The additive hazard estimator is consistent for continuous-time marginal structural models

Abstract

Marginal structural models (MSMs) allow for causal analysis of longitudinal data. The standard MSM is based on discrete time models, but the continuous-time MSM is a conceptually appealing alternative for survival analysis. In applied analyses, it is often assumed that the theoretical treatment weights are known, but these weights are usually unknown and must be estimated from the data. Here we provide a sufficient condition for continuous-time MSM to be consistent even when the weights are estimated, and we show how additive hazard models can be used to estimate such weights. Our results suggest that continuous-time weights perform better than IPTW when the underlying process is continuous. Furthermore, we may wish to transform effect estimates of hazards to other scales that are easier to interpret causally. We show that a general transformation strategy can be used on weighted cumulative hazard estimates to obtain a range of other parameters in survival analysis, and explain how this strategy can be applied on data using our R packages ahw and transform.hazards.

Appendices

Appendix: proofs

We need some lemmas to prove Theorem 1.

Lemma 1

Suppose that $\{V^i\}_i$ are processes on [0, T] such that $\sup _ i E\big [ \sup _{s} | V^i_s | \big ] < \infty$, then

$$\begin{aligned} \lim _{a \rightarrow \infty } \sup _n P \left( \sup _s \big | \frac{1}{n} \sum _{i=1}^n V^{i}_{s} \big | \ge a \right) = 0. \end{aligned}$$

(25)

Proof

By Markov’s inequality, we have for every $a > 0$ that

$$\begin{aligned}&P \left( \sup _s \big | \frac{1}{n} \sum _{i=1}^n V^{i}_{s} \big | \ge a \right) \le \frac{1}{n a} \sum _{i=1}^n E_P \bigg [\sup _s \big | V^{i}_{s} \big |\bigg ], \end{aligned}$$

which proves the claim. $\square$

Lemma 2

(A perturbed law of large numbers) Suppose

• (I) $p^{-1} + q^{-1} = 1$, $p < \infty$,

• (II) $\{V_i\}_i \subset L^p(P)$, $\{ S_i\}_i \subset L^q(P)$ such that $\{ (V_{i},S_{i})\}_i$ is i.i.d., and $V_i$, $S_i$ are measurable with respect to a $\sigma$-algebra ${\mathcal {F}}_i$,

• (III) Triangular array $\{S_{(i,n)}\}_{n,i \le n }$ such that

  $$\begin{aligned} \lim \limits _{n\longrightarrow \infty } P \big ( |S_{(1,n)}- S_1| \ge \epsilon \big ) = 0 \end{aligned}$$

  (26)

  for every $\epsilon > 0$, and there exists a ${{\tilde{S}}} \in L^q(P)$ such that ${{\tilde{S}}} \ge |S_{(1,n)}|$ for every n,

• (IV) The conditional density of $S_{(i,n)}$ given ${\mathcal {F}}_i$ does not depend on i.

This implies that

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } E\bigg [ \bigg | \frac{1}{n} \sum _{i = 1} ^n S_{(i,n)} V_i -E_{ P} [S_1 V_1] \bigg | \bigg ] = 0. \end{aligned}$$

(27)

Proof

From the triangle inequality and condition (IV) we have that

$$\begin{aligned}&E \bigg [\bigg | \frac{1}{n} \sum _{i = 1}^n S_{(i,n)} V_i - \frac{1}{n} \sum _{i = 1}^n S_{i} V_i \bigg | \bigg ] \le \frac{1}{n} \sum _{i = 1}^n E \big [\big | \big ( S_{(i,n)} - S_{i} \big )V_i \big | \big ] \\&\quad = E \big [\big | \big ( S_{(1,n)} - S_{1} \big )V_1 \big |\big ]. \end{aligned}$$

The dominated convergence theorem implies that the last term converges to 0. Finally, the weak law of large numbers and the triangle inequality yields

$$\begin{aligned}&\lim \limits _{n\longrightarrow \infty } E\bigg [ \bigg | \frac{1}{n} \sum _{i = 1} ^n S_{(i,n)} V_i -E_{ P} [S_1 V_1] \bigg | \bigg ]\\&\quad \le \lim \limits _{n\longrightarrow \infty } E \bigg [\bigg | \frac{1}{n} \sum _{i = 1}^n S_{(i,n)} V_i - \frac{1}{n} \sum _{i = 1}^n S_{i} V_i \bigg | \bigg ]+ E \bigg [\bigg | \frac{1}{n} \sum _{i = 1}^n S_{i} V_i - E[S_{1} V_1] \bigg | \bigg ] = 0. \end{aligned}$$

$\square$

Lemma 3

$\{V_i\}_i$ i.i.d. non-negative variables in $L^2(P)$, then

$$\begin{aligned} \lim _{n \rightarrow \infty } P \bigg ( \frac{1}{n} \max _{ i \le n} V_i \ge \epsilon \bigg ) = 0 \end{aligned}$$

(28)

for every $\epsilon > 0$.

Proof

Note that

$$\begin{aligned} P \bigg ( \frac{1}{n} \max _{ i \le n} V_i> \epsilon \bigg )&= 1- P \bigg ( \max _{ i \le n} V_i \le \epsilon n \bigg ) = 1- P \bigg ( V_1\le \epsilon n \bigg )^n \\&= 1- \bigg (1 - P \big ( V_1 > \epsilon n \big )\bigg )^n \end{aligned}$$

If $n > \Vert V_1 \Vert _2\epsilon ^{-1}$, we therefore have by Chebyshev’s inequality that

$$\begin{aligned} P \bigg ( \frac{1}{n} \max _{ i \le n} V_i > \epsilon \bigg ) \le 1- \bigg (1 - \frac{E[V_1^2]}{n^2 \epsilon ^2} \bigg )^n, \end{aligned}$$

where the last term converges to 0 when $n \rightarrow \infty$ since $\lim \nolimits _{n\longrightarrow \infty } n \log \big ( 1 - \frac{E[V_1^2]}{n^2 \epsilon ^2} \big ) = 0$ for every $\epsilon > 0$. $\square$

Lemma 4

Define $\gamma _s^i := Y_s^{i,D} \pmb X^i_s \pmb b_s$, where $\pmb X^i_{s}$ is the i’th row of $\pmb X^{(n)}_{s}$. If the assumptions of Theorem 1 are satisfied, then

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } P \bigg ( \sup _t \bigg | \int _0^t \pmb \varGamma ^{(n)-1} \frac{1}{n} \sum _{i=1}^n R^{(i,n)}_{s-}\pmb X_{s-}^{i \intercal } ( \lambda ^{i,D}_s - \gamma ^i_s ) ds \bigg | \ge \delta \bigg )= 0 \end{aligned}$$

(29)

for every $\delta > 0$.

Proof

Assumption (III) from Theorem 1 and Lemma 1 implies that

$$\begin{aligned} \lim _{ J \rightarrow \infty } \inf _n P\bigg ( \sup _t \big | \varGamma _{t}^{(n)-1} \frac{1}{n} \sum _{i=1}^n R^{(i,n)}_{t-}\pmb X_{t-}^{i \intercal } ( \lambda ^{i,D}_t - \gamma ^i_t )\big | > J \bigg ) = 0. \end{aligned}$$

(30)

Moreover, Lemma 2 implies that

$$\begin{aligned} \frac{1}{n} \sum _{i=1}^n R^{(i,n)}_{t-}\pmb X_{t-}^{i \intercal } ( \lambda ^{i,D}_t - \gamma ^i_t ) \end{aligned}$$

converges in probability to

$$\begin{aligned} E_P \big [R^1_{t-} \pmb X_{t-} ^{1 \intercal } ( \lambda ^{1,D}_t - \gamma _t^1 ) \big ] \end{aligned}$$

However, from the innovation theorem we have that this equals

$$\begin{aligned} E_{{{\tilde{P}}}} \big [\pmb X_{t-} ^{1 \intercal } ( \lambda ^{1,D}_t - \gamma _t^1 )\big ] = E_{{{\tilde{P}}}} \big [\pmb X_{t-} ^{1 \intercal } (E_{ {{\tilde{P}}}} [ \lambda ^{1,D}_t |{\mathcal {F}}_{t-}^{1 ,{\mathcal {V}}_0} ] - \gamma _t^1 )\big ] = 0, \end{aligned}$$

since $\pmb X_{t-}^1$ and $\gamma _t^1$ are ${\mathcal {F}}_{t-}^{1,{\mathcal {V}}_0}$ measurable. This and (30) enables us to apply Andersen et al. (1993, Lemma II.5.3) to obtain (29). $\square$

Lemma 5

Suppose that (II) and (III) from Theorem 1 are satisfied and let $\pmb M^{(n)}_t := \begin{pmatrix}N_t^{1,D} - \int _0^t \lambda _s^{1,D} ds , \dots , N_t^{n,D} - \int _0^t \lambda _s^{n,D} ds \end{pmatrix}^\intercal$. Then

$$\begin{aligned} \pmb \varXi ^{(n)}_t := \frac{1}{n} \int _0^t \pmb \varGamma ^{(n)-1}_s \pmb X_{s-}^{(n)\intercal } \pmb Y^{(n),D}_{s} d \pmb M_s^{(n)} \end{aligned}$$

(31)

defines a square integrable local martingale with respect to the filtration ${\mathcal {F}}_{s}^{1,{\mathcal {V}}_0 \cup {\mathcal {L}}} \otimes \dots \otimes \mathcal F_{s}^{n,{\mathcal {V}}_0 \cup {\mathcal {L}}}$ and

$$\begin{aligned} \lim _{n \rightarrow \infty } P\bigg ( {{\,\mathrm{Tr}\,}}( \langle \pmb \varXi ^{(n)}\rangle _T) \ge \delta \bigg ) = 0 \end{aligned}$$

(32)

for every $\delta >0$.

Proof

Writing $\pmb \lambda ^{(n)}$ for the diagonal matrix with i’th diagonal element equal to $\lambda ^{i,D}$, we have that

$$\begin{aligned} {{\,\mathrm{Tr}\,}}(\langle \pmb \varXi ^{(n)} \rangle _T) =&\int _0^T \frac{1}{n^2} {{\,\mathrm{Tr}\,}}\bigg (\pmb \varGamma ^{(n)-1}_s \pmb X_{s-}^{(n)\intercal } \pmb Y^{(n),D}_{s} \pmb \lambda _s^{(n)} \pmb Y^{(n),D}_{s} \pmb X_{s-}^{(n)} \pmb \varGamma ^{(n)-1}_s\bigg ) ds. \end{aligned}$$

(33)

Moreover,

$$\begin{aligned}&\frac{1}{n^2} {{\,\mathrm{Tr}\,}}\bigg (\pmb \varGamma ^{(n)-1}_s \pmb X_{s-}^{(n)\intercal } \pmb Y^{(n),D}_{s} \pmb \lambda _s^{(n)} \pmb Y^{(n),D}_{s} \pmb X_{s-}^{(n)} \pmb \varGamma ^{(n)-1}_s\bigg ) \end{aligned}$$

(34)

$$\begin{aligned}&\quad \le \frac{1}{n^2} {{\,\mathrm{Tr}\,}}\bigg (\pmb \varGamma ^{(n)-1}_s \pmb X_{s-}^{(n)\intercal } \pmb Y^{(n),D}_{s} \pmb X_{s-}^{(n)} \pmb \varGamma ^{(n)-1}_s\bigg ) \max _{i \le n} Y_s^{i,D} R^{(i,n)}_{s-} \lambda _s^{i,D} \end{aligned}$$

(35)

$$\begin{aligned}&\quad \le {{\,\mathrm{Tr}\,}}\bigg (\pmb \varGamma ^{(n)-1}_s \bigg ) \big (\frac{1}{n} \max _{i \le n} \lambda _s^{i,D} \big ) \Vert R^{(i,n)}\Vert _\infty \end{aligned}$$

(36)

$$\begin{aligned}&\quad \le {{\,\mathrm{Tr}\,}}\bigg (\pmb \varGamma ^{(n)-1}_s \bigg ) \big (\frac{1}{n} \sum _{i \le n} \lambda _s^{i,D} \big ) \Vert R^{(i,n)}\Vert _\infty \end{aligned}$$

(37)

Now, (III), (37) and Lemma 1 implies that

$$\begin{aligned} \lim _{a \rightarrow \infty } \inf _n P \bigg ( \sup _s \frac{1}{n^2} {{\,\mathrm{Tr}\,}}\bigg (\pmb \varGamma ^{(n)-1}_s \pmb X_{s-}^{(n)\intercal } \pmb Y^{(n),D}_{s} \pmb \lambda _s^{(n)} \pmb Y^{(n),D}_{s} \pmb X_{s-}^{(n)} \pmb \varGamma ^{(n)-1}_s \bigg ) \ge a \bigg ) = 0. \end{aligned}$$

On the other hand, Lemma 3, (36) and (III) gives us that

$$\begin{aligned} \lim _{n\rightarrow \infty } P \bigg ( \frac{1}{n^2} {{\,\mathrm{Tr}\,}}\bigg (\pmb \varGamma ^{(n)-1}_s \pmb X_{s-}^{(n)\intercal } \pmb Y^{(n),D}_{s} \pmb \lambda _s^{(n)} \pmb Y^{(n),D}_{s} \pmb X_{s-}^{(n)} \pmb \varGamma ^{(n)-1}_s \bigg ) \ge \delta \bigg ) = 0 \end{aligned}$$

for every s and $\delta >0$, so Andersen et al. (1993, Propositon II.5.3) implies that (31) also holds. $\square$

Proof of Theorem 1

We have the following decomposition:

$$\begin{aligned} \pmb B^{(n)}_t -\pmb B_t =&\int _0^t (\pmb X_{s-}^{(n)\intercal } \pmb Y^{(n),D}_{s}\pmb X_{s-}^{(n)})^{-1} \big ( \pmb X_{s-}^{(n)\intercal } \pmb Y^{(n),D}_{s} \pmb \lambda ^{(n)}_s - \pmb X_{s-}^{(n)\intercal }\pmb Y^{(n),D}_{s} \pmb X_{s-}^{(n)} b_s \big ) ds \\&+ \int _0^t (\pmb X_{s-}^{(n)\intercal } \pmb Y^{(n),D}_{s} \pmb X_{s-}^{(n)})^{-1} \pmb X_{s-}^{(n)\intercal } \pmb Y^{(n),D}_{s} d \pmb M_s^{(n)} \\ =&\int _0^t \pmb \varGamma ^{(n)-1} \frac{1}{n} \sum _{i=1}^n R^{(i,n)}_{s-}\pmb X_{s-}^{i \intercal } ( \lambda ^{i,D}_s - \gamma ^i_s ) ds + \pmb \varXi _t^{(n)}. \end{aligned}$$

Leglarts inequality (Jacod and Shiryaev 2003, Lemma I.3.30) together with Lemma 5 implies that $\pmb \varXi ^{(n)}$ converges uniformly in probability to 0. Moreover, Lemma 4 implies that $\int _0^\cdot \pmb \varGamma ^{(n)-1} \frac{1}{n} \sum _{i=1}^n R^{(i,n)}_{s-}\pmb X_{s-}^{i \intercal } (\lambda ^{i,D}_s - \gamma ^i_s)ds$ converges in same sense to 0, which proves the consistency.

To see that $\pmb B^{(n)}$ is P-UT, note that it coincides with the sum of $\pmb B_t$, $\pmb \varXi ^{(n)}$ and $\int _0^\cdot \pmb \varGamma _s^{(n)-1} \frac{1}{n} \sum _{i=1}^n R^{(i,n)}_{s-}\pmb X_{s-}^{i \intercal } ( \lambda ^i_s - \gamma ^i_s)ds$. According to Ryalen et al. (2018b, Lemma 1), the latter is P-UT since (III) and Lemma 1 implies (7). Moreover, $\pmb B_t = \int _0^\cdot \pmb b_s ds$ is clearly P-UT, since $\pmb b_t$ is uniformly bounded. $\pmb \varXi ^{(n)}$ is also P-UT since Lemma 5 implies that (8) is satisfied. Finally, as $\pmb B^{(n)}$ is a sum of three processes that are P-UT, it is necessarily P-UT itself. $\square$

Proof of Theorem 2

Lemma 6

Suppose that c. and d. from Theorem 2 are satisfied, and that

• (I)
  $$\begin{aligned} \lim _{a \rightarrow \infty } \sup _n P \bigg ( \sup _t \big | \theta ^{(i,n)}_t \big | \ge a \bigg ) = 0, \end{aligned}$$

• (II) $\theta _{t-}^{(i,n)}$ converges to $\theta ^i_t$ in probability for each i and t.

Then we have that $K^{(i,n)}$ is predictably uniformly tight (P-UT) and

$$\begin{aligned} \lim _{n} P\bigg ( \sup _t \big | K_t^{(i,n)} - K_t^i \big | \ge \epsilon \bigg ) = 0 \end{aligned}$$

(38)

for every i and $\epsilon > 0$.

Proof

Note that

$$\begin{aligned} K_t^{(i,n)} - K_t^i= & {} \int _0^t ( \theta ^{(i,n)}_{s-} - \theta _s^i ) dN_s^{i,A} + n^{-1/2} \int _0^t Y_s ^i \pmb Z_{s-}^{i\intercal } d\pmb W^{(n)}_s \nonumber \\&- n^{-1/2} \int _0^t Y_s^{i,A} \pmb {{{\tilde{Z}}}}_{s-}^{i\intercal } d \pmb {{{\tilde{W}}}}^{(n)}_s, \end{aligned}$$

(39)

where $\pmb W_t^{(n)} := n^{1/2}(\pmb H^{(n)}_t - \pmb H_t)$ and $\pmb {{{\tilde{W}}}}_t^{(n)} :=n^{1/2}( \pmb {{{\tilde{H}}}}^{(n)}_t - \pmb {{{\tilde{H}}}}_t)$ are square-integrable martingales with respect to ${\mathcal {F}}_{t}^{1,{\mathcal {V}}_0 \cup {\mathcal {L}} } \otimes \cdots \otimes {\mathcal {F}}_{t}^{n,{\mathcal {V}}_0 \cup {\mathcal {L}} }$ and ${\mathcal {F}}_{t}^{1,{\mathcal {V}}_0 } \otimes \cdots \otimes {\mathcal {F}}_{t}^{n,{\mathcal {V}}_0 }$ respectively.

Let $\tau$ be an optional stopping time and note that

$$\begin{aligned}&E\bigg [ \bigg | \int _0^\tau (\theta ^{(i,n)}_{s-} - \theta ^i_s) dN_s^{i,A} \bigg | \bigg ] \le E\bigg [ \int _0^\tau \big |\theta ^{(i,n)}_{s-} - \theta ^i_s \big | dN_s^{i,A} \bigg ] \\&\quad = E\bigg [ \int _0^\tau \big |\theta ^{(i,n)}_{s-} - \theta ^i_s \big | \lambda _s^{i,A} ds \bigg ], \end{aligned}$$

so by Lenglarts inequality, (Jacod and Shiryaev 2003, I.3.30), we see that

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } P \bigg ( \sup _{t \le T} \bigg | \int _0^t (\theta ^{(i,n)}_{s-} - \theta ^i_s) dN_s^{i,A} \bigg | \ge \epsilon \bigg ) = 0 \end{aligned}$$

(40)

for every $\epsilon > 0$ if

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } P \bigg ( \int _0^T \big | \theta ^{(i,n)}_{s-} - \theta ^i_s \big | \lambda _s^{i,A} ds \ge \epsilon \bigg ) = 0, \end{aligned}$$

(41)

for every $\epsilon > 0$. The latter property holds due to (I), (II) and Andersen et al. (1993, Proposition II.5.3).

Since $\{ \int _0^t Y_s^{i,A} \pmb Z_{s-}^{i\intercal } d\pmb W^{(n)}_s \}_n$ converges in the skorokhod topology, we have that $\{ \sup _{t \le T}|\int _0^t Y_s^{i,A} \pmb Z_{s-}^{i\intercal } d\pmb W^{(n)}_s | \}_n$ is tight (Jacod and Shiryaev 2003, Theorem VI.3.21). Therefore, we also get that

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } P\bigg ( \sup _{t \le T} |n ^{-1/2} \int _0^t Y_s^{i,A} \pmb Z_{s-}^{i\intercal } d\pmb W^{(n)}_s |\ge \epsilon \bigg ) = 0 \end{aligned}$$

(42)

for every $\epsilon > 0$. For the same reason we also have

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } P\bigg ( \sup _{t \le T} | n ^{-1/2} \int _0^t Y_s^{i,A} \pmb {{{\tilde{Z}}}}_{s-}^{i \intercal } d\pmb {{{\tilde{W}}}}^{(n)}_s |\ge \epsilon \bigg ) = 0. \end{aligned}$$

(43)

By combining (42), (43) and (40), we obtain that

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } P\bigg ( \sup _{t \le T} | K_t ^{(i,n)} - K_t^i |\ge \epsilon \bigg ) = 0 \end{aligned}$$

(44)

for every $\epsilon > 0$.

To see that $K^{(i,n)}$ is P-UT, note that the compensator of $\int _0^\cdot (\theta ^{(i,n)}_{s-} - 1) dN_s^{i,A}$ equals $\int _0^\cdot (\theta ^{(i,n)}_{s-} - 1) \lambda _s^{i,A} ds$ and

$$\begin{aligned} \left\langle \int _0^\cdot (\theta ^{(i,n)}_{s-} - 1) dN_s^{i,A} - \int _0^\cdot (\theta ^{(i,n)}_{s-} - 1) \lambda _s^{i,A} ds \right\rangle _T = \int _0^T (\theta ^{(i,n)}_{s-} - 1)^2 \lambda _s^{i,A} ds. \end{aligned}$$

The assumptions (I) in this Lemma and c) together with Ryalen et al. (2018b, Lemma 1) therefore imply that $\int _0^\cdot (\theta ^{(i,n)}_{s-} - 1) dN_s^{i,A}$ is P-UT.

To see that $\int _0^\cdot Y_s ^i \pmb {{{\tilde{Z}}}}_{s-}^{i\intercal } d \pmb {{{\tilde{H}}}}^{(n)}_s$ is P-UT, note that

$$\begin{aligned} \int _0^\cdot Y_s ^i \pmb {{{\tilde{Z}}}}_{s-}^{i\intercal } d \pmb {{{\tilde{H}}}}^{(n)}_s = n^{-1/2} \int _0^\cdot Y_s ^i \pmb {{{\tilde{Z}}}}_{s-}^{i\intercal } d \pmb {{{\tilde{W}}}}^{(n)}_s + \int _0^\cdot Y_s ^i \pmb {{{\tilde{Z}}}}_{s-}^{i\intercal } d \pmb {{{\tilde{H}}}}_s. \end{aligned}$$

(45)

An analogous decompositon yields that $\int _0^\cdot Y_s ^i \pmb Z_{s-}^{i\intercal } d \pmb H^{(n)}_s$ is P-UT. This means that $K^{(i,n)}$ is a sum of three processes that are P-UT, and must therefore be P-UT itself. $\square$

Lemma 7

Suppose that

• (I) $\{\kappa _n\}_n$ increasing sequence of positive numbers such that

  $$\begin{aligned} \lim \limits _{n\longrightarrow \infty } \kappa _n = \infty \text { and } \sup _n \frac{\kappa _n }{\sqrt{n}} < \infty , \end{aligned}$$

• (II) $\pmb h_t$ is a bounded and continuous vector valued function,

• (III) $\pmb Z^i$ is caglad with $E[\sup _{t\le T} |\pmb Z_t^i|^3_3 ] < \infty$,

• (IV)
  $$\begin{aligned} \lim _{J \rightarrow \infty } \sup _n P \bigg ( {{\,\mathrm{Tr}\,}}\Big ( \big ( \frac{1}{n} \pmb Z^{(n)\intercal }_{t-} \pmb Y^{(n),A}_t \pmb Z^{(n)}_{t-})^{-1} \Big ) \ge J \bigg ) = 0 \end{aligned}$$

  (46)

• (V) $Y^{i,A} \pmb Z_{\cdot -}^{i\intercal } \pmb h$ defines the intensity for $N^{i,A}$ with respect to P and ${\mathcal {F}}_\cdot ^{i,{\mathcal {V}}_0}$. Now,

  $$\begin{aligned} \lim \limits _{n\longrightarrow \infty } P \bigg ( \sup _{ 1/\kappa _n \le t \le T} \Big | \kappa _n \int _{t - 1/\kappa _n } ^t Y_s^{i,A} \pmb Z_{s-}^{i\intercal } d\pmb H_s^{(n)} - Y_t^{i,A} \pmb Z_{t-}^{i \intercal } \pmb h_t \Big | \ge \epsilon \bigg ) = 0.\qquad \end{aligned}$$

  (47)

Proof

Note that

$$\begin{aligned}&\kappa _n \int _{t-1/\kappa _n}^t Y_s^{i,A} \pmb Z_{s-}^{i \intercal } d\pmb H^{(n)}_s - Y_t^{i,A} \pmb Z_{t-}^{i \intercal } \pmb h_t \end{aligned}$$

(48)

$$\begin{aligned}&\quad = \frac{\kappa _n }{ \sqrt{n} } \int _0^t Y_s^{i,A} \pmb Z_{s-}^{i\intercal } d\pmb W^{(n)}_s - \frac{\kappa _n }{ \sqrt{n} } \int _0^{t - 1/\kappa _n } Y_s^{i,A} \pmb Z_{s-}^{i\intercal } d\pmb W^{(n)}_s \end{aligned}$$

(49)

$$\begin{aligned}&\quad \quad + \kappa _n \int _{t-1/\kappa _n }^t Y_s^{i,A} \pmb Z_{s-}^{i \intercal } \pmb h_s ds - Y_t^{i,A} \pmb Z_{t-}^{i\intercal } \pmb h_t. \end{aligned}$$

(50)

The martingale central limit theorem implies that $\{\pmb W^{(n)} \}$ is a sequence of martingales that converges in law to a continuous Gaussian processes with independent increments, see Andersen et al. (1993). Moreover, Ryalen et al. (2018b, Proposition 1) says that $\{\pmb W^{(n)}\}_n$ is P-UT.

Therefore Jacod and Shiryaev (2003, Theorem VI 6.22) implies that $\int _0 ^\cdot Y_s^{i,A} \pmb Z_{s-}^{i\intercal } d \pmb W^{(n)}_s$ converges in law to a continuous process, so it is C-tight. Moreover, from Jacod and Shiryaev (2003, Proposition VI.3.26) we have that

$$\begin{aligned} \lim _{n \longrightarrow \infty } P \bigg ( \sup _{1/\kappa _n \le t \le T} \Big | \int _0 ^t Y_s^{i,A} \pmb Z_{s-}^{i\intercal } d \pmb W^{(n)}_s - \int _0 ^{t - 1/\kappa _n } Y_s^{i,A} \pmb Z_{s-}^{i\intercal } d \pmb W^{(n)}_s \Big | \ge \epsilon \bigg ) = 0\nonumber \\ \end{aligned}$$

(51)

for every $\epsilon > 0$. The mean value theorem of elementary calculus implies that

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } \sup _{1/\kappa _n \le t \le T} \Big | \kappa _n \int _{ t - 1/\kappa _n } ^t Y_s^{i,A} \pmb Z_{s-}^{i\intercal } \pmb h_s ds - Y_t^{i,A} \pmb Z_{t-}^{i\intercal } \pmb h_t \Big | = 0 \end{aligned}$$

(52)

P a.s. Combining (51) and (52) yields the claim. $\square$

Proof of Theorem 2

Combining (16) and the decomposition in the proof of Lemma 7, we see that

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } P \bigg ( \sup _{ 1/ \kappa _n \le t \le T} \bigg | \kappa _n \int _{t- 1/\kappa _n} ^t Y_s^{i,A} \pmb {{{\tilde{Z}}}}_{s-}^{i\intercal } d \pmb {{{\tilde{H}}}}^{(n)}_s /{{\tilde{\lambda }}}_t^{i,A} - 1 \bigg | \ge \epsilon \bigg ) = 0. \end{aligned}$$

(53)

Combining (16) and a. we also have

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } P \bigg ( \sup _{ 1/ \kappa _n \le t \le T} \bigg | \kappa _n \int _{t- 1/\kappa _n} ^t Y_s^{i,A} \pmb Z_{s-}^{i\intercal } d \pmb H^{(n)}_s / \lambda _t^{i,A} - 1 \bigg | \ge \epsilon \bigg ) = 0. \end{aligned}$$

(54)

Whenever $t \ge 1/ \kappa _n$, we have that by the continuous mapping theorem that

$$\begin{aligned}&\lim \limits _{n\longrightarrow \infty } P \bigg ( \sup _{1/\kappa _n \le t \le T } \big |\theta ^{(i,n)}_t- \theta _t^i \big | \ge \epsilon \bigg ) \\&\quad = \lim \limits _{n\longrightarrow \infty } P \left( \sup _{1/\kappa _n \le t \le T } \big | \theta _t^i \left( \frac{ \kappa _n \int _{t- 1/\kappa _n} ^t Y_s^{i,A} \pmb {{{\tilde{Z}}}}_{s-}^{i\intercal } d \pmb {{{\tilde{H}}}}^{(n)}_s /{{\tilde{\lambda }}}_t^{i,A} }{ \kappa _n \int _{t- 1/\kappa _n} ^t Y_s^{i,A} \pmb Z_{s-}^{i\intercal } d \pmb H^{(n)}_s / \lambda _t^{i,A} } - 1\right) \big | \ge \epsilon \right) \\&\quad = 0. \end{aligned}$$

Since $\theta ^i$ is right-continuous at $t= 0$, we have that

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty } P \bigg ( \sup _{0 \le t \le T } \big |\theta ^{(i,n)}_t- \theta _t^i \big | \ge \epsilon \bigg ) = 0. \end{aligned}$$

(55)

Finally, Jacod and Shiryaev (2003, Corollary VI 3.33) implies that $\{( R^{(i,n)}_0,K^{(i,n)})\}_n$ converges to $(R_0^i, K^i)$ in probability. Since $K^{(i,n)}$ is P-UT,

$$\begin{aligned} R_t^{(i,n)} = 1 + \int _0^t R_{s-}^{(i,n)} dK_{s}^{(i,n)} \end{aligned}$$

and

$$\begin{aligned} R_t^{i} = 1 + \int _0^t R_{s-}^{i} dK_{s}^{i} \end{aligned}$$

Jacod and Shiryaev (2003, Theorem IX 6.9) implies that $R^{(i,n)}$ converges to $R^{i}$ in probability. $\square$