Let

B

be the integral operator with kernel

β

. Since

Y = BX + ε

and the centered error is independent of

X

E [a_{k} b_{l}] = E [a_{k} ⟨ BX, u_{l} ⟩] .

β_{l k} = ⟨ B ϕ_{k}, u_{l} ⟩

and using

E [a_{k} a_{j}] = λ_{k} 1_{{j = k}}

gives

E [a_{k} b_{l}] = λ_{k} β_{l k} .

Expanding the function-on-function linear model kernel in the product basis therefore yields

β (t, s) = k = 1 \sum \infty l = 1 \sum \infty \frac{E [ a _{k} b _{l} ]}{λ _{k}} ϕ_{k} (s) u_{l} (t) .

ii

Solution

Replacing

ϕ_{k}

- ϕ_{k}

also replaces

a_{k}

- a_{k}

, so both

E [a_{k} b_{l}]

and

ϕ_{k} (s)

change sign and their product is unchanged. Replacing

u_{l}

- u_{l}

similarly replaces

b_{l}

- b_{l}

. Every summand, and hence

β

, is independent of all eigenfunction sign choices.

Applying the regression operator from part i gives

E [Y (t) ∣ X] = \sum_{k = 1}^{\infty} a_{k} {\sum_{l = 1}^{\infty} \frac{E [ a _{k} b _{l} ]}{λ _{k}} u_{l} (t)} .

The uncorrelated scores satisfy

Var (a_{k}) = λ_{k}

, while

Var (Y (t)) = \sum_{l} γ_{l} u_{l} (t)^{2}

. Hence

R (t) = \frac{k = 1 \sum \infty \frac{1}{λ _{k}} ( l = 1 \sum \infty E [ a _{k} b _{l} ] u _{l} ( t ) ) ^{2}}{l = 1 \sum \infty γ _{l} u _{l} ( t ) ^{2}} .

The sign cancellations from part ii leave every squared numerator term unchanged; changing the sign of

u_{l}

also leaves each denominator term unchanged. Thus

R (t)

is sign invariant.

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