= Solution
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 $\beta-\beta_K\ne0$.
Back to article page