approach uses a single dataset and exploits non-linearity in the function $$F$$ instead. For the remainder of this discussion I will assume that $$\alpha_0(X) + \alpha_1(X) < 1$$ or $$\alpha_0 + \alpha_1 < 1$$ if the mis-classification probabilities are fixed.

## Method 1: Auxiliary Data

Let’s start by making life simple: assume fixed mis-classification. Now suppose that we observe two random samples from the same population. In the first, we observe pairs $$(Y_i,X_i)$$ for $$i = 1, ..., n$$ and in the second we observe pairs $$(Y_j, Y^*_j)$$ for $$j = 1, ..., m$$. Notice that neither dataset contains observations of $$X$$ and $$Y^*$$ for the same individual. Using the $$(Y_i,X_i)$$ observations we can estimate $$\mathbb{P}(Y=1|X)$$, and using the $$(Y_j, Y^*_j)$$ observations we can estimate $\alpha_0 = \mathbb{P}(Y=1|Y^*=0), \quad \alpha_1 = \mathbb{P}(Y=0|Y^*=1).$ This gives us everything we need to determine $$F(X'\beta)$$ as a function of $$X$$ and hence $$\beta$$. The observations of $$(Y_i, Y^*_i)$$ are called an auxiliary dataset. In theory, auxiliary data provide a simple and general solution to measurement error problems of all stripes. Suppose, for example, that we were uncomfortable with the assumption of fixed mis-classification. If we observed an auxiliary dataset of triples $$(Y_j, Y_j^*, X_j)$$ then we could directly estimate $$\alpha_0(X)$$ and $$\alpha_1(X)$$. Of course, we if we observed $$(Y_j, Y_j^*, X_j)$$ for a random sample drawn from the population of interest we could estimate $$\mathbb{P}(Y^*=1|X)$$ directly without the need to account for measurement error! And here lies the fundamental tension of the auxiliary data approach: if we had sufficiently rich auxiliary data we wouldn’t really have a measurement error problem in the first place. More typically, we either observe $$(Y_j, Y_j^*, X_j)$$ for a different population, or only observe a subset of these variables for our population of interest. Either way we need to rely on modeling assumptions to bridge the gap. For example, fixed mis-classification and an auxiliary dataset of $$(Y^*_j, Y_j)$$ suffice to solve the measurement error problem but only if $$\alpha_0(X)$$ and $$\alpha_1(X)$$ do not in fact depend on $$X$$.

## Method 2: Nonlinearity of $$F$$

The auxiliary data approach is very general in principle but relies on information that we simply may not have in practice: a second dataset from the same population. An alternative approach uses only one dataset, $$(Y_i, X_i)$$ for $$i = 1, ..., n$$, and instead exploits the shape of the function $$F$$. This second approach is a bit less general but doesn’t require any outside sources of information.

To begin, suppose that the mis-classification probabilities are fixed and that $$F$$ is a known function, e.g. the standard logistic CDF. Suppose further that $$F$$ is strictly increasing and hence invertible. Then, applying $$F^{-1}$$ to both sides of our expression for $$F(X'\beta)$$ from above, $X'\beta = F^{-1} \left[\frac{\mathbb{P}(Y = 1|X) - \alpha_0}{1 - \alpha_0 - \alpha_1}\right]$ and thus, pre-multiplying both sides by $$X$$ and taking expectations, $\mathbb{E}[XX']\beta = \mathbb{E}\left\{X F^{-1} \left[\frac{\mathbb{P}(Y = 1|X) - \alpha_0}{1 - \alpha_0 - \alpha_1}\right]\right\}.$ Therefore, if $$\mathbb{E}[XX']$$ is invertible, $\beta = \mathbb{E}[XX']^{-1}\mathbb{E}\left\{X F^{-1} \left[\frac{\mathbb{P}(Y = 1|X) - \alpha_0}{1 - \alpha_0 - \alpha_1}\right]\right\}.$ Since $$\mathbb{P}(Y=1|X)$$ depends only on the observed data $$(Y,X)$$, this function is point identified. Since $$F$$ is assumed to be a known function, it follows that $$\beta$$ is point identified whenever $$\mathbb{E}[XX']$$ is invertible and $$(\alpha_0, \alpha_1)$$ are known.4 So if we can find a way to point identify $$\alpha_0$$ and $$\alpha_1$$, we will immediately identify $$\beta$$.

Easier said than done! How can we possibly learn $$\alpha_0$$ and $$\alpha_1$$ without auxiliary data? Nonlinearity is the key. If $$F$$ is a cumulative distribution function, then $$\lim_{t\rightarrow \infty} F(t) = 1$$ and $$\lim_{t\rightarrow -\infty} F(t) = 0$$. Now suppose that $$X$$ contains at least one covariate, call it $$V$$, that is continuous and has “large support,” i.e. takes on values in a very wide range. Without loss of generality, suppose that the coefficient $$\beta_v$$ on $$V$$ is positive. (If it’s negative, then apply the following argument to $$-V$$ instead.) For $$V$$ large and positive $$X'\beta$$ is large and positive so that $$F(X'\beta)$$ is close to one. In this case \begin{aligned} \alpha_0 + (1 - \alpha_0 - \alpha_1) F(X'\beta) &\approx \alpha_0 + (1 - \alpha_0 - \alpha_1) \times 1 \\ &= (1 - \alpha_1). \end{aligned} For $$V$$ large and negative, on the other hand, $$X'\beta$$ is large and negative, $$F(X'\beta)$$ is close to zero, and \begin{aligned} \alpha_0 + (1 - \alpha_0 - \alpha_1) F(X'\beta) &\approx \alpha_0 + (1 - \alpha_0 - \alpha_1) \times 0\\ &= \alpha_0. \end{aligned} Intuitively, by examining values of $$X_i$$ for which $$F(X_i'\beta)$$ is close to one we can learn $$(1 - \alpha_1)$$ and by examining values for which $$F(X_i'\beta)$$ is close to zero we can identify $$\alpha_0$$.

You may object that the preceding identification argument sounds suspiciously circular: doesn’t this idea at least implicitly require us to know $$\beta$$? Fortunately the answer is no. We only need to know the signs of $$\beta$$. Under the assumption that $$\alpha_0 + \alpha_1 < 1$$ these are the same as the signs of the observed partial effects $$\partial \mathbb{P}(Y=1|X) /\partial \beta_j$$. An example may help. Suppose $$Y=1$$ means “graduated from college.” Under fixed misclassification, we would learn $$\alpha_0$$ from the observations of $$(Y_i, X_i)$$ for people who almost certainly didn’t graduate from college, based on their covariates, and $$(1 - \alpha_1)$$ from observations of $$(Y_i, X_i)$$ for people who almost certainly did. By first estimating $$\mathbb{P}(Y=1|X)$$ we learn attenuated versions of the true partial effects $$F(X'\beta) \beta_j$$. In other words, we learn how reported education varies with $$X$$. But this information suffices to show us how to make $$F(X'\beta)$$ close to zero or one.

The preceding argument crucially relies on the assumption that $$F$$ is nonlinear. To see why, consider the linear probability model $$F(X'\beta) = X'\beta$$ and let $$X' = (1, X_1')$$ and $$\beta' = (\beta_0, \beta_1')$$. Then, \begin{aligned} \mathbb{P}(Y=1|X) &= \alpha_0 + \left(1 - \alpha_0 - \alpha_1\right) F(X'\beta)\\ &= \alpha_0 + \left(1 - \alpha_0 - \alpha_1\right)(X'\beta) \\ &= \alpha_0 + \left(1 - \alpha_0 - \alpha_1\right)(\beta_0 + X_1' \beta_1) \\ &= \alpha_0 + (1 - \alpha_0 - \alpha_1) \beta_0 + X_1' (1 - \alpha_0 - \alpha_1) \beta_1. \end{aligned} Now, defining $$\widetilde{\beta}_0 \equiv \alpha_0 + (1 - \alpha_0 - \alpha_1)\beta_0$$, $$\widetilde{\beta}_1 = (1 - \alpha_0 - \alpha_1) \beta_1$$, and $$\widetilde{\beta}' = (\widetilde{\beta}_0, \widetilde{\beta}_1')$$ we have $\mathbb{P}(Y=1|X) = \alpha_0 + (1 - \alpha_0 - \alpha_1) X'\beta = X'\widetilde{\beta}.$ This shows that a linear probability model with coefficient vector $$\beta$$ and mis-classification probabilities $$(\alpha_0, \alpha_1)$$ is observationally equivalent to a linear probability model with no mis-classification and coefficient $$\widetilde{\beta}$$. To put it another way: there is no way to tell whether mis-classification is present or absent in a linear model. Doing so requires non-linearity.

So how can we use these results in practice? If $$(\alpha_0, \alpha_1, \beta)$$ are identified and $$F$$ is assumed known, we can proceed via garden-variety maximum likelihood estimation. The log-likelihood function is only slightly more complicated than in the standard binary outcome setting, in particular: \begin{aligned} \ell_n(\alpha_0, \alpha_1, \beta) &= \frac{1}{n} \sum_{i=1}^n \log\left\{ \mathbb{P}(Y_i=1|X)^{\mathbb{1}(Y_i=1)}\mathbb{P}(Y_i=0|X_i)^{\mathbb{1}(Y_i=0)} \right\} \\ &= \frac{1}{n} \sum_{i=1}^n Y_i \log\left\{ \alpha_0 + (1 - \alpha_0 - \alpha_1) F(X_i'\beta) \right\} + (1 - Y_i) \log\left\{ 1 - \alpha_0 - (1 - \alpha_0 - \alpha_1) F(X_i'\beta) \right\}. \end{aligned} If $$F$$ is unknown, estimation is more complicated but the intuition from above continues to hold: a regressor $$V$$ with “large support” allows us to identify the mis-classification probabilities, and hence $$F$$. Indeed, we can even allow $$\alpha_0$$ and $$\alpha_1$$ to depend covariates, as long as they don’t depend on $$V$$ itself. For more details on the “identification by nonlinearity” approach, see Hausman et al. (1998) and Lewbel (2000).

# Coming Attractions

That’s more than enough about measurement error for now! When I return to this topic in a few weeks time, I’ll consider the problem of a mis-measured binary regressor. In my next installment, however, I’ll put measurement error to one side and revisit a classic problem from introductory statistics: constructing a confidence interval for a population proportion. Sometimes the easiest things turn out to be much harder than they first appear.

1. For more details of these models, see my lecture notes.↩︎

2. Below we’ll examine a simpler special case in which $$\alpha_0$$ and $$\alpha_1$$ are fixed probabilities that do not depend on $$X$$.↩︎

3. Alternatively, you could say that the only way to identify $$(\alpha_0, \alpha_1)$$ is by making an assumption about the sign of one component of $$\beta$$.↩︎

4. We also need $$\alpha_0 + \alpha_1$$ to avoid division by zero!↩︎