Mathematical and Statistical Properties of Decomposition Techniques. The Splines Method

Abstract

As discussed in Chap. 1, the theoretical Divisia index is calculated upon the basis of the continuous time paths of the observed variables. However, only a finite number of discrete observations is available in practice.

Appendix I: Mathematical Proofs of Chapter 2

The following two lemmas are required for the proof of Proposition 2.

Lemma A.1

Under Assumptions 1 and 2, there exists a constant ${B < \infty }$ such that the following holds for ${j = 1 ,\ldots ,r}$ and ${ 0\le \alpha \le 1}$:

(a) ${\max_{t \in [0,1 ]} \left| {D^{\alpha } \hat{e}_{j} \left( t \right) - D^{\alpha } e_{j} \left( t \right)} \right| \le B\left( {n - 1} \right)^{{ - \left( {2 - \alpha } \right)}} }$, and

(b) ${{ \hbox{max} }_{{t \in [\text{0,1} ]}} \left| {D^{\alpha } \hat{y}_{j} \left( t \right) - D^{\alpha } y_{j} \left( t \right)} \right| \le B\left( {n - 1} \right)^{{ - \left( { 2- \alpha } \right)}} }$.

Proof

The Lemma is a direct consequence of Proposition 1.

Lemma A.2

Under Assumptions 1 and 2, there exist constants $B_{k} < \infty ,k = 1, \ldots ,4$ such that, for each ${j = 1 ,\ldots ,r}$ , the following holds for ${ 0\le t \le 1}$ and ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$ , where B is as in Lemma A.1 above and does not depend on t:

(a) ${\left| {\hat{I}_{j} \text{(}t\text{)} - I_{j} \text{(}t\text{)}} \right| \le B_{ 1} \text{(}n - 1\text{)}^{ - 2} },$

(b) $\left| {\hat{S}_{j} \left( t \right) - S_{j} \left( t \right)} \right| \le B_{2} \left( {n - 1} \right)^{ - 2},$

(c) ${\left| {D^{ 1} \hat{I}_{j} \text{(}t\text{)} - D^{ 1} I_{j} \text{(}t\text{)}} \right| \le B_{ 3} \left( {n - 1} \right)^{ - 1} },$

(d) ${\left| {D^{1}\hat{S}_{j} \text{(}t\text{)} - D^{1} S_{j} \text{(}t\text{)}} \right| \le B_{ 4} \left( {n - 1} \right)^{ - 1} }$.

Proof

In the proof of this Lemma (and throughout the rest of the Appendix) we will rely on the fact that any continuous function in $[0,1]$ is bounded in that interval. This implies that, under Assumptions 1(i) and 2(i), there exists a constant $M < \infty$ such that ${\max_{0 \le \alpha \le 1} \max_{0 \le t \le 1} \left| {D^{\alpha } e_{j} \left( t \right)} \right| \le M}$ and ${\max_{0 \le \alpha \le 1} \max_{0 \le t \le 1} \left| {D^{\alpha } y_{j} \left( t \right)} \right| \le M}$ for each $j = 1, \ldots ,r$. The same is true for the aggregate consumption and production functions (respectively, $e (t )$ and $y(t)$).

Let us select an arbitrary point ${t \in \left[ { 0 , 1} \right]}$. Regarding statement (a), we have:

$\left| {\hat{I}_{j} \left( t \right) - I_{j} \left( t \right)} \right| = \left| {\frac{{\hat{e}_{j} \left( t \right)}}{{\hat{y}_{j} \left( t \right)}} - \frac{{e_{j} \left( t \right)}}{{y_{j} \left( t \right)}}} \right| \le A_{I} + A_{II}$, where ${A_{I} = \left| {\frac{{\hat{e}_{j} \left( t \right)}}{{\hat{y}_{j} \left( t \right)}} - \frac{{e_{j} \left( t \right)}}{{\hat{y}_{j} \left( t \right)}}} \right|}$ and ${A_{II} = \left| {\frac{{e_{j} \left( t \right)}}{{\hat{y}_{j} \left( t \right)}} - \frac{{e_{j} \left( t \right)}}{{y_{j} \left( t \right)}}} \right|}$. Let ${A_{III} = \hat{y}_{j} \left( t \right) - y_{j} \left( t \right)}$. Lemma A.1 ensures both ${\left| {\hat{e}_{j} \left( t \right) - e_{j} \left( t \right)} \right| \le B\left( {n - 1} \right)^{ - 2} }$ and ${\left| {\hat{y}_{j} \left( t \right) - y_{j} \left( t \right)} \right| \le B\left( {n - 1} \right)^{ - 2} }$. Thus, arbitrarily small ${\left| {\hat{e}_{j} \left( t \right) - e_{j} \left( t \right)} \right|}$ and ${\left| {A_{III} } \right|}$ can be obtained for large n. In particular, if ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$ we have ${A_{III} < m/ 2}$. Therefore, Assumption 2(ii) implies that, for any ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$, it holds ${\hat{y}_{j} \left( t \right) = y_{j} \left( t \right) + A_{III} \ge m - m/ 2= m/ 2> 0}$, so ${A_{I} = \left| {\hat{y}_{j} \left( t \right)} \right|^{ - 1} \left| {\hat{e}_{j} \left( t \right) - e_{j} \left( t \right)} \right| \le \frac{B}{m/ 2}\left( {n - 1} \right)^{ - 2} }$.

As for ${A_{II} }$ we have:

$${A_{II} = \left| {e_{j} \left( t \right)} \right| \cdot \left| {\frac{{y_{j} \left( t \right) - \hat{y}_{j} \left( t \right)}}{{y_{j} \left( t \right)\hat{y}_{j} \left( t \right)}}} \right|}$$

As ${\left| {e_{j} \left( t \right)} \right| \le M < \infty }$ by continuity in $[0,1]$, Lemma A.1(b) ensures that, for ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$, it holds ${A_{II} \le \frac{M}{{m^{ 2} / 2}}B\left( {n - 1} \right)^{ - 2} }$.

Therefore, ${A_{I} + A_{II} \le B_{ 1} \left( {n - 1} \right)^{ - 2} }$ for all n large enough and some finite $B_{1}$. As t was arbitrary, uniform convergence is obtained, which completes the proof of statement (a).

Regarding (b) we have, for any $t \in \left[0,1\right]$, ${\left| {\hat{S}_{j} \text{(}t\text{)} - S_{j} \text{(}t\text{)}} \right| = \left| {\frac{{\hat{y}_{j} \left( t \right)}}{{\hat{y}\left( t \right)}} - \frac{{y_{j} \left( t \right)}}{y\left( t \right)}} \right| \le A_{I} + A_{II} }$, where ${A_{I} = \frac{{\left| {\hat{y}_{j} \left( t \right) - y_{j} \left( t \right)} \right|}}{{\hat{y}\left( t \right)}}}$ and ${A_{II} = \left| {y_{j} \left( t \right)\frac{{y\left( t \right) - \hat{y}\left( t \right)}}{{y\left( t \right)\hat{y}\left( t \right)}}} \right|}$.

For ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$ it holds ${\hat{y}\left( t \right) = \sum {_{j = 1}^{r} \hat{y}_{j} } \left( t \right) = \sum {_{j = 1}^{r} y_{j} \left( t \right) + \sum {_{j = 1}^{r} \left( {\hat{y}_{j} \left( t \right) - y_{j} \left( t \right)} \right)} } \ge rm - rm/ 2= rm/ 2}$.

Thus ${A_{I} \le \frac{ 2}{rm}B\left( {n - 1} \right)^{ - 2} }$ and ${A_{II} \le \frac{{ 2 {\textit{M}}}}{{r^{ 2} m^{ 2} }}B\left( {n - 1} \right)^{ - 2} }$, so for ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$ we have ${\left| {\hat{S}_{j} \text{(}t\text{)} - S_{j} \text{(}t\text{)}} \right| \le B_{ 2} \left( {n - 1} \right)^{ - 2} }$. Again, as t is arbitrary, convergence is uniform.

Regarding (c), select an arbitrary t in $[0,1]$ and apply the quotient rule for derivatives. We have ${\left| {D^{ 1} \hat{I}_{j} \left( t \right) - D^{ 1} I_{j} \left( t \right)} \right| = \left| {A_{I} - A_{II} } \right|}$, where

$${A_{I} = \frac{{D^{ 1} \hat{e}_{j} \left( t \right)\hat{y}_{j} \left( t \right){-}\hat{e}_{j} \left( t \right)D^{ 1} \hat{y}_{j} \left( t \right)}}{{\left( {\hat{y}_{j} \left( t \right)} \right)^{ 2} }}}$$

and

${A_{II} = \frac{{D^{ 1} e_{j} \left( t \right)y_{j} \left( t \right){-}e_{j} \left( t \right)D^{ 1} y_{j} \left( t \right)}}{{\left( {y_{j} \left( t \right)} \right)^{ 2} }}}$.

By the triangle inequality

${\left| {A_{I} - A_{II} } \right| \le \left| {A_{I} - A_{III} } \right| + \left| {A_{III} - A_{II} } \right|}$, where ${A_{III} = \frac{{D^{ 1} e_{j} \left( t \right)y_{j} \left( t \right){-}e_{j} \left( t \right)D^{ 1} y_{j} \left( t \right)}}{{\left( {\hat{y}_{j} \left( t \right)} \right)^{ 2} }}}$.

It is directly obtained that, for ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$, there exists ${B_{ 5} < \infty }$ (not depending on n) such that ${\left| {A_{I} - A_{II} } \right| \le B_{ 5} \left( {n - 1} \right)^{ - 1} }$.

Analogously, continuity (and so boundedness in $[0,1]$) of the first derivatives of ${y_{j} }$ and ${e_{j} }$ implies that, for some ${M < \infty }$, ${\left| {D^{1} e_{j} \left( t \right)y_{j} \left( t \right){-}e_{j} \left( t \right)D^{1} y_{j} \left( t \right)} \right| \le \left| {D^{1} e_{j} \left( t \right)} \right| \cdot \left| {y_{j} \left( t \right)} \right| + \left| {e_{j} \left( t \right)} \right| \cdot \left| {D^{1} y_{j} \left( t \right)} \right| \le 2M^{2} }$.

So it follows, for ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$ and some ${B_{ 6} < \infty }$ (not depending on n), ${\left| {A_{II} - A_{III} } \right| \le \frac{{ 4 {\textit{M}}^{ 2} }}{{m^{ 2} }}B_{6} \left( {n - 1} \right)^{ - 2} }.$

Therefore, ${\left| {A_{I} - A_{II} } \right| \le \left| {A_{I} - A_{III} } \right| + \left| {A_{III} - A_{II} } \right| \le B_{ 3} \left( {n - 1} \right)^{ - 1} }$ for ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$ and some $B_{3}$ finite, with convergence being uniform.

As for (d), a similar procedure is applied. We have ${\left| {D^{ 1} \hat{S}_{j} \left( t \right) - D^{ 1} S_{j} \left( t \right)} \right| \le \left| {A_{I} - A_{III} } \right| + \left| {A_{III} - A_{II} } \right|}$, where ${A_{I} = \frac{{D^{ 1} \hat{y}_{j} \left( t \right)\hat{y}\left( t \right){-}\hat{y}_{j} \left( t \right)D^{ 1} \hat{y}\left( t \right)}}{{\left( {\hat{y}\left( t \right)} \right)^{ 2} }}}$, ${A_{II} = \frac{{D^{ 1} y_{j} \left( t \right)y\left( t \right){-}y_{j} \left( t \right)D^{ 1} y\left( t \right)}}{{\left( {y\left( t \right)} \right)^{ 2} }}}$ and ${A_{III} = \frac{{D^{ 1} y_{j} \left( t \right)y\left( t \right){-}y_{j} \left( t \right)D^{ 1} y\left( t \right)}}{{\left( {\hat{y}\left( t \right)} \right)^{ 2} }}}$.

Again we obtain that, for ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$, there exists ${B_{ 6} < \infty }$ (not depending on n) so that ${\left| {A_{I} - A_{II} } \right| \le B_{ 6} \left( {n - 1} \right)^{ - 1} }$.

As ${\left| {D^{ 1} y_{j} \left( t \right)y\left( t \right){-}y_{j} \left( t \right)D^{ 1} y\left( t \right)} \right| \le \left| {D^{ 1} y_{j} \left( t \right)} \right| \cdot \left| {y\left( t \right)} \right| + \left| {y_{j} \left( t \right)} \right| \cdot \left| {D^{ 1} y\left( t \right)} \right| \le rM^{2} }$ for some ${M < \infty }$, and since ${\left| {\left( {y\left( t \right)} \right)^{ 2} - \left( {\hat{y}\left( t \right)} \right)^{ 2} } \right| = \left| {y\left( t \right) + \hat{y}\left( t \right)} \right| \cdot \left| {y\left( t \right) - \hat{y}\left( t \right)} \right| \le \left( { 2 {\textit{M}} + B\left( {n - 1} \right)^{ - 2} } \right)B\left( {n - 1} \right)^{ - 2} }$, we obtain for ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$:

$$\begin{gathered} \left| {A_{III} - A_{II} } \right| \le \frac{{\left| {D^{1} y_{j} \left( t \right)y\left( t \right) - y_{j} \left( t \right)D^{1} y\left( t \right)} \right| \cdot \left| {\left( {y\left( t \right)} \right)^{2} - (\hat{y}\left( t \right))^{2} } \right|}}{{\left( {y\left( t \right)} \right)^{2} \left( {\hat{y}\left( t \right)} \right)^{2} }} \hfill \\ \le \frac{{4rM^{2} \left( {2M + 1} \right)}}{{r^{4} m^{4} }}\left( {2M + B\left( {n - 1} \right)^{ - 2} } \right)B\left( {n - 1} \right)^{ - 2} \hfill \\ \end{gathered}$$

Therefore, for each ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$ and ${ 0\le t \le 1}$, there exists ${B_{ 4} < \infty }$ (not depending on n or t), such that ${\left| {A_{I} - A_{II} } \right| \le \left| {A_{I} - A_{III} } \right| + \left| {A_{III} - A_{II} } \right| \le B_{ 3} \left( {n - 1} \right)^{ - 1} }$.

Lemma A.3

Let ${D^{ 1} I\left( t \right) = \sum {_{j = 1}^{r} \left( {D^{ 1} I_{j} \left( t \right)S_{j} \left( t \right) + I_{j} \left( t \right)D^{ 1} S_{j} \left( t \right)} \right)} }$ and ${D^{ 1} \hat{I}\left( t \right) = \sum {_{j = 1}^{r} \left( {D^{ 1} \hat{I}_{j} \left( t \right)\hat{S}_{j} \left( t \right) + \hat{I}_{j} \left( t \right)D^{ 1} \hat{S}_{j} \left( t \right)} \right)} }$ . Under Assumptions 1 and 2, there exists a constant ${B_{ 5} < \infty }$ such that, for each ${ 0\le t \le 1}$ and ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$ , with B being as in Lemma A.1, it holds ${\left| {D^{ 1} \hat{I}_{j} \left( t \right) - D^{ 1} I_{j} \left( t \right)} \right| \le B_{ 5} \left( {n - 1} \right)^{ - 1} }$.

Proof

It straightforwardly derives from Lemma A.2, which establishes that each component of ${D^{ 1} \hat{I}\left( t \right)}$ converges uniformly to its analogue in ${D^{ 1} I\left( t \right)}$. In order to obtain uniform convergence we will rely on the fact that all the components appearing in ${D^{ 1} I\left( t \right)}$ are uniformly bounded in ${ 0\le t \le 1}$. In particular: ${\left| {I_{j} \left( t \right)} \right| = \left| {\frac{{e_{j} \left( t \right)}}{{y_{j} \left( t \right)}}} \right| \le \frac{M}{m}}$, ${\left| {S_{j} \left( t \right)} \right| = \left| {\frac{{y_{j} \left( t \right)}}{{y\text{(}t\text{)}}}} \right| \le \frac{M}{rm}}$, ${\left| {D^{ 1} I_{j} \left( t \right)} \right| = \frac{{\left| {D^{ 1} e_{j} \left( t \right)y_{j} \left( t \right){-}e_{j} \left( t \right)D^{ 1} y_{j} \left( t \right)} \right|}}{{\left( {y_{j} \left( t \right)} \right)^{ 2} }} \le \frac{{ 2 {\textit{M}}^{ 2} }}{{m^{ 2} }}}$ and ${\left| {D^{ 1} S_{j} \left( t \right)} \right| = \frac{{\left| {D^{ 1} y_{j} \left( t \right)y\left( t \right){-}y_{j} \left( t \right)D^{ 1} y\left( t \right)} \right|}}{{\left( {y\left( t \right)} \right)^{ 2} }} \le \frac{{ 2 {\textit{M}}^{ 2} }}{{r^{ 2} m^{ 2} }}}$.

Similar uniform bounds are obtained for the components of ${D^{ 1} \hat{I}\left( t \right)}$, since according to Lemma A.2, as ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$ we have, for each ${ 0\le t \le 1}$, ${\left| {\hat{I}_{j} \left( t \right)} \right| \le \left| {I_{j} \left( t \right)} \right| + \left| {\hat{I}_{j} \left( t \right) - I_{j} \left( t \right)} \right| \le \frac{M}{m} + B_{1} \left( {n - 1} \right)^{ - 2} }$, ${\left| {\hat{S}_{j} \left( t \right)} \right| \le \left| {S_{j} \left( t \right)} \right| + \left| {\hat{S}_{j} \left( t \right) - S_{j} \left( t \right)} \right| \le \frac{M}{rm} + B_{2} \left( {n - 1} \right)^{ - 2} }$, ${\left| {D^{ 1} \hat{I}_{j} \left( t \right)} \right| \le \left| {D^{ 1} I_{j} \left( t \right)} \right| + \left| {D^{ 1} \hat{I}_{j} \left( t \right) - D^{ 1} I_{j} \left( t \right)} \right| \le \frac{{ 2 {\textit{M}}^{ 2} }}{{m^{ 2} }} + B_{3} \left( {n - 1} \right)^{ - 1} }$

and

${\left| {D^{ 1} \hat{S}_{j} \left( t \right)} \right| \le \left| {D^{ 1} S_{j} \left( t \right)} \right| + \left| {D^{ 1} \hat{S}_{j} \left( t \right) - D^{ 1} S_{j} \left( t \right)} \right| \le \frac{{ 2 {\textit{M}}^{ 2} }}{{r^{ 2} m^{ 2} }} + B_{4} \left( {n - 1} \right)^{ - 1} }$.

By using the decomposition ${D^{ 1} \hat{I}\left( t \right) - D^{ 1} I\left( t \right) = A_{I} + A_{II} }$, where

$${A_{I} = \sum {_{j = 1}^{r} \left( {D^{ 1} \hat{I}_{j} \left( t \right)\hat{S}_{j} \left( t \right) + \hat{I}_{j} \left( t \right)D^{ 1} \hat{S}_{j} \left( t \right)} \right)} - \sum {_{j = 1}^{r} \left( {D^{ 1} \hat{I}_{j} \left( t \right)S_{j} \left( t \right) + \hat{I}_{j} \left( t \right)D^{ 1} S_{j} \left( t \right)} \right)} }$$

and

$${A_{II} = \sum {_{j = 1}^{r} \left( {D^{ 1} \hat{I}_{j} \left( t \right)S_{j} \left( t \right) + \hat{I}_{j} \left( t \right)D^{ 1} S_{j} \left( t \right)} \right)} - \sum {_{j = 1}^{r} \left( {D^{ 1} I_{j} \left( t \right)S_{j} \left( t \right) + I_{j} \left( t \right)D^{ 1} S_{j} \left( t \right)} \right)} }.$$

and applying the above bounds, it is readily obtained that there exists some ${B_{ 5} < \infty }$ such that, for each t in $[0,1]$ and ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$, it holds ${\left| {D^{ 1} \hat{I}\left( t \right) - D^{ 1} I\left( t \right)} \right| \le \left| {A_{I} } \right| + \left| {A_{II} } \right| \le B_{ 5} \left( {n - 1} \right)^{ - 1} }$.

Proof of Proposition 2

It is a direct consequence of Lemma A.3. It is readily checked that ${D^{ 1} I\left( t \right)}$ is continuous in $[0,1]$, so it is also bounded in that interval. In particular, for each t in $[0,1]$, it holds ${\left| {D^{ 1} I\left( t \right)} \right| = \frac{{\left| {D^{ 1} e\left( t \right)y\left( t \right) - e\left( t \right)D^{ 1} y\left( t \right)} \right|}}{{\left( {y\left( t \right)} \right)^{ 2} }} \le \frac{{ 2 {\textit{M}}^{ 2} }}{{m^{ 2} }}}$. An analogous result is obtained for ${D^{ 1} \hat{I}\left( t \right)}$ when ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$, as ${\hat{y}\left( t \right) \ge rm/ 2> 0}$ in that case, so ${D^{ 1} \hat{I}\left( t \right)}$ is a ratio of continuous functions, with strictly positive denominator (having a positive lower bound not depending on n).

Applying Lemma 3, we obtain

$${\left| {\hat{T}E_{n} {-}TE} \right| = \left| {\int\limits_{ 0}^{ 1} {D^{ 1} \hat{I}\left( t \right){\text{d}}t} - \int\limits_{ 0}^{ 1} {D^{ 1} I\left( t \right){\text{d}}t} } \right| \le \int\limits_{ 0}^{ 1} {\left| {D^{ 1} \hat{I}\left( t \right) - D^{ 1} I\left( t \right)} \right|{\text{d}}t} \le B_{ 5} \left( {n - 1} \right)^{ - 1} \int\limits_{0}^{1} {{\text{d}}t} = B_{5} \left( {n - 1} \right)^{ - 1} }$$

for ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$, which proves part (b), and therefore part (a). □

Proof of Proposition 3

It is a consequence of Lemma A.3. Let us select an arbitrary point t in $[0,1]$ and apply the decomposition ${\left| {\frac{{D^{ 1} \hat{I}\left( t \right)}}{{\hat{I}\left( t \right)}} - \frac{{D^{ 1} I\left( t \right)}}{I\left( t \right)}} \right| \le A_{I} + A_{II} }$, where ${A_{I} = \frac{{\left| {D^{ 1} \hat{I}\left( t \right) - D^{ 1} I\left( t \right)} \right|}}{{\hat{I}\left( t \right)}}}$ and ${A_{II} = \left| {\frac{{D^{ 1} I\left( t \right)}}{{\hat{I}\left( t \right)}} - \frac{{D^{ 1} I\left( t \right)}}{I\left( t \right)}} \right|}$.

The inequality ${\left| {D^{ 1} \hat{I}\left( t \right) - D^{ 1} I\left( t \right)} \right| \le B_{ 5} \left( {n - 1} \right)^{ - 1} }$ was obtained in the proof of Lemma A.3 above, for each t in $[0,1]$ and ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$. In addition, ${\frac{m}{M} \le \frac{ 1}{I\left( t \right)} = \frac{y\left( t \right)}{e\left( t \right)} \le \frac{M}{m}}$.

It is readily shown, by a procedure analogous to that used in Lemma A.2 (a), that ${\left| {\hat{I}\left( t \right) - I\left( t \right)} \right| \le B_{ 6} \left( {n - 1} \right)^{ - 2} }$ for each t in $[0,1]$ and ${n > 1+ \left( { 2 {\textit{B}}/m} \right)^{ 1/ 2} }$, with ${B_{ 6} < \infty }$ not depending on t.

Since ${\frac{ 1}{{\hat{I}\left( t \right)}} = \frac{ 1}{I\left( t \right)} + \frac{{I\left( t \right) - \hat{I}\left( t \right)}}{{I\left( t \right)\hat{I}\left( t \right)}}}$, and as ${\left| {\hat{I}\left( t \right) - I\left( t \right)} \right| \le B_{ 6} \left( {n - 1} \right)^{ - 2} \le \frac{m}{{ 2 {\textit{M}}}}}$ for all large enough n (for this, it suffices to select ${n \ge \sqrt {B_{ 6} \frac{{ 2 {\textit{M}}}}{m} - 1} }$), we arrive at ${\hat{I}\left( t \right) = I\left( t \right) + \left( {\hat{I}\left( t \right) - I\left( t \right)} \right) \ge \frac{m}{M} - \frac{m}{{ 2 {\textit{M}}}} = \frac{m}{{ 2 {\textit{M}}}}}$.

Therefore, it eventually holds:

$${\left| {\frac{ 1}{{\hat{I}\left( t \right)}}} \right| = \left| {\frac{ 1}{I\left( t \right)} + \frac{{I\left( t \right) - \hat{I}\left( t \right)}}{{I\left( t \right)\hat{I}\left( t \right)}}} \right| \le \frac{M}{m} + \frac{{ 2 {\textit{M}}^{ 2} }}{{m^{ 2} }}B_{ 6} \left( {n - 1} \right)^{ - 2} }$$

So, for large enough n, we obtain ${A_{I} = \frac{{\left| {D^{ 1} \hat{I}\left( t \right) - D^{ 1} I\left( t \right)} \right|}}{{\hat{I}\left( t \right)}} \le \left( { 1+ \frac{M}{m}} \right)B_{ 5} \left( {n - 1} \right)^{ - 1} }$.

As for ${A_{II} }$, we have ${A_{II} = \frac{{\left| {D^{ 1} I\left( t \right)} \right| \cdot \left| {I\left( t \right) - \hat{I}\left( t \right)} \right|}}{{I\left( t \right)\hat{I}\left( t \right)}}}$. As shown in the proof of Proposition 2, it holds ${\left| {D^{ 1} I\left( t \right)} \right| \le \frac{{2 {\textit{M}}^{ 2} }}{{m^{ 2} }}}$ for all large enough n, and by applying the bounds obtained for ${A_{I} }$ above the following inequality is readily obtained:

$${A_{II} = \frac{{\left| {D^{ 1} I\left( t \right)} \right| \cdot \left| {I\left( t \right) - \hat{I}\left( t \right)} \right|}}{{I\left( t \right)\hat{I}\left( t \right)}} \le \frac{{\frac{{ 2 {\textit{M}}^{ 2} }}{{m^{ 2} }} \times M}}{m}\left( { 1+ \frac{M}{m}} \right)B_{ 6} \left( {n - 1} \right)^{ - 2} }$$

Therefore, there exists ${B_{ 7} < \infty }$ such that, for each t in $[0,1]$, ${\left| {\frac{{D^{ 1} \hat{I}\left( t \right)}}{{\hat{I}\left( t \right)}} - \frac{{D^{ 1} I\left( t \right)}}{I\left( t \right)}} \right| \le B_{ 7} \left( {n - 1} \right)^{ - 1} }$.

By the same procedure as in Proposition 2 it is shown that, for each t in $[0,1]$, it holds ${I\left( t \right) > 0}$, and the same is true for ${\hat{I}\left( t \right)}$ for large enough n. This ensures that

$$\begin{gathered} \left| {\hat{L}TE_{n} - LTE} \right| = \left| {\int\limits_{0}^{1} {\left( {D^{1} \ln \hat{I}\left( t \right) - D^{ 1} { \ln }I\left( t \right)} \right)} \,{\text{d}}t} \right| \le \int\limits_{0}^{1} {\left| {D^{1} \ln \hat{I}\left( t \right) - D^{ 1} \ln I\left( t \right)} \right|} \,{\text{d}}t \hfill \\ \le B_{ 7} \left( {n - 1} \right)^{ - 1} \int\limits_{0}^{1} {{\text{d}}t} = B_{5} \left( {n - 1} \right)^{ - 1} \hfill \\ \end{gathered}$$

for sufficiently large n, which completes the proof of both parts of the proposition.

Proof of Proposition 4

First, we will prove (a), that for each n large enough and each (fixed) set of knots ${N_{n} = \left\{ {t_{i} = {{\left( {i - 1} \right)} \mathord{\left/ {\vphantom {{\left( {i - 1} \right)} {\left( {n - 1} \right)}}} \right. \kern-0pt} {\left( {n - 1} \right)}},\,i = 1 ,\ldots ,n} \right\}}$, the vector of interpolated time paths, $\hat{Z}_{n} = \left( {\hat{z}_{ 1 ,n} ,\ldots ,\hat{z}_{{ 2 {\textit{r,}}n}} } \right)$, is ${{\mathbf{B}}\left( S \right)}$-measurable. We shall use symbol ${\left( .\right)_{n} }$ to denote the operator that associates with each path in $C^{1} \left[ {0,1} \right]$ its natural spline interpolant, with n knots located at $N_{n}$. Thus, ${\hat{z}_{j,n} = \left( {z_{j} } \right)_{n} }$ is the natural spline that interpolates path ${z_{j} }$ at ${N_{n} }$.

It suffices to show that ${\left( .\right)_{n} }$ is a continuous mapping of ${C^{ 1} \left[ {\text{0,1}} \right]}$ into ${C^{ 1} \left[ {\text{0,1}} \right]}$, which implies that it is also continuous as a vector function of S into S, and therefore ${\hat{Z}_{n} = \left( Z \right)_{n} }$ is ${{\mathbf{B}}\left( S \right)}$-measurable. Select two arbitrary functions ${f,f' \in C^{ 1} \left[ {\text{0,1}} \right]}$, with that function space endowed with the norm ${\left| {\left| .\right|} \right|}$ particularized to the case of a single trajectory. We will show that, given n fixed (and therefore, a fixed set of n knots), ${\left| {\left| {f - f'} \right|} \right| \to 0}$ implies ${\left| {\left| {\hat{f} - \hat{f}'} \right|} \right| \to 0}$. This stems, as we shall see, from the fact that ${\left( .\right)_{n} }$ is a continuous linear operator.

Indeed it can be shown that, provided that $n \ge 2$, the set of natural splines (having degree 3, continuous derivatives up to order 1 and knots at ${N_{n} }$) is a vector space of dimension n. This implies that the natural spline interpolant ${\hat{f}_{n} \left( t \right)}$ for ${f \in C^{ 1} \left[ {\text{0,1}} \right]}$ is unique (Powell 1981, Chap. 23, Theorem 23.1.) and has the following expression:

$${\hat{f}_{n} (t) = \sum\limits_{i = 1}^{n} {\hat{\beta }_{i,n} \varphi_{i,n} (t)} }$$

where ${\left( {\varphi_{ 1 ,n} , \ldots ,\varphi_{n,n} } \right)}$ is a vector of n linearly independent functions in ${C^{ 1} \left[ { 0 , 1} \right]}$. The coefficient vector ${\hat{\beta }_{n} = \left( {\hat{\beta }_{1,n} , \ldots ,\hat{\beta }_{n,n} } \right)^{T} }$ is obtained by imposing the n conditions of interpolation at ${N_{n} }$, i.e., ${f\left( {t_{i} } \right) = \hat{f}\left( {t_{i} } \right),\,\,i = 1 ,\ldots ,n}$. This is equivalent to solving the system of linear equations ${f_{n} = \varPhi_{n} \hat{\beta }_{n} }$, where $f_{n} = \left( {f\left( {t_{1} } \right), \ldots ,f\left( {t_{n} } \right)} \right)^{T}$ and ${\varPhi_{n} = \left[ {c_{i,k} } \right]}$, $i,k = 1, \ldots ,n$, is a squared matrix whose elements are ${c_{i,k} = \varphi_{i,n} \left( {t_{k} } \right)}$. Uniqueness of the solution of this problem implies that matrix ${\varPhi_{n} }$ is nonsingular, so ${\hat{\beta }_{n} = \varPhi_{n}^{ - 1} f_{n} }$.

Now consider another function ${f^{\prime} \in C^{ 1} \left[ {\text{0,1}} \right]}$, and let ${\hat{f}_{n} '(t) = \sum\limits_{i = 1}^{n} {\hat{\beta }'_{i,n} \varphi_{i,n} (t )} }$ be its natural spline interpolant at ${N_{n} }$, with ${\hat{\beta }^{\prime}_{n} = \left( {\hat{\beta }^{\prime}_{ 1 ,n} , \ldots ,\hat{\beta }^{\prime}_{n,n} } \right)^{T} }$. For any integer ${ 0\le \alpha \le 1}$, select an arbitrary point ${ 0\le t \le 1}$. We have

$$\begin{gathered} \left| {D^{\alpha } f'_{n} (t) - D^{\alpha } f_{n} \text{(}t\text{)}} \right| = \left| {\sum\limits_{i = 1}^{n} {D^{\alpha } \varphi_{i,n} \text{(}t\text{)}} \left( {\beta '_{i,n} - \beta_{i,n} } \right)} \right| \le \sum\limits_{i = 1}^{n} {\left| {D^{\alpha } \varphi_{i,n} \text{(}t\text{)}} \right| \cdot } \left| {\beta '_{i,n} - \beta_{i,n} } \right| \le \hfill \\ \mathop {\hbox{max} }\limits_{0 \le \alpha \le 1} \;\mathop {\hbox{max} }\limits_{i = 1, \ldots ,n} \left| {D^{\alpha } \varphi_{i,n} \text{(}t_{i} \text{)}} \right| \cdot \sum\limits_{i = 1}^{n} {\left| {\hat{\beta }'_{i,n} - \hat{\beta }_{i,n} } \right|} \le \mathop {\hbox{max} }\limits_{0 \le \alpha \le 1} \;\mathop {\hbox{max} }\limits_{i=1,...,n} \;\mathop {\hbox{max} }\limits_{t \in [0,1]} \left| {D^{\alpha } \varphi_{i,n} \text{(}t\text{)}} \right|\,n\,n^{ - 1} \sum\limits_{i = 1}^{n} {\left| {\hat{\beta }'_{i,n} - \hat{\beta }_{i,n} } \right|} \le \hfill \\ \mathop {\hbox{max} }\limits_{0 \le \alpha \le 1} \; \mathop {\hbox{max} }\limits_{i=1,...,n} \; \mathop {\hbox{max} }\limits_{t \in [0,1]} \left| {D^{\alpha } \varphi_{i,n} \text{(}t\text{)}} \right|n\sqrt {n^{ - 1} \sum\limits_{i = 1}^{n} {\left( {\hat{\beta }'_{i,n} - \hat{\beta }_{i,n} } \right)}^{2} } \hfill \\ \end{gathered}$$

As ${\hat{\beta }'_{n} - \hat{\beta }_{n} = \varPhi_{n}^{ - 1} e_{n} }$, with $e_{n} = f^{\prime}_{n} - f_{n}$, we have ${\sqrt {\left( {\hat{\beta }'_{n} - \hat{\beta }_{n} } \right)^{T} \left( {\hat{\beta }'_{n} - \hat{\beta }_{n} } \right)} = \sqrt {e_{n}^{T} \left( {\varPhi_{n}^{ - 1} } \right)^{T} \varPhi_{n}^{ - 1} e_{n} } }$.

Since matrix ${\left( {\varPhi_{n}^{ - 1} } \right)^{T} \varPhi_{n}^{ - 1} }$ is symmetric and positive definite, it admits the orthogonal decomposition ${\left( {\varPhi_{n}^{ - 1} } \right)^{T} \varPhi_{n}^{ - 1} = P_{n}^{T} \varLambda_{n} P_{n} }$, where ${P_{n}^{T} P_{n} = I_{n} }$, with ${I_{n} }$ being the unit matrix of order n and ${\varLambda_{n} }$ being a diagonal matrix with the eigenvalues of ${\left( {\varPhi_{n}^{ - 1} } \right)^{T} \varPhi_{n}^{ - 1} }$ on its main diagonal. It is readily obtained that ${\left[ {e_{n}^{T} \left( {\varPhi_{n}^{ - 1} } \right)^{T} \varPhi_{n}^{ - 1} e_{n} } \right]^{ 1/ 2} = \left[ {e_{n}^{T} P_{n}^{T} \varLambda_{n} P_{n} e_{n} } \right]^{ 1/ 2} \le \lambda_{ 1}^{ 1/ 2} \left[ {e_{n}^{T} e_{n} } \right]^{ 1/ 2} }$, where ${\lambda_{ 1} }$ is the largest eigenvalue of ${\left( {\varPhi_{n}^{ - 1} } \right)^{T} \varPhi_{n}^{ - 1} }$.

Therefore, we obtain

$$\begin{aligned} \sqrt {\left( {\hat{\beta}_{n} - \hat{\beta}_{n} } \right)^{T} \left( {\hat{\beta}_{n} - \hat{\beta}_{n} } \right) }= \sqrt {e_{n}^{T} \left( {\phi_{n}^{ - 1} } \right)^{T} \phi_{n}^{ - 1} e_{n} } \le \sqrt {\lambda_{1} } \sqrt {e_{n}^{T} e_{n} } = \sqrt {\lambda_{1} } \sqrt {\sum\limits_{i = 1}^{n} {\left( {f^{\prime}\left( {t_{i} } \right) - f\left( {t_{i} } \right)} \right)^{2} } } \hfill \\ \quad \le \sqrt {\lambda_{1} n} \;\mathop {\hbox{max} }\limits_{i = 1, \ldots ,n} |f^{\prime}\left( {t_{i} } \right) - f\left( {t_{i} } \right)| \le \sqrt {\lambda_{1} n} \cdot ||f^{\prime} - f|| \hfill \\ \end{aligned}$$

and it holds ${\left| {D^{\alpha } \hat{f}'_{n} \left( t \right) - D^{\alpha } \hat{f}_{n} \left( t \right)} \right| \le \mathop { \hbox{max} }\limits_{0 \le \alpha \le 1} \; \mathop { \hbox{max} }\limits_{i=1,...,n} \; \mathop { \hbox{max} }\limits_{t \in [0,1 ]} \left| {D^{\alpha } \varphi_{i,n} (t)} \right|\,n\sqrt {\lambda_{1} } \,\;\left\| {f' - f} \right\|}$. Since both $\alpha$ and t are arbitrary the above bound is uniform, so ${\left| {\left| {\hat{f}'_{n} - \hat{f}_{n} } \right|} \right| \to 0}$ as ${\left| {\left| {f' - f} \right|} \right| \to 0}$, which ensures continuity—in terms of the distance induced by the norm ${\left| {\left| f \right|} \right| = \max_{0 \le \alpha \le 1} \max_{0 \le t \le 1} \left| {D^{\alpha } f\left( t \right)} \right|}$ in ${C^{ 1} \left[ {\text{0,1}} \right]}$)—of the natural spline interpolant ${\hat{f}_{n} }$. This evidently implies that the mapping from S into S defined by the natural spline interpolation operator with knots at ${N_{n} }$, applied element by element of ${Z = \left( {z_{ 1} ,\ldots ,z_{2r} } \right)}$, i.e., ${\hat{Z}_{n} = \left( Z \right)_{n} = \left( {\left( {z_{ 1} } \right)_{n} ,\ldots ,\left( {z_{{ 2 {\textit{r}}}} } \right)_{n} } \right)}$, is continuous with respect to the metric induced by the norm ${\left| {\left| Z \right|} \right| = \max_{j = 1, \ldots ,2r} \max_{0 \le \alpha \le 1} \max_{0 \le t \le 1} \left| {D^{\alpha } z_{j} \left( t \right)} \right|}$. (For brevity we use the same symbol, ${\left| {\left| .\right|} \right|}$, for the norms of $C^{1} \left[ {0,1} \right]$ and S; the proper interpretation will be clear in each case depending on the context.)

Since we have assumed that Z—the vector of time paths—is ${\mathbf{B}}\left( S \right)$-measurable, the approximant vector $\hat{Z}_{n}$, generated by natural spline interpolation, is also ${\mathbf{B}}\left( S \right)$-measurable, as it is obtained by a continuous (and so, measurable) transformation of Z.

Part (b) of the statement is obtained directly. For $n \ge 2$ the natural spline interpolant ${\hat{Z}_{n} }$ is unique and ${{\mathbf{B}}\left( S \right)}$-measurable, as established in part (a). Select an arbitrary point ${\omega \in \varOmega }$. By Assumption 1′.(i) each component of the observed path vector ${Z\left( { .,\omega } \right)}$ belongs to ${W^{ 2} \left[ {\text{0,1}} \right]}$, and Lemma A.1 establishes, given ${\omega \in \varOmega }$, uniform convergence with respect to t, i.e., ${\left| {\left| {\hat{Z}_{n} \left( { .,\omega } \right) - Z\left( { .,\omega } \right)} \right|} \right| \to 0}$ as ${n \to \infty }$. Since Assumption 1′.(ii) imposes that, for some ${m > 0}$ and each ${\omega \in \varOmega }$, it holds ${\min_{0 \le t \le 1} D^{1} z_{j} \left( {t,\omega } \right) \ge m}$, it is then obtained that, for each j and all large enough n (possibly depending on $\omega$), it holds (uniformly in t) ${D^{ 1} \hat{z}_{j} \left( {t,\omega } \right) = D^{1} z_{j} \left( {t,\omega } \right) + \left( {D^{ 1} \hat{z}_{j} \left( {t,\omega } \right) - D^{ 1} z_{j} \left( {t,\omega } \right)} \right) \ge m - m/ 2= m' > 0}$. This is a consequence of ${\min_{j = 1, \ldots ,2r} \min_{0 \le t \le 1} D^{1} z_{j} \left( {t,\omega } \right) \ge m}$, which is ensured by Assumption 1′.(ii) and the fact that ${max_{0 \le t \le 1} \left| {D^{ 1} \hat{z}_{j} \left( {t,\omega } \right) - D^{ 1} z_{j} \text{(}t,\omega \text{)}} \right| \to 0}$ as ${n \to \infty }$ by Lemma A.1. Therefore, for each ${\omega \in \varOmega }$ and n large enough, it holds $\min_{j = 1, \ldots ,2r} \min_{0 \le t \le 1} D^{1} \hat{z}_{j} \left( {t,\omega } \right) \ge m^{\prime} > 0$, that is, ${\hat{Z}_{n} \left( { .,\omega } \right) \in A_{m'} }$.

It is easily checked that, for any ${m' > 0}$, the mappings ${g\left( Z \right)}$ and ${h\left( Z \right)}$ defining, respectively, ${TE}$ and ${LTE}$, are continuous (and thus ${{\mathbf{B}}\left( S \right)}$-measurable) in $A_{m'}$, which in turn is a closed subset of S with nonempty interior, made up of all the vector functions Z in S with coordinates belonging to ${W^{ 2} \left[ {\text{0,\;1}} \right]}$ and having $\min_{j = 1, \ldots ,2r} \min_{0 \le t \le 1} D^{1} \hat{z}_{j} \left( {t,\omega } \right) \ge m^{\prime}$ for some m′ > 0 fixed a priori and not depending on $\omega$.

So, for each ${\omega \in \varOmega }$ and n large enough, the sample realization of the interpolant for Z also has first derivative that is (uniformly in [0,1]) bigger than some $m^{\prime} > 0$, i.e., $\hat{Z}_{n} \left( \omega \right) \in A_{{m^{\prime}}}$. Therefore, with probability 1, it holds ${\hat{Z}_{n} \in A_{m'} }$ as ${n \to \infty }$, and ${\hat{T}E_{n} = g\left( {\hat{Z}_{n} } \right)}$ and ${\hat{L}TE_{n} = h\left( {\hat{Z}_{n} } \right)}$ are continuous (and so ${{\mathbf{B}}\text{(}\Re \text{)}}$-measurable) functions of ${\hat{Z}_{n} }$ for each n large enough.

Once we have established that, for large enough n, ${\hat{T}E_{n} }$ and ${\hat{L}TE_{n} }$ are random variables, convergence with probability 1 to TE y LTE, respectively, is derived by the same procedure as in Propositions 2 y 3, applied to an arbitrary realization ${\omega \in \varOmega }$. For each ${\omega \in \varOmega }$, the integrals appearing in the definitions of TE and LTE are classical Riemann integrals, so the proofs of Lemmas A.1–A.3 and Propositions 2 and 3 readily extend (for each fixed ${\omega \in \varOmega }$) to the random case, with the only inconsequential issue that the Lipschitz bounds (${B_{ 1} , \ldots ,B_{6} }$) will generally range with each realization $\omega$ of the random experiment.