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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2021

Paper 3

Let
$$
H=X(X^TX)^{-1}X^T
$$
be the <hat matrix>. Then
$$
\widehat\beta=(X^TX)^{-1}X^T(HY)
$$
depends only on the orthogonal projection $HY$, while
$$
n\widehat\sigma^2=\|(I-H)Y\|^2
$$
depends only on the orthogonal residual projection $(I-H)Y$.

The random vectors $HY$ and $(I-H)Y$ are jointly <multivariate normal distribution>[multivariate normal], and their cross-covariance is
$$
\sigma^2H(I-H)=0.
$$
Uncorrelated jointly Gaussian vectors are <independent random variables>[independent]. Therefore
$$
\boxed{\widehat\beta\ \text{and}\ \widehat\sigma^2
\text{ are independent}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3 / 18H / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3 / 18H / ii / Solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3 / 18H / ii / Solution

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