A regressor measured with noise
A signal has variance , and the target is with noise independent of . You cannot observe ; you observe , where has mean , variance , and is independent of everything else.
With a very large sample, what slope does the regression of on (with an intercept) give?
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
The slope converges to .
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. Noise in the regressor inflates its variance but adds nothing to its covariance with . The slope is shrunk towards zero by the share of the regressor's variance that is signal.
Step 1: covariance. .
Step 2: variance. .
Step 3. The slope converges to instead of .
In general. The estimated slope is the true slope times , here . More data does not fix it: the estimator is consistent for the wrong value. This is why noisy signals look weaker in a regression than they are, and why noise in (which only widens the error bars) is much less harmful than noise in .