Beyond Classical Measurement Error

Pop Quiz: If \(D^*\) and \(D\) are binary random variables and \(D\) is a noisy measure of \(D^*\), is it possible for the measurement error \(W \equiv D - D^*\) to be classical? Explain why or why not. (Answer below)

Classical Measurement Error

Classical measurement error is a problem that is easy to understand and relatively easy to address. Roughly speaking, classical measurement error refers to a situation in which the variable we observe equals the truth plus noise \[ \text{Observed} = \text{Truth} + \text{Noise} \] where the noise is unrelated to the truth and “everything else.” (I’ll be precise about the meaning of “unrelated” and “everything else” in a moment.) Mis-measuring a regressor \(X\) in this way biases the OLS slope estimator towards zero (attenuation bias) but we can correct for this with a valid instrument. Mis-measuring the outcome \(Y\) increases standard errors but doesn’t bias the OLS estimator. You can find all the details in your favorite introductory econometrics textbook, but in the interest of making this post self-contained, here’s a quick review.

Least Squares Attenuation Bias

Suppose that we want to learn the slope coefficient from a population linear regression of \(Y\) on \(X^*\): \[ \beta \equiv \frac{\text{Cov}(Y,X^*)}{\text{Var}(X^*)}. \] Unfortunately we observe not \(X^*\) but a noisy measure \(X = X^* + W_X\) where \(W_X\) is uncorrelated with both \(X^*\) and \(Y\). Then \[ \begin{aligned} \text{Cov}(Y, X) &= \text{Cov}(Y, X^* + W_X) = \text{Cov}(Y, X^*)\\ \text{Var}(X) &= \text{Var}(X^* + W_X) = \text{Var}(X^*) + \text{Var}(W_X). \end{aligned} \] Now, define the reliability ratio \(\lambda\) as follows: \[ \lambda \equiv \frac{\text{Var}(X^*)}{\text{Var}(X^*) + \text{Var}(W_X)}. \] Measurement error means that \(\text{Var}(W_X)\) is positive. Since variances can’t be negative, this implies \(0 < \lambda < 1\). Combining our definition of \(\lambda\) with the expressions for \(\text{Cov}(Y,X)\) and \(\text{Var}(X)\) from above, \[ \begin{aligned} \frac{\text{Cov}(Y,X)}{\text{Var}(X)} &= \frac{\text{Cov}(Y, X^*)}{\text{Var}(X^*) + \text{Var}(W_X)} \\ &=\frac{\text{Var}(X^*)}{\text{Var}(X^*) + \text{Var}(W_X)}\cdot \frac{\text{Cov}(Y, X^*)}{\text{Var}(X^*)}\\ &= \lambda \beta \end{aligned} \] so we see that regressing \(Y\) on \(X\) gives \(\lambda \beta\) rather than \(\beta\). Since \(0 < \lambda < 1\), this phenomenon is called least squares attenuation bias: \(\lambda \beta\) has the same sign as \(\beta\) but is smaller in magnitude. The greater the extent of measurement error, the larger the variance of \(W_X\) and the smaller that \(|\lambda \beta|\) becomes.

Instrumental variables to the rescue

Suppose that \(Y = \alpha + \beta X^* + U\) where \(X = X^* + W_X\) as above. Now suppose that we can find a variable \(Z\) that is correlated with \(X^*\) but uncorrelated with \(U\) and \(W_X\). Then \[ \begin{aligned} \text{Cov}(Y,Z) &= \text{Cov}(\alpha + \beta X^* + U, Z) = \beta\text{Cov}(X^*,Z)\\ \text{Cov}(X,Z) &= \text{Cov}(X^* + W_X, Z) = \text{Cov}(X^*,Z) \end{aligned} \] so that \(\beta = \text{Cov}(Y, Z) / \text{Cov}(X,Z)\). If \(X^*\) is measured with classical measurement error, a simple instrumental variables regression solves the problem of attenuation bias.1 Notice that we haven’t said anything about \(U\) in relation to \(X^*\). If \(\beta\) is the population linear regression slope, then \(U\) is uncorrelated with \(X^*\) by definition. But this derivation still goes through if \(Y = \alpha + \beta X^* + U\) is a causal model in which \(X^*\) is correlated with \(U\), e.g. if \(Y\) is wage and \(X^*\) is years of schooling, in which case \(U\) might be “unobserved ability.” In this way, a single valid instrument can serve “double-duty,” eliminating both attenuation bias and selection bias.

Measurement error in the outcome

Now suppose that \(X^*\) is observed but the true outcome \(Y^*\) is not: we only observe a noisy measure \(Y = Y^* + W_Y\). If \(W_Y\) is uncorrelated with \(X^*\), \[ \frac{\text{Cov}(Y,X^*)}{\text{Var}(X^*)} = \frac{\text{Cov}(Y^* + W_Y, X^*)}{\text{Var}(X^*)} = \frac{\text{Cov}(Y^*,X^*)}{\text{Var}(X^*)} \]

so we’ll obtain the same slope from a regression of \(Y\) on \(X^*\) as we would from a regression of \(Y^*\) on \(X^*\). Classical measurement error in the outcome variable doesn’t introduce a bias.

Solution to the Pop Quiz

Now that we’ve refreshed our memories about classical measurement error, let’s a take a look at my pop quiz question from above:

If \(D^*\) and \(D\) are binary random variables and \(D\) is a noisy measure of \(D^*\), is it possible for the measurement error \(W \equiv D - D^*\) to be classical? Explain why or why not.

If \(W\) is a classical measurement error then, among other things, it must be uncorrelated with \(D^*\). But this is impossible if both \(D^*\) and \(D\) are binary. By the definition of \(W\), \(D = D^* + W\). If \(D^* = 1\) then \(D = 1 + W\). To ensure that \(D\) takes on a value in \(\{0, 1\}\), this means that \(W\) must be either \(0\) or \(-1\). If instead \(D^* = 0\), then \(D = W\), so \(W\) must be either \(0\) or \(1\). Hence, unless \(W\) always equals zero, in which case there’s no measurement error, \(W\) must always be negatively correlated with \(D^*\). In other words, measurement error in a a binary variable can never be classical. The same basic logic applies whenever \(X\) and \(X^*\) are bounded: to ensure that \(X\) stays within its bounds, any measurement error must be correlated with \(X^*\).

Non-Differential Measurement Error

Classical measurement error, as we’ve seen, is a very special case. Or to put it another way, non-classical measurement error isn’t as exotic as it sounds. Because discrete random variables cannot be subject to classical measurement error, non-classical measurement error should be on any applied economist’s radar. My next few posts will provide an overview of the simplest case: non-differential measurement error in a binary variable. This assumption allows \(D^*\) to be correlated with \(W\), but assumes that conditioning on \(D^*\) is sufficient to break the dependence between \(W\) and everything else. Even in this relatively simple case, everything we’ve learned about classical measurement error goes out the window:

  1. Non-differential measurement error does not necessarily cause attenuation.
  2. The IV estimator doesn’t correct for non-differential measurement error, and a single instrument cannot serve “double-duty.”
  3. Non-classical measurement error in the outcome variable generally does introduce bias.

The good news is that there are methods to address non-differential measurement error. In my next post, I’ll start by considering the case of a mis-measured binary outcome.

  1. Strictly speaking I haven’t used the assumption that \(W_X\) is uncorrelated with \(X^*\) in this derivation, but it’s implicit in the assumption that \(Z\) is correlated with \(X^*\) but not with \(W_X\).↩︎