Question
A glassmaker varies furnace temperature around its ideal setting. Let X be the temperature deviation, in degrees Celsius, and suppose X is uniformly distributed over \([-1,1]\). The finished glass has strength
\[Y=100-12X^{2}+\varepsilon\] megapascals,
where the measurement and manufacturing noise \(\varepsilon\) is independent of X, has mean zero, and has variance 1 (megapascal)². A report fits the best population linear regression with an intercept, predicting strength using X alone.
What slope and intercept will it find, and what is its mean squared prediction error? Compare those results with a regression that is allowed to use \(X^{2}\) as a predictor. Does a zero fitted linear slope imply temperature deviation carries no useful predictive information?
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Hint 1
Symmetry makes odd moments of X zero. Use \(\frac{\operatorname{Cov}(X,Y)}{\operatorname{Var}(X)}\) to find the linear slope.
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Solution
For X uniform on \([-1,1]\), symmetry gives \(\mathbb{E}[X]=\mathbb{E}[X^{3}]=0\), while direct integration gives \(\mathbb{E}[X^{2}]=\frac{1}{3}\) and \(\mathbb{E}[X^{4}]=\frac{1}{5}\). Therefore \(\operatorname{Var}(X)=\frac{1}{3}\) and
\[\operatorname{Cov}(X,Y)=\operatorname{Cov}(X,100-12X^{2}+\varepsilon)=-12\big(\mathbb{E}[X^{3}]-\mathbb{E}[X]\mathbb{E}[X^{2}]\big)+\operatorname{Cov}(X,\varepsilon)=0\].
The population least-squares slope on X is \(\beta_{1}=\frac{\operatorname{Cov}(X,Y)}{\operatorname{Var}(X)}=0\). Its intercept is \(\beta_{0}=\mathbb{E}[Y]-\beta_{1}\mathbb{E}[X]=100-\frac{12}{3}=96\) MPa, so the best linear predictor is the constant 96 MPa.
With an intercept, the linear model's MSE is \(\operatorname{Var}(Y)\) because its prediction is \(\mathbb{E}[Y]\). Independence of \(\varepsilon\) gives \(\operatorname{Var}(Y)=144\operatorname{Var}(X^{2})+\operatorname{Var}(\varepsilon)\). But \(\operatorname{Var}(X^{2})=\frac{1}{5}-\left(\frac{1}{3}\right)^{2}=\frac{4}{45}\), hence \(\operatorname{MSE}_{\mathrm{linear}}=144\left(\frac{4}{45}\right)+1=\frac{64}{5}+1=\frac{69}{5}=13.8\) (MPa)².
If \(X^{2}\) is included as a feature, the conditional mean \(\mathbb{E}[Y\mid X]=100-12X^{2}\) is representable exactly; coefficients are intercept 100 MPa and −12 MPa per (°C)² on \(X^{2}\), with zero contribution from any added X term. The residual is \(\varepsilon\), so \(\operatorname{MSE}_{\mathrm{quadratic}}=\operatorname{Var}(\varepsilon)=1\) (MPa)². The absence of a straight-line relationship comes from two symmetric directions of effect cancelling in \(\operatorname{Cov}(X,Y)\), even though departures in either direction lower expected strength. The improvement of 12.8 (MPa)² in MSE quantifies the missed nonlinear predictive structure.
Assumptions
X is continuous Uniform[−1,1] degrees Celsius and has nonzero variance.
Noise \(\varepsilon\) is independent of X, with zero mean and variance 1 (MPa)².
Population mean-squared-error optimal regressions include an unrestricted intercept.
Final Answer
Linear predictor \(\hat{Y}=96\) MPa with slope 0 and MSE \(\frac{69}{5}=13.8\) (MPa)². With \(X^{2}\), the predictor is \(100-12X^{2}\) MPa and MSE 1 (MPa)²; zero linear slope does not imply no predictive signal.