# Riddle: Estimate effect of x on y if you only have two noisy measures of x.

26 Jan 2022

Consider the following data generating process:

set.seed(1)
n = 100000
x = rnorm(n)

eta1 = rnorm(n) # measurement error1
noisy1 = x + eta1

eta2 = rnorm(n) # measurement error2
noisy2 = x + eta2

u = rnorm(n)
beta0=0; beta1 = 1
y = beta0+beta1*x + u


Can you solve the following statistics riddle? We want to consistently estimate the causal effect beta1 = 1 of x on y. We don’t observe x but only noisy1 and noisy2, which are noisy versions of x whose unobserved measurement errors eta1 and eta2 are independently distributed from each other. (My answer consists of one line of R code using a common econometrics package.)

## Discussion and Solution

For starters let us just regress y on noisy1:

coef(lm(y~noisy1))

## (Intercept)      noisy1
## -0.00285417  0.50197301


Our estimator for beta1 in this OLS regression is biased towards 0. That is the well known attenuation bias.

Here a short explanation for the bias. The true relationship is

y = beta0 + beta1 * x + u

But we estimate the regression

y = beta0 + beta1 * noisy1 + eps

Since the right hand side of our estimating equation must be the same as in the true relationship, we know that

eps = beta1 * x - beta1 *noisy1 + u
= beta1 * (x - noisy1) + u
= -beta1 * eta1 + u


Our OLS estimator is consistent only if our explanatory variable noisy1 is uncorrelated with the error term eps. Yet, since noisy1 is positively correlated with eta1, it is negatively correlated with eps. The bias of our OLS estimator for beta1 has the same sign as the correlation between noisy1 and eps. Here, this correlation always has the opposite sign of beta1 and our estimator is thus biased towards 0.

Solution of the riddle:

A consistent estimator of beta1 comes from a pretty popular method in applied econometrics to overcome endogeneity problems: instrumental variable regression. We can consistently estimate beta1 in the regression equation

y = beta0 + beta1 * noisy1 + eps

if we have an instrumental variable that is correlated with noisy1 and uncorrelated with the error term eps = -beta1 * eta1 + u. Well it happens that our second noisy measure noisy2 = x + eta2 fulfills both conditions. Obviously it is correlated with noisy1 since both are noisy measures of x and since the measurement errors eta1 and eta2 are independent in our data generating process, noisy2 is also uncorrelated with eps. Let’s run the instrumental variable regression using the ivreg function from the AER package.

AER::ivreg(y~noisy1 | noisy2)

##
## Call:
## AER::ivreg(formula = y ~ noisy1 | noisy2)
##
## Coefficients:
## (Intercept)       noisy1
##   -0.002249     1.000015


Yep, looks like a consistent estimator!

Somehow I like this observation: both noisy1 and noisy2 are absolutely symmetric but one gets the role of explanatory variable, the other the role of instrument. Of course, we could also just swap instrument and explanatory variable:

AER::ivreg(y ~ noisy2 | noisy1)

##
## Call:
## AER::ivreg(formula = y ~ noisy2 | noisy1)
##
## Coefficients:
## (Intercept)       noisy2
##   -0.001177     0.997248


(You can try out yourself: If noisy1 and noisy2 have different precision, does swapping their roles systematically affect the precision of the resulting estimate?)

I don’t know whether the result is of much practical importance, though. How often do we have two noisy measures whose measurement errors are independent from each other?

If the measurement errors are correlated with each other, the procedure does not yield a consistent estimator. Here is an illustration for positively correlated measurement errors.

eta2 = 0.5*eta1+ 0.5*rnorm(n)
noisy2 = x + eta2
y = beta0+beta1*x + u
AER::ivreg(y~noisy1 | noisy2)

##
## Call:
## AER::ivreg(formula = y ~ noisy1 | noisy2)
##
## Coefficients:
## (Intercept)       noisy1
##   -0.002649     0.671033


Now, an attenuation bias still remains. It is just reduced a bit compared to the OLS estimator.

Published on 26 Jan 2022