Why doesn't measurement error in the dependent variable bias the results?

Question

When there is measurement error in the independent variable I have understood that the results will be biased against 0. When the dependent variable is measured with error they say it just affects the standard errors but this doesn't make much sense to me because we are estimating the effect of $X$ not on the original variable $Y$ but on some other $Y$ plus an error. So how does this not affect the estimates? In this case can I also use instrumental variables to remove this problem?

Answer 1

When you want to estimate a simple model like $$Y_i = \alpha + \beta X_i + \epsilon_i$$ and instead of the true $Y_i$ you only observe it with some error $\widetilde{Y}_i = Y_i + \nu_i$ which is such that it is uncorrelated with $X$ and $\epsilon$, if you regress $$\widetilde{Y}_i = \alpha + \beta X_i + \epsilon_i$$ your estimated $\beta$ is $$ \begin{align} \widehat{\beta} &= \frac{Cov(\widetilde{Y}_i,X_i)}{Var(X_i)} \newline &= \frac{Cov(Y_i + \nu_i,X_i)}{Var(X_i)} \newline &= \frac{Cov(\alpha + \beta X_i + \epsilon_i + \nu_i,X_i)}{Var(X_i)} \newline &= \frac{Cov(\alpha ,X_i)}{Var(X_i)} + \beta\frac{Cov(X_i,X_i)}{Var(X_i)} + \frac{Cov(\epsilon_i,X_i)}{Var(X_i)} + \frac{Cov(\nu_i,X_i)}{Var(X_i)} \newline &= \beta \frac{Var(X_i)}{Var(X_i)} \newline &= \beta \end{align} $$ because the covariance between a random variable and a constant ($\alpha$) is zero as well as the covariances between $X_i$ and $\epsilon_i, \nu_i$ since we assumed that they are uncorrelated.

So you see that your coefficient is consistently estimated. The only worry is that $\widetilde{Y}_i = Y_i + \nu_i = \alpha + \beta X_i + \epsilon_i + \nu_i$ gives you an additional term in the error which reduces the power of your statistical tests. In very bad cases of such measurement error in the dependent variable you may not find a significant effect even though it might be there in reality. Generally, instrumental variables will not help you in this case because they tend to be even more imprecise than OLS and they can only help with measurement error in the explanatory variable.

Answer 2

Regression analysis answers the question, "What is the AVERAGE Y value for those who have given X values?" or, equivalently, "How much is Y predicted to change ON AVERAGE if we change X by one unit?" Random measurement error doesn't change the average values of a variable, or the average values for subsets of individuals, so random error in the dependent variable will not bias regression estimates.

Let's say you have height data on a sample of individuals. These heights are very precisely measured, accurately reflecting everyone's true stature. Within the sample, the average for men is 175 cm and the average for women is 162 cm. If you use regression to calculate how well gender predicts height, you estimate the model

$\mathit{HEIGHT = CONSTANT + β * GENDER + RESIDUAL}$

If women are coded as 0 and men as 1, $\mathit{CONSTANT}$ is the female average, or 162 cm. The regression coefficient $\mathit{β}$ shows how much height changes ON AVERAGE when you change $\mathit{GENDER}$ by one unit (from 0 to 1). $\mathit{β}$ equals 13 because people whose value for $\mathit{GENDER}$ is 0 (women) have a mean height of 162 cm while people whose value for $\mathit{GENDER}$ is 1 (men) have a mean height of 175 cm; $\mathit{β}$ estimates the average difference between men's and women's heights, which is 13 cm. ($\mathit{RESIDUAL}$ reflects the within-gender variance in height.)

Now, if you randomly add -1 cm or +1 cm to everyone's true height, what will happen? Individuals whose actual height is, say, 170 cm will now be reported as being 169 or 171 cm. However, the average of the sample, or any subsample, will not change. Those whose actual height is 170 cm will average 170 cm in the new, erroneous dataset, women will average 162 cm, etc. If you rerun the regression model specified above using this new dataset, the (expected) value of $\mathit{β}$ will not change because the average difference between men and women is still 13 cm, regardless of the measurement error. (The standard error of $\mathit{β}$ will be larger than before because the variance of the dependent variable is now larger.)

If there's measurement error in the independent variable rather than the dependent variable, $\mathit{β}$ will be a biased estimate. This is easy to understand when you consider the height example. If there's random measurement error in the $\mathit{GENDER}$ variable, some men will be erroneously coded as female and vice versa. The effect of this is to reduce apparent gender differences in height, because moving males to the female group will make the female mean is larger while moving females to the male group will make the male mean smaller. With measurement error in the independent variable, $\mathit{β}$ will be lower than the unbiased value of 13 cm.

While I used a categorical independent variable ($\mathit{GENDER}$) for simplicity here, the same logic applies to continuous variables. For example, if you used a continuous variable like birth height to predict adult height, the expected value of $\mathit{β}$ would be the same regardless of the amount of random error in adult height measurements.

Answer 3

Another way to see this is an M-estimation argument. In general, OLS is consistent and asymptotically normal for data $(Y_i, X_i)$ coming from a model that satisfies $\mathbb E(Y_i|X_i) = X \beta$ and some mild regularity conditions, with $\hat \beta_\text{ols} \to \beta$. See Chapter 7 of [1], for example. Using Andy's notation from above, since $\mathbb E(\tilde Y_i|X_i) = X \beta$, we're in luck and the standard results apply.

[1] Boos, Dennis D, and L. A Stefanski. Essential Statistical Inference. Vol. 120. Springer Texts in Statistics. New York, NY: Springer New York, 2013. https://doi.org/10.1007/978-1-4614-4818-1.