We shall define the vector Z_{i} y_{i} / _{i} and the matrix F_{ij} f_{j}(x_{i}) / _{i}.
Note that H = F^{T}^{ . }F by Eq. (27),
(33) |
Then
(34) |
where the unstarred is used for _{0}.
using Eq. (34). The second term on the right is zero because of Eq. (33).
(34) |
Note that
If q_{i} is an eigenvalue of Q, it must be equal q_{i}^{2}, an eigenvalue of Q^{2}. Thus q_{i} = 0 or 1. The trace of Q is
Since the trace of a matrix is invariant under a unitary transformation, the trace always equals the sum of the eigenvalues of the matrix. Therefore M of the eigenvalues of Q are one, and (p - M) are zero. Let U be the unitary matrix which diagonalizes Q (and also (1 - Q)). According to Eq. (35),
Thus
where S* is the square of the radius vector in (p - M)-dimensional space. By definition (see Section 16) this is the ^{2} distribution with (p - M) degrees of freedom.