A flat line through a curve · Problem 213

s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.

Idea. By symmetry, $X$ and $X^{2}$ are uncorrelated, so the best line is flat. Then the mean squared error is simply the variance of $Y$ , which is large even though $Y$ is almost a deterministic function of $X$ .

Step 1: the slope. $Cov (X, X^{2}) = E [X^{3}] - E [X] E [X^{2}] = 0$ , since both $X$ and $X^{3}$ have mean 0. So $b = 0$ and $a = E [Y] = 100 - 12 \cdot \frac{1}{3} = 96$ .

Step 2: the error. The mean squared error is $Var (Y) = 144 Var (X^{2}) + 1$ . With $E [X^{2}] = \frac{1}{3}$ and $E [X^{4}] = \frac{1}{5}$ , $Var (X^{2}) = \frac{1}{5} - \frac{1}{9} = \frac{4}{45} .$

Step 3. $MSE = 144 \cdot \frac{4}{45} + 1 = 12.8 + 1 = \frac{69}{5} .$

The lesson. A zero slope, or a zero correlation, does not mean that $X$ is useless for predicting $Y$ . Adding $X^{2}$ as a regressor brings the error down to the noise variance, 1.