Put
$$
\xi_{ik}=\langle X_i,B_k\rangle,
\qquad
\beta_K=\sum_{k=1}^Kc_kB_k,
\qquad
r_i=\langle X_i,\beta-\beta_K\rangle.
$$
With the $n$ by $K$ <design matrix> $\Xi=(\xi_{ik})$, the <scalar-on-function linear model> becomes
$$
Y=\Xi c+r+\varepsilon.
$$
The <ordinary least squares> estimator based on the truncated model is
$$
\widehat c=(\Xi^{\mathsf T}\Xi)^{-1}\Xi^{\mathsf T}Y
=c+(\Xi^{\mathsf T}\Xi)^{-1}\Xi^{\mathsf T}r
+(\Xi^{\mathsf T}\Xi)^{-1}\Xi^{\mathsf T}\varepsilon.
$$
Thus the omitted tail produces the conditional bias $(\Xi^{\mathsf T}\Xi)^{-1}\Xi^{\mathsf T}r$.

Let
$$
\Sigma_{jk}=\mathbb E[\xi_{1j}\xi_{1k}]
=\langle C_XB_j,B_k\rangle,
\qquad
g_j=\mathbb E[\xi_{1j}r_1]
=\langle C_XB_j,\beta-\beta_K\rangle.
$$
The supplied <weak law of large numbers> and noise limit give
$$
\widehat c\xrightarrow{p}c+\Sigma^{-1}g.
$$
For a general fixed basis, $g$ need not vanish, so the retained coefficients are asymptotically biased and the estimator is inconsistent even for $c$. Moreover, with fixed $K$ it cannot recover the full slope $\beta$ when $\beta-\beta_K\ne0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 225 / 4 / a / i / Solution

A scalar-on-function linear model has $Y=\alpha+\langle X,\beta\rangle+\varepsilon$, where the response is scalar and the predictor and slope are functions. Basis truncation turns it into a finite-dimensional linear regression.


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