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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2023

Paper 4

Write
$$
\bar X_m=\frac1m\sum_{i=1}^mX_i,
\qquad
\bar Y_m=\frac1m\sum_{i=1}^mY_i.
$$
The <ordinary least squares estimators> give
$$
\widehat f_m(x)=\widehat\alpha_m+\widehat\beta_mx
=\bar Y_m+\widehat\beta_m(x-\bar X_m),
$$
where
$$
\widehat\beta_m
=\frac{\sum_{i=1}^m(X_i-\bar X_m)(Y_i-\bar Y_m)}
{\sum_{i=1}^m(X_i-\bar X_m)^2}
=\frac{\sum_{i=1}^mX_iY_i-m\bar X_m\bar Y_m}
{\sum_{i=1}^mX_i^2-m\bar X_m^2}.
$$
The denominator is positive because the sorted inputs are distinct and $m\geq2$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 30J / c / Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 30J / c / Solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 30J / c / Solution

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