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Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 13J / Solution

This is attenuation bias from classical measurement error, also called regression dilution. The latent predictor is

Z

, but

X_{1}

and

X_{2}

are noisy proxies. The noisier proxy produces a slope closer to zero.

For a regression with an intercept, the population slope is

\frac{Cov ( X , Y )}{Var ( X )} .

Here all variables are centred and the noises are independent. Therefore

Cov (Z, Y) = β σ_{z}^{2}, Var (Z) = σ_{z}^{2},

so regressing

Y

on

Z

has slope

β = 2

. For

X_{j} = Z + η_{j},

we have

Cov (X_{j}, Y) = β σ_{z}^{2}, Var (X_{j}) = σ_{z}^{2} + σ_{x_{j}}^{2} .

Thus

β_{slope} (Y \sim X_{j}) = β \frac{σ _{z}^{2}}{σ _{z}^{2} + σ _{x_{j}}^{2}} .

With the stated values, the slopes are

2 \frac{1}{1 + 0. 5 ^{2}} = 1.6, 2 \frac{1}{1 + 1 ^{2}} = 1,

matching the simulation.

Under this independent additive measurement-error model, the magnitude of the

Y

-on-

X_{1}

slope is generally smaller than the magnitude of the

Y

-on-

Z

slope whenever

σ_{x_{1}}^{2} > 0

. Increasing

σ_{y}

does not change the population slope, because response noise contributes neither to

Cov (X_{1}, Y)

nor to

Var (X_{1})

. It increases residual variance and the sampling variability of the estimate; with one million observations, doubling it should leave the displayed slope close to

1.6

while increasing its standard error.

Solved by gpt-5.6-sol high.

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