2.1. Existence of guided proposals

The guiding term of $X^\circ$ involves $\tilde{r}\colon [0,\infty) \times {\mathbb{R}}^d \to {\mathbb{R}}$ . In the uniformly elliptic case it is easily verified that this mapping is well-defined. This need not be the case in the hypo-elliptic setting.

Let $\Phi(t)$ denote the fundamental matrix solution of the ODE ${\mathrm{d}} \Phi(t) = \tilde{B}(t) \Phi(t) \,{\mathrm{d}} t$ , $\Phi(0)=I$ and set $\Phi(t,s)=\Phi(t)\Phi(s)^{-1}$ . Define

\begin{equation*} L(t) \coloneqq L \Phi(T,t).\end{equation*}

Assumption 2.3. The matrix

$$ \int_t^T \Phi(T,s) \tilde{a}(s) \Phi(T,s)^{\prime} \,{\mathrm{d}} s $$

is strictly positive definite for $t<T$ .

In the uniformly elliptic setting, this assumption is always satisfied. Under this assumption, the matrix

$$ {M^\dagger}(t) \coloneqq \int_t^T L(s) \tilde{a}(s) L(s)^{\prime} \,{\mathrm{d}} s$$

is also strictly positive definite for all $t\in [0,T)$ and, in particular, invertible.

Lemma 2.1. If Assumption 2.3 holds, then

(2.2) \begin{equation}\tilde{r}(t,x)=L(t)^{\prime} M(t) ( v -\mu(t)-L(t)x), \quad t\in [0,T],\end{equation}

where

$$\mu(t)=\int_t^T L(s) \tilde\beta(s) \,{\mathrm{d}} s$$

and

\begin{equation*} M(t) = [{M^\dagger}(t)]^{-1}.\end{equation*}

Proof.The solution to the SDE for $\tilde{X}_u$ is given by

$$\tilde{X}_u = \Phi(u,t) x + \int_t^u \Phi(u,s) \tilde\beta(s) \,{\mathrm{d}} s + \int_t^u \Phi(u,s) \tilde{\sigma}(s) \,{\mathrm{d}} W_s, \quad u\ge t ,\quad \tilde{X}_t =x.$$

See [23, Theorem 4.10]. The result now follows directly upon taking $u=T$ , multiplying both sides by L, and using the definition of L(t). □

In Appendix A easily verifiable conditions for the existence of $\tilde{p}$ are given for the case $L=I$ .

Since $t\mapsto \mu(t)$ and $t\mapsto M(t)$ are continuous and $x\mapsto \tilde{r}(t,x)$ is linear in x for fixed t, the process $X^\circ$ is well-defined on intervals bounded away from T.

Throughout, without explicitly stating it in lemmas and theorems, we will assume that Assumptions 2.1, 2.2, and 2.3hold true.

2.2. Behaviour of guided proposals near the endpoint

Let $\Delta(t)$ be an invertible $m\times m$ diagonal matrix-valued measurable function on [0,T). Define

(2.3) \begin{equation}{Z_{\Delta,t}}=\Delta(t) ( v- \mu(t)-L(t) X^\circ_t )\end{equation}

and

(2.4) \begin{equation}{L_{\Delta}}(t)=\Delta(t) L(t) {,} \quad {M_{\Delta}}(t)=\Delta(t)^{-1} M(t) \Delta(t)^{-1}.\end{equation}

Note that for $t\approx T$ we have $\Phi(T,t)\approx I$ and hence ${Z_{\Delta,t}} \approx \Delta(t)(v-L X^\circ_t)$ . The matrix $\Delta(t)$ is a scaling matrix which in the hypo-elliptic case incorporates the difference in rate of convergence for smooth and rough components of $L X^\circ_t$ to v, when $t\uparrow T$ . In the uniformly elliptic case, we can always take $\Delta(t)=I_{m}$ .

The following assumption is of key importance.

Assumption 2.4. There exists an invertible $m\times m$ diagonal matrix-valued function $\Delta(t)$ , which is measurable on [0,T), a $t_0<T$ , $\alpha \in (0,1]$ and positive constants $\underline{c}$ , $\overline{c}$ , $c_1$ , $c_2$ , and $c_3$ such that, for all $t \in [t_0,T)$ ,

(2.5) \begin{align}\underline{c}\, (T-t)^{-1}\le {\lambda_{\rm min}}({M_{\Delta}}(t)) &\le {\lambda_{\rm max}}({M_{\Delta}}(t)) \le \overline{c}\, (T-t)^{-1}, \notag\\[3pt]\| {L_{\Delta}}(t) (\tilde{b}(t,x) - b(t,x)) \| &\le c_1 {,} \notag \\[3pt] \mathrm{tr}({L_{\Delta}}(t)\, a(t,x)\, {L_{\Delta}}(t)^{\prime} )&\le c_2 {,} \notag\\[3pt] \|{L_{\Delta}}(t) (\tilde{a}(t)-a(t,x)) {L_{\Delta}}(t)^{\prime}\| &\le c_3 (T-t)^\alpha.\end{align}

Proposition 2.2. Under Assumption 2.4, there exists a positive number C such that

$$\limsup_{t\uparrow T} \frac{\|{Z_{\Delta,t}}\|}{\sqrt{(T-t) \log(1/(T-t))}} \le C \quad \text{a.s.}$$

The uniformly elliptic case is particularly simple.

The proofs of Theorem 2.2 and Corollary 2.1 are given in Section 5.

In Section 3 we give a set of tractable hypo-elliptic models for which the conclusion of Theorem 2.2 is valid. The appropriate choice of the scaling matrix $\Delta(t)$ is really problem-specific. Moreover, the assumptions on the auxiliary process depend on the choice of L.

In most cases it will not be possible to satisfy the fourth inequality of Assumption 2.4 when the diffusion coefficient is state-dependent. The following example shows an exception.

Example 2.1. Suppose the diffusion is uniformly elliptic and $L=[ \underline{L}\ \ 0_{m\times k'}]$ , where $L \in {\mathbb{R}}^{m\times k}$ and $d=k+k'$ . Now suppose a(t, x) is of block form

$$a(t,x) =\begin{bmatrix} a_{11}(t) & 0_{k\times k'} \\ 0_{k'\times k} & \quad a_{22}(t,x) \end{bmatrix}{,}$$

and that we take $\tilde{a}$ to be of the same block form. Upon taking $\tilde{B} =0_{d\times d}$ and $\Delta(t)=I_d$ , we see that ${L_{\Delta}}(t)=L$ and hence

$${L_{\Delta}}(t) (\tilde{a}(t)-a(t,x)) {L_{\Delta}}(t)^{\prime} = \underline{L} (\tilde{a}_{11}(t)-a_{11}(t)) \underline{L}^{\prime}.$$

Therefore, if we choose $\tilde{a}_{11}(t)$ to be equal to $a_{11}(t)$ the fourth inequality in Assumption 2.4 is trivially satisfied.

Empirically, however, it appears that Assumption 2.4 is stronger than needed for valid guided proposals; see Example 4.4.

2.3. Absolute continuity

The following theorem gives sufficient conditions for absolute continuity of ${\mathbb{P}}_T^\star$ with respect to ${\mathbb{P}}_T^\circ$ . First we introduce an assumption.

Assumption 2.5. There exists a constant C such that

\begin{equation*} p(s,x;\ t,y) \le C \tilde{p}(s,x;\ t,y){,} \quad 0\le s<t<T\end{equation*}

for all $x, y \in {\mathbb{R}}^d$ .

Theorem 2.6. Assume there exists a positive $\delta$ such that $|\Delta(t)| \lesssim (T-t)^{-\delta}$ . If Assumptions 2.4 and 2.5 hold true, then

(2.6) $$\frac{{\mathrm{d}} {\mathbb{P}}_T^\star}{{\mathrm{d}} {\mathbb{P}}_T^\circ}(X^\circ) = \frac{\tilde \rho(0,x_0)}{ \rho(0,x_0)}\Psi_T(X^\circ),$$

where

\begin{equation}\Psi_t(X^\circ)=\exp\bigg(\int_0^t \mathcal{G}(s,X^\circ_s) \,{\mathrm{d}} s\bigg),\end{equation}

\begin{equation*} \mathcal{G}(s,x) = (b(s,x) - \tilde b(s,x))^{\prime} \tilde r(s,x) - \frac12 \mathrm{tr}([a(s,x) - \tilde a(s)] [\tilde H(s)-\tilde{r}(s,x)\tilde{r}(s,x)^{\prime}]){,}\end{equation*}

and $\tilde{H}(s)=L(s)^{\prime} M(s) L(s)$ .

The proof is given in Section 6.

The following lemma is useful for verifying Assumption 2.5. Its proof is located in Section 6.

Lemma 2.3. Assume $\eta(s,x)$ satisfies the equation

$${\sigma}(s,x) \eta(s,x) = b(s,x)-\tilde{b}(s,x)$$

and that $\eta$ is bounded. Then there exists a constant C such that

$$p(s,x;\ t,y) \le C \tilde{p}(s,x;\ t,y)$$

for all $x, y \in {\mathbb{R}}^d$ and $0\le s<t\le T $ .

4.1. Numerical checks on the validity of guided proposals

In this section we first investigate the quality of guided proposals over long time spans. Next, we empirically demonstrate that the conditions of our main theorem, especially Assumption 2.4, are stronger than actually needed. In each numerical experiment we compare two histogram estimators for $v \mapsto \rho(0,x_0;\ T,v)$ . The first estimator is obtained by making a histogram of a large number of forward simulations of the unconditioned diffusion process. Let $\{A_k\}$ denote the bins of this histogram. A second estimator is obtained by using the equality

$$\rho(0,x_0;\ T, v) = \tilde \rho(0, x_0;\ T, v) {{\mathbb{E}}}[{\Psi_T(X^{\circ, T, v}) }] {,}$$

which is a direct consequence of Theorem 2.6. Note that we extended the notation to highlight that $\rho$ , $\tilde \rho$ , and $X^\circ$ depend on T and v. We use the relation in the previous display as follows: for each bin $A_k$ ,

$$\int_{A_k} \rho(0,x_0;\ T, v) \,{\mathrm{d}} v ={{\mathbb{E}}}\bigg[{{\textbf{1}}_{A_k}(\tilde V) \frac{\tilde \rho(0, x_0;\ T, \tilde V)}{q( \tilde V)} \Psi_T(X^{T, \tilde V,\circ}) }\bigg]{,}$$

where $\tilde V$ is sampled from the density q. Hence, $\int_{A_k} \rho(0,x_0;\ T, v) \,{\mathrm{d}} v$ can be approximated using importance sampling, where repeatedly first the endpoint v is sampled from q and subsequently a guided proposal is simulated that is conditioned to hit v at time T. In our experiments we took the importance sampling density q to be the Gaussian density with mean and covariance obtained from the unconditioned forward-simulated endpoint values.

Note that the set-up is such that this is feasible, at least when estimating the entire histogram, but of course it would be prohibitively expensive to use forward sampling to compute the density in a single small bin or at a single point.

Example 4.4. It is interesting to ask if – numerically speaking – the change of measure is successful in cases where $\sigma$ depends on x and the fourth inequality of Assumption 2.4 cannot be verified. For that purpose, we slightly adjust the setting of the previous example by now taking $L = [ 1\ \ 0 ]$ and $\sigma(t, x) = [ 0;\ 2 + \frac12\cos(x_2) ]$ and repeating the experiment. In this case we chose $\tilde \sigma = \sigma(0, [ 0;\,0])$ . As the problem is more difficult, we took fewer bins ( $K = 50$ ) and set $h = 0.005$ (otherwise keeping our previous choices). The resulting Figure 9 shows no indication of lack of absolute continuity or loss of probability mass. This strongly indicates that guided proposals can perform perfectly well for the present complex setting that includes state-dependent diffusion coefficient and hypo-ellipticity.

Figure 8.

Dark orange: histogram baseline estimate of the density of observation $V = L X_T + Z$ , $Z \sim N(0, 10^{-6})$ from forward simulation. Dashed blue: observation density estimate using weighted histogram of points $\tilde V$ sampled from Gaussian distribution weighted with importance weights from guided proposals steered towards those points. Top: $T = 4\pi$ . Bottom: $T = 40\pi$ . Pink: difference between histograms.

Figure 9.

As Figure 8, but estimates for the model with observation operator $L =[1\ \ 0]$ and $\sigma(t, x) = [ 0;\ 2 + \frac12\cos(x_2) ]$ , at $T = 4\pi$ .

However, care is needed. In Figure 10 we show the result for the same experiment but with L changed to $L = [ 1\ \ 1]$ . Here, the loss of probability mass indicates violation of absolute continuity. We conjecture that $L a(T,v) L^{\prime} = L \tilde{a}(T) L^{\prime}$ may be the ‘right’ restriction on choosing $\tilde{a}(T)$ . To obtain empirical evidence, we redid the experiment with $L = [ 1\ \ 1]$ but now $\sigma(t, x) = [ 0;\ 2 + \frac12\cos(Lx) ]$ . In this case one can match the diffusivity at time T by taking $\tilde\sigma = [ 0;\ 2 + \frac12\cos(v) ]$ . The resulting figure (Figure 11) indicates no loss of absolute continuity, supporting the conjecture.

Figure 10.

As Figure 8, but estimates for the model with observation operator $L =[1\ \ 1]$ and $\sigma(t, x) = [ 0;\ 2 + \frac12\cos(x_2) ]$ , at $T = 4\pi$ . Note the loss of probability mass indicating lack of absolute continuity.

Figure 11.

As Figure 8, but estimates for the model with observation operator $L =[1\ \ 1]$ and $\sigma(t, x) = [ 0; \ 2 + \frac12\cos(x_1 + x_2) ]$ , at $T = 4\pi$ .

5.1. Centering and scaling of the guided proposal

To reduce notational overheads, we write $a_t\equiv a(t,X^\circ_t)$ . Then $\tilde{b}_t$ , $b_t$ , and ${\sigma}_t$ are defined similarly. Our starting point is the expression for $\tilde{r}$ in (2.2).

Lemma 5.1. If we define

$$ Z_t = v- \mu(t)-L(t) X^\circ_t ,$$

then

$$ {\mathrm{d}} Z_t = L(t) (\tilde{b}_t-b_t)\,{\mathrm{d}} t + L(t){\sigma}_t \,{\mathrm{d}} W_t - L(t)a_t L(t)^{\prime} M(t) Z_t \,{\mathrm{d}} t.$$

Proof.We have

$${\mathrm{d}} Z_t = -\bigg(\frac{{\mathrm{d}} }{{\mathrm{d}} t} L(t)\bigg) X^\circ_t \,{\mathrm{d}} t -\frac{{\mathrm{d}} }{{\mathrm{d}} t} \mu(t)-L(t) \,{\mathrm{d}} X^\circ_t.$$

The results now follow because the first two terms on the right-hand side together equal $L(t) \tilde{b}(t,X^\circ_t)$ .

Lemma 5.2. We have

\begin{align*} \frac12 {\mathrm{d}} (Z_t^{\prime} M(t) Z_t) & = \frac12 Z_t^{\prime} M(t) L(t) (\tilde{a}(t)-a_t) L(t)^{\prime} M(t) Z_t \,{\mathrm{d}} t \\[3pt] & \quad\, +Z_t^{\prime} M(t) L(t) (\tilde{b}_t-b_t) \,{\mathrm{d}} t \\[3pt]& \quad\, + Z_t^{\prime} M(t) L(t) {\sigma}_t \,{\mathrm{d}} W_t -\frac12 Z_t^{\prime} M(t) L(t) a_t L(t)^{\prime} M(t) Z_t \,{\mathrm{d}} t \\[3pt]& \quad\, +\mathrm{tr}(L(t) a_t L(t)^{\prime} M(t))\,{\mathrm{d}} t.\end{align*}

Proof.By Itô’s lemma,

$$\frac12 {\mathrm{d}} (Z_t^{\prime} M(t) Z_t) = \frac12 Z_t^{\prime} \frac{{\mathrm{d}} M(t)}{{\mathrm{d}} t} Z_t \,{\mathrm{d}} t +Z_t^{\prime} M(t) \,{\mathrm{d}} Z_t +\mathrm{tr}(L(t) a_t L(t)^{\prime} M(t))\,{\mathrm{d}} t.$$

Next, substitute the SDE for $Z_t$ from Lemma 5.1 and use

$$\frac{{\mathrm{d}} M(t)}{{\mathrm{d}} t} = -M(t) \frac{{\mathrm{d}} M(t)^{-1}}{{\mathrm{d}} t} M(t) =M(t) L(t) \tilde{a}(t) L(t)^{\prime} M(t).$$

The final equality follows from the fact that ${M^\dagger}(t)=M(t)^{-1}$ satisfies the ordinary differential equation ${\mathrm{d}} {M^\dagger}(t)=- L(t) \tilde{a}(t) L(t)^{\prime}\,{\mathrm{d}} t.$ The result follows upon reorganising terms.

Whereas in the uniformly elliptic case all elements of $Z_t$ and M(t) behave in the same way as a function of $T-t$ , this is not so in the hypo-elliptic case. For this reason, we introduce a diagonal scaling matrix $\Delta(t)$ .

Lemma 5.3. Let $\Delta(t)$ be an invertible $m\times m$ diagonal matrix. If ${Z_{\Delta,t}}$ , ${L_{\Delta}}(t)$ and ${M_{\Delta}}(t)$ are as defined in (2.3) and (2.4), then

(5.1) \begin{align}\frac12 {\mathrm{d}} ({Z_{\Delta,t}^{\prime}} {M_{\Delta}}(t) {Z_{\Delta,t}})& = \frac12 {Z_{\Delta,t}^{\prime}} {M_{\Delta}}(t) {L_{\Delta}}(t) (\tilde{a}(t)-a_t) {L_{\Delta}}(t)^{\prime} {M_{\Delta}}(t) {Z_{\Delta,t}} \,{\mathrm{d}} t \notag \\[3pt] & \quad\, + {Z_{\Delta,t}^{\prime}} {M_{\Delta}}(t) {L_{\Delta}}(t) (\tilde{b}_t-b_t) \,{\mathrm{d}} t \notag \\[3pt] & \quad\, + {Z_{\Delta,t}^{\prime}} {M_{\Delta}}(t) {L_{\Delta}}(t) {\sigma}_t \,{\mathrm{d}} W_t -\frac12 {Z_{\Delta,t}^{\prime}} {M_{\Delta}}(t) {L_{\Delta}}(t) a_t {L_{\Delta}}(t)^{\prime} {M_{\Delta}}(t) {Z_{\Delta,t}} \,{\mathrm{d}} t \notag \\* & \quad\, +\mathrm{tr}({L_{\Delta}}(t) a_t {L_{\Delta}}(t)^{\prime} {M_{\Delta}}(t))\,{\mathrm{d}} t.\end{align}

Moreover,

(5.2) \begin{equation}\tilde{r}(t,X^\circ_t)={L_{\Delta}}(t)^{\prime} {M_{\Delta}}(t) {Z_{\Delta,t}}{.}\end{equation}

Proof.This is a straightforward consequence of Lemma 5.2. The expression for $\tilde{r}$ follows from equation (2.2).

5.2. Recap on notation and results

For clarity we summarise our notation, some of which was already defined in Section 1.6. The auxiliary process is defined by the SDE ${\mathrm{d}} \tilde{X}_t = (\tilde{B}(t) \tilde{X}_t + \tilde\beta(t)) \,{\mathrm{d}} t + \tilde{\sigma}(t) \,{\mathrm{d}} W_t$ . The matrix $\Phi(t)$ satisfies the ODE ${\mathrm{d}} \Phi(t) =\tilde{B}(t) \Phi(t) \,{\mathrm{d}} t$ and we set $\Phi(T,t)=\Phi(T) \Phi(t)^{-1}$ . A realisation v of $V=LX_T$ is observed. The scaled process is defined by

$${Z_{\Delta,t}}=\Delta(t) \bigg( v- \int_t^T L(\tau) \tilde\beta(\tau) \,{\mathrm{d}} \tau-L(t) X^\circ_t \bigg),$$

where $ L(t)= L \Phi(T,t)$ and ${L_{\Delta}}(t)=\Delta(t)L(t). $ Furthermore, we defined

$$ M(t)= \bigg(\int_t^T L(\tau) \tilde{a}(\tau) L(\tau)^{\prime} \,{\mathrm{d}} \tau\bigg)^{-1}\quad\text{and}\quad {M_{\Delta}}(t) = \Delta(t)^{-1} M(t) \Delta(t)^{-1},$$

where $\tilde{a}(t)=\tilde{\sigma}(t) \tilde{\sigma}(t)^{\prime}$ . Finally, the guiding term in the SDE for the guided proposal $X^\circ_t$ is given by $a(t,X^\circ_t) {L_{\Delta}}(t)^{\prime} {M_{\Delta}}(t) {Z_{\Delta,t}}$ . The process $t\mapsto {Z_{\Delta,t}}$ is the key object to be studied in this section.

5.3. Proof of Proposition 2.2

The line of proof is exactly as suggested in [Reference Mao25, page 341], as follows.

1. (1) Start with the Lyapunov function $V(t,{Z_{\Delta,t}})=\frac12 {Z_{\Delta,t}}{\prime} {M_{\Delta}}(t) {Z_{\Delta,t}}$ .

2. (2) Apply Itô’s lemma to $V(t,{Z_{\Delta,t}})$ .

3. (3) Use martingale inequalities to bound the stochastic integral.

4. (4) Apply a Gronwall-type inequality.

We bound all terms appearing in equation (5.1). Note that the first term on the right-hand side vanishes. We start with the Wiener integral term. To this end, fix $t_0 \in [0,T)$ and let

$$N_t= \int_{t_0}^t {Z_{\Delta,s}}^{\prime} {M_{\Delta}}(s) {L_{\Delta}}(s) {\sigma} \,{\mathrm{d}} W_s.$$

Then

$$\int_{t_0}^t {Z_{\Delta,s}}^{\prime} {M_{\Delta}}(s) {L_{\Delta}}(s) {\sigma}_s \,{\mathrm{d}} W_s -\frac12 \int_{t_0}^t {Z_{\Delta,s}}^{\prime} {M_{\Delta}}(s) {L_{\Delta}}(s)\, a_s\, {L_{\Delta}}(s)^{\prime} {M_{\Delta}}(s) {Z_{\Delta,s}} \,{\mathrm{d}} s =N_t -\frac12[N]_t.$$

Now $N_t$ can be bounded using an exponential martingale inequality. Let $\{{\gamma}_n\}$ be a sequence of positive numbers. For $n\in {\mathbb{N}}$ , define $t_n=T-1/n$ and

$$E_n=\bigg\{ \sup_{0\le t \le t_{n+1}} \bigg( N_t- \frac{1}{2} [N]_t \bigg) > {\gamma}_n\bigg\}.$$

By the exponential martingale inequality of [Reference Mao26, Theorem 1.7.4], we obtain that $P(E_n) \le {\mathrm{e}}^{-{\gamma}_n}$ . If we assume $\sum_{n=1}^\infty {\mathrm{e}}^{-{\gamma}_n}<\infty$ , then by the Borel–Cantelli lemma $ P(\limsup_{n\to \infty} E_n)=0. $ Hence, for almost all $\omega$ , there exists $ n_0(\omega)$ such that, for all $n\ge n_0(\omega)$ ,

(5.3) \begin{equation} \sup_{t_0\le t \le t_{n+1}} \bigg( N_t -\frac{1}{2}\int_0^t [N]_t \bigg)\le{\gamma}_n.\end{equation}

Let ${\varepsilon}>0$ . Upon taking ${\gamma}_n=(1+2{\varepsilon}) \log n$ , we get

$$\sum_{n=1}^\infty {\mathrm{e}}^{-{\gamma}_n}=\sum_{n=1}^\infty n^{-1-2{\varepsilon}} <\infty.$$

Since ${M_{\Delta}}(t)$ is strictly positive definite,

$${\lambda_{\rm min}}({M_{\Delta}}(t)) \|{Z_{\Delta,t}}\|^2 \le {Z_{\Delta,t}^{\prime}} {M_{\Delta}}(t) {Z_{\Delta,t}}.$$

Assume $t_0 < t< t_{n+1}$ . Combining the inequality of the above display with Lemma 5.3 and substituting the bound in (5.3), we obtain, for any ${\varepsilon}>0$ ,

\begin{align*}\frac12{\lambda_{\rm min}}({M_{\Delta}}(t)) \|{Z_{\Delta,t}}\|^2 &\le \frac12 {Z_{\Delta,t_0}} {M_{\Delta}}(t_0) {Z_{\Delta,t_0}}\\ &\quad\, +\int_{t_0}^t \|{Z_{\Delta,s}}\| \| {M_{\Delta}}(s)\| \| {L_{\Delta}}(s) (\tilde{b}_s-b_s)\| \,{\mathrm{d}} s \\ & \quad\, +{\gamma}_n + \int_{t_0}^t \mathrm{tr}({L_{\Delta}}(s)\, a_s\, {L_{\Delta}}(s)^{\prime} {M_{\Delta}}(s)) \,{\mathrm{d}} s \\ & \quad\, + \frac12 \int_{t_0}^t{Z_{\Delta,s}}^{\prime} {M_{\Delta}}(s) {L_{\Delta}}(s) (\tilde{a}(s)-a_s) {L_{\Delta}}(s)^{\prime} {M_{\Delta}}(s) {Z_{\Delta,s}} \,{\mathrm{d}} s {.}\end{align*}

Recall that for positive semidefinite matrices A and C we have

$$|\mathrm{tr}(AC)| \le \mathrm{tr}(A) \mathrm{tr}(C)\le \mathrm{tr}(A) p {\lambda}_{\rm max}(C)$$

if $C \in {\mathbb{R}}^{p\times p}$ . Hence

(5.4) \begin{equation}\mathrm{tr}({L_{\Delta}}(s)\, a_s\, {L_{\Delta}}(s)^{\prime} {M_{\Delta}}(s)) \le\mathrm{tr}({L_{\Delta}}(s)\, a_s\, {L_{\Delta}}(s)^{\prime}) m {\lambda}_{\rm max}({M_{\Delta}}(s)) {.}\end{equation}

Furthermore, as

(5.5) \begin{equation}\|{M_{\Delta}}(s)\|=\sqrt{ {\lambda}_{\rm max}({M_{\Delta}}(s)^2)} = {\lambda}_{\rm max}({M_{\Delta}}(s)){,}\end{equation}

we can combine the preceding three inequalities to obtain

\begin{align*} \frac12{\lambda_{\rm min}}({M_{\Delta}}(t)) \|{Z_{\Delta,t}}\|^2 &\le \frac12 {Z_{\Delta,t_0}} {M_{\Delta}}(t_0) {Z_{\Delta,t_0}} \notag\\[3pt] &\quad\, +\int_{t_0}^t \|{Z_{\Delta,s}}\| {\lambda}_{\rm max}({M_{\Delta}}(s)) \| {L_{\Delta}}(s) (\tilde{b}_s-b_s)\| \,{\mathrm{d}} s \notag \\[3pt] & \quad\, +{\gamma}_n + m\int_{t_0}^t \mathrm{tr}({L_{\Delta}}(s)\, a_s\, {L_{\Delta}}(s)^{\prime}){\lambda}_{\rm max}({M_{\Delta}}(s)) \,{\mathrm{d}} s \notag \\[3pt] & \quad\, + \frac12 \int_{t_0}^t{Z_{\Delta,s}}^{\prime} {M_{\Delta}}(s) {L_{\Delta}}(s) (\tilde{a}(s)-a_s) {L_{\Delta}}(s)^{\prime} {M_{\Delta}}(s) {Z_{\Delta,s}} \,{\mathrm{d}} s .\end{align*}

Upon substituting the bounds in (2.5), for certain positive constants $C_0$ , $C_1$ , $C_2$ , $C_3$ , and $C_4$ we obtain

(5.6) \begin{align} (T-t)^{-1} \|{Z_{\Delta,t}}\|^2 &\le C_0+ C_1 \int_{t_0}^t \|{Z_{\Delta,s}}\| (T-s)^{-1} \,{\mathrm{d}} s \notag \\[3pt]& \quad\, +C_2 {\gamma}_n +C_3 \int_{t_0}^t (T-s)^{-1} \,{\mathrm{d}} + C_4 \int_{t_0}^t \|{Z_{\Delta,s}}\|^2 (T-s)^{\alpha-2}.\end{align}

If we define $\xi_t=(T-t)^{-1} \|{Z_{\Delta,t}}\|^2$ , then this inequality can be rewritten as

\begin{align*} \xi_t &\le C_0 +C_2 {\gamma}_n + C_3 \log\bigg(\frac{T-t_0}{T-t}\bigg) \notag \\[3pt] & \quad\, + C_1 \int_{t_0}^t (T-s)^{-1/2} \sqrt{\xi_s}\,{\mathrm{d}} s + C_4 \int_{t_0}^t (T-s)^{\alpha-1} \xi_s \,{\mathrm{d}} s .\end{align*}

By Lemma B.1 in the Appendix this implies

\begin{align*} \xi_t & \leq \bigg( \sqrt{C_0 +C_2 {\gamma}_n + C_3 \log\bigg(\!\frac{T-t_0}{T-t}\!\bigg)} + \frac12 C_1 \big( \sqrt{T- t_0} - \sqrt{T-t} \big)\bigg)^2 \exp\bigg(\!C_4 \int_{t_0}^t (T-s)^{\alpha-1}\!\bigg).\end{align*}

Now divide both sides of this inequality by $\log(1/(T-t))$ and consider $t_n < t < t_{n+1}$ . Then $\log n \le \log(1/(T-t))$ . It then follows that

$$\limsup_{t\uparrow T} \frac{\|{Z_{\Delta,t}}\|^2}{(T-t)\log(1/(T-t))} \le C_2(1+2{\varepsilon}) +C_3.$$

Now let ${\varepsilon}\downarrow 0$ .

5.4. Proof of Corollary 2.1

As $\Delta(t)=I_m$ it is easy to see that $M(t)={\mathrm{O}} (1/(T-t)$ and ${L_{\Delta}}(t)={\mathrm{O}} (1)$ . This behaviour of M(t) is also contained in the first inequality of [Reference Schauer, van der Meulen and van Zanten35, Lemma 8] (in that paper, $\tilde{H}$ corresponds to M as defined here). Now it is easy to see that the conditions of Theorem 2.2 are satisfied.

6.1. Proof of Theorem 2.6

We start with a result that gives the Radon–Nikodým derivative of ${\mathbb{P}}^\star_t$ relative to ${\mathbb{P}}^\circ_t$ for $t<T$ .

Proposition 6.1. For $t<T$ we have

\begin{equation*} \frac{{\mathrm{d}} {\mathbb{P}}_t^\star}{{\mathrm{d}} {\mathbb{P}}_t^\circ}(X^\circ) = \frac{\tilde \rho(0,x_0)}{ \rho(0,x_0)}\frac{ \rho(t,X^\circ_t)}{\tilde \rho(t,X^\circ_t)} \Psi_t(X^\circ),\end{equation*}

where $\Psi_t$ is defined in (2.6).

Proof.Although this result is not a special case of [35, Proposition 1] (where it is assumed that $L=I$ and that the diffusion is uniformly elliptic), the arguments for deriving the likelihood ratio of ${\mathbb{P}}^\star_t$ with respect to ${\mathbb{P}}^\circ_t$ are the same and therefore omitted. The only thing that needs to be checked is that $\tilde\rho(t,x)$ satisfies the Kolmogorov backward equation associated with $\tilde X$ . This can be proved along the lines of Lemma 3.4 and Corollary 3.5 of [Reference van der Meulen and Schauer40]. Let $\tilde{\mathcal{F}}_t ={\sigma}(\tilde{X}_s,\, 0\le s \le t)$ and set $\tilde{Y}_t =\tilde\rho(t,\tilde{X}_t)$ . Now

\begin{align*} {{\mathbb{E}}}[{\tilde{Y}_t \mid \tilde{\mathcal{F}}_s}] & = \int_{{\mathbb{R}}^d} \tilde\rho(t,x) \tilde{p}(s,\tilde{X}_s;\ t,x) \,{\mathrm{d}} x \\[3pt] & = \int_{{\mathbb{R}}^d} \tilde{p}(s,\tilde{X}_s;\ t,x) \int_{{\mathbb{R}}^{d-m}} \Bigg(\tilde{p}t,x;\ T, \sum_{j=1}^d \xi_j f_j\Bigg) \,{\mathrm{d}} \xi_{m+1},\ldots, {\mathrm{d}} \xi_d \\[3pt] & =\int_{{\mathbb{R}}^{d-m}} \tilde{p}\Bigg(s,\tilde{X}_s;\ T, \sum_{j=1}^d \xi_j f_j\Bigg)\,{\mathrm{d}} \xi_{m+1},\ldots, {\mathrm{d}} \xi_d = \tilde\rho(s,\tilde{X}_s)=\tilde{Y}_s.\end{align*}

That is, ${(\tilde{Y}_t, \tilde{\mathcal{F}}_t)}$ is a martingale. If $\tilde{\mathcal{L}}$ denotes the infinitesimal generator of $\tilde{X}_t$ , then $\mathcal{K}=\partial / (\partial t) + \tilde{\mathcal{L}}$ is the infinitesimal generator of the space–time process ${(t,\tilde{Y}_t)}$ . Since $\tilde{Y}_t$ is a martingale, the mapping ${(t,x) \mapsto \tilde\rho(t,x)}$ is space–time harmonic. Then by Proposition 1.7 in Chapter VII of [Reference Revuz and Yor32], $\mathcal{K} \tilde\rho(t,x)=0$ . That is, $\tilde{\rho}(t,x)$ satisfies Kolmogorov’s backward equation.

This absolute continuity result is only useful for simulating conditioned diffusions if it can be shown to hold in the limit $t\uparrow T$ as well. The main line of proof is the same as in the proof of [Reference Schauer, van der Meulen and van Zanten35, Theorem 1], where at various places p and $\tilde{p}$ need to be replaced with $\rho$ and $\tilde\rho$ . However, some of the auxiliary results that are used require new arguments in the present setting. Moreover, the assumed Aronson-type bounds are not suitable for hypo-elliptic diffusions.

6.2. Proof of Theorem 2.6

We start by introducing some notation. Define the mapping $g_{\Delta}\colon [0,\infty) \times {\mathbb{R}}^d \to {\mathbb{R}}^m$ by

$$g_{\Delta}(t,x) = \Delta(t) ( v-\mu(t) - L(t) x)$$

and note that ${Z_{\Delta,t}} = g_{\Delta}(t,X^\circ_t)$ . For a diffusion process Y we define the stopping time

$${\sigma}_k(Y) = T \wedge \inf_{t\in [0,T]} \big\{ \|g_{\Delta}(t,Y_t)\| \ge k \sqrt{(T-t) \log(1/(T-t))}\big\},$$

where $k\in {\mathbb{N}}$ . We write

$${\sigma}_k^\circ = {\sigma}_k(X^\circ) \quad {\sigma}_k = {\sigma}_k(X)\quad {\sigma}_k^\star = {\sigma}_k(X^\star).$$

Define $\bar\rho = \tilde\rho(0,x_0)/\rho(0,x_0)$ . By Proposition 6.1, for any $t<T$ and bounded, $\mathcal F_t$ -measurable f, we have

(6.1) \begin{equation}{{\mathbb{E}}}\bigg[{ f(X^\star) \frac{ \tilde \rho(t, X^\star_t)}{\rho(t, X^\star_t)}}\bigg] ={{\mathbb{E}}}[{ f(X^\circ) \bar{\rho} \: \Psi_t(X^\circ)}].\end{equation}

By taking $f_t(x)={\textbf{1}}\{t\le {\sigma}_k(x)\}$ , we get

(6.2) \begin{equation}\bar\rho\, {\mathbb{E}}[ \Psi_t(X^\circ) {\textbf{1}}\{t\le {\sigma}^\circ_k\}] = {\mathbb{E}}\bigg[ \frac{\tilde\rho(t,X^\star_t)}{\rho(t,X^\star_t)} {\textbf{1}}\{t\le {\sigma}^\star_k\}\bigg].\end{equation}

Next, we take $\lim_{k\to\infty} \lim_{t\uparrow T}$ on both sides. We start with the left-hand side. By Lemma 6.1, for each $k\in{\mathbb{N}}$ , $\sup_{0\le t\le T}\Psi_t(X^\circ)$ is uniformly bounded on the event $\{T={\sigma}_k^\circ\}$ . Hence, by the dominated convergence theorem we obtain

$$\lim_{k\to\infty} \lim_{t\uparrow T}{\mathbb{E}}[ \Psi_t(X^\circ) {\textbf{1}}\{t\le {\sigma}^\circ_k\}] = \lim_{k\to \infty} {\mathbb{E}}[ \Psi_T(X^\circ) {\textbf{1}}\{T\le {\sigma}^\circ_k\}].$$

Since by definition ${\sigma}^\circ_k\le T$ , we have $\{T\le {\sigma}^\circ_k\}=\{T= {\sigma}^\circ_k\}$ . Furthermore,

$${\textbf{1}}\{T={\sigma}^\circ_k\} = {\textbf{1}}\big\{\|Z^\circ_{\Delta,t}\| \le k \sqrt{(T-t) \log(1/(T-t))} \big\} \uparrow 1 \quad \text{as } k \to \infty,$$

by Proposition 2.2. Therefore, by monotone convergence,

$$\lim_{k\to\infty} \lim_{t\uparrow T}{\mathbb{E}}[ \Psi_t(X^\circ) {\textbf{1}}\{t\le {\sigma}^\circ_k\}] = {\mathbb{E}}[ \Psi_T(X^\circ)].$$

It remains to show that the right-hand side of (6.2) tends to 1. We write

\begin{align*} \rho(0,x_0) {\mathbb{E}}\bigg[ \frac{\tilde\rho(t,X^\star_t)}{\rho(t,X^\star_t)} {\textbf{1}}\{t\le {\sigma}^\star_k\}\bigg] &= {\mathbb{E}}[ \tilde\rho(t,X_t) {\textbf{1}}\{t\le {\sigma}_k\}] \\[3pt] & ={\mathbb{E}}[ \tilde\rho(t,X_t)] - {\mathbb{E}}[ \tilde\rho(t,X_t) {\textbf{1}}\{t> {\sigma}_k\}]{.}\end{align*}

By Lemma 6.3 the first of the terms on the right-hand side tends to $\rho(0,x_0)$ when $t\uparrow T$ . The second term tends to zero by Lemma 6.4.

To complete the proof we note that by equation (6.1) and Lemma 6.3 we have $\bar \rho\, {{\mathbb{E}}}[{\Psi_t(X^\circ)}] \to 1$ as $t\uparrow T$ . In view of the preceding and Scheffé’s lemma this implies that $\Psi_t(X^\circ) \to \Psi_T(X^\circ)$ in the $L^1$ -sense as $t \uparrow T$ . Hence for $s< T$ and a bounded, $\mathcal{F}_s$ -measurable, continuous functional g,

$${{\mathbb{E}}}[{g(X^\circ) \bar \rho \Psi_T(X^\circ)}] = \lim_{t \uparrow T} {{\mathbb{E}}}\bigg[{g(X^\star)\frac{\tilde \rho(t, X^\star_{t})}{\rho(t, X^\star_{t})}}\bigg].$$

By Lemma 6.3 this converges to ${\mathbb{E}}\, g (X^\star)$ as $t \uparrow T$ and we find that ${\mathbb{E}}\, g(X^\circ) \bar \rho \Psi_T(X^\circ) = {\mathbb{E}}\, g (X^\star)$ .

Lemma 6.1. Under Assumption 2.4there exists a positive constant K (not depending on k) such that

$$\Psi_t(X^\circ) {\textbf{1}}_{t\le {\sigma}^\circ_m} \le \exp(Kk^2).$$

Proof.To bound $\Psi_t(X^\circ)$ , we will first rewrite $\mathcal{G}(s,X^\circ)$ in terms of ${Z_{\Delta,t}}$ , ${L_{\Delta}}(t)$ and ${M_{\Delta}}(t)$ , as defined in (2.3) and (2.4). By (5.2), we have

$$\tilde{r}(t,X^\circ_t)={L_{\Delta}}(t)^{\prime} {M_{\Delta}}(t){Z_{\Delta,t}} \quad \text{and} \quad \tilde{H}(t)={L_{\Delta}}(t)^{\prime} {M_{\Delta}}(t) {L_{\Delta}}(t).$$

Here, the expression for $\tilde{H}(t)$ was obtained from

$$\tilde{H}(t)= -{\,\mathrm D} L(t)^{\prime} M(t) (v-\mu(t)-L(t)x)={\,\mathrm D} (L(t)^{\prime} M(t) L(t) x)= L(t)^{\prime} M(t) L(t).$$

Hence

\begin{align*} \mathcal{G}(s,X^\circ_s) &= (b(s,X^\circ_s) - \tilde b(s,X^\circ_s))^{\prime} {L_{\Delta}}(s)^{\prime} {M_{\Delta}}(s){Z_{\Delta,s}} \notag \\[3pt] & \quad\, - \frac12 \mathrm{tr}([a_s - \tilde a(s)] {L_{\Delta}}(s)^{\prime} {M_{\Delta}}(s) {L_{\Delta}}(s)) \notag \\[3pt] & \quad\, +\frac12 {Z_{\Delta,s}}^{\prime} {M_{\Delta}}(s) {L_{\Delta}}(s) [a_s - \tilde a(s)] {L_{\Delta}}(s)^{\prime} {M_{\Delta}}(s){Z_{\Delta,s}} .\end{align*}

On the event $\{t \le {\sigma}_k^\circ\}$ we have

$$\|{Z_{\Delta,t}}\| \le k \sqrt{(T-t) \log(1/(T-t))}.$$

The absolute value of the first term of $\mathcal{G}$ can be bounded by

\begin{align*} & \| {M_{\Delta}}(s)\|\!\| {L_{\Delta}}(s)( \tilde{b}(s,X^\circ_s)-b(s,X^\circ_s) )\|\!\|{Z_{\Delta,s}}\| \\[3pt] & \quad \le (T-s)^{-1} \|{Z_{\Delta,s}}\| \\[3pt] &\quad \le c_1 m (T-s)^{-1/2} \sqrt{\log(1/(T-s))}.\end{align*}

Here we bounded $\|{M_{\Delta}}(t)\| \le {\lambda_{\rm max}}({M_{\Delta}}(t))$ , as in (5.5). The absolute value of twice the second term of $\mathcal{G}$ can be bounded by

$$\mathrm{tr}({L_{\Delta}}(s) (a_s-\tilde{a}(s)) {L_{\Delta}}(s)^{\prime} ) k {\lambda}_{\rm max}({M_{\Delta}}(s)),$$

just as in (5.4). As for a $p\times p$ matrix A we have $\mathrm{tr}(A)\le p {\lambda_{\rm max}}(A) = p \|A\|^2$ (recall we assume the spectral norm on matrices throughout), this can be bounded by

$$m\|{L_{\Delta}}(s) (a_s-\tilde{a}(s)) {L_{\Delta}}(s)^{\prime}\|^2 m {\lambda}_{\rm max}({M_{\Delta}}(s)) \le m^2 c_3 \bar{c}(T-s)^{2\alpha-1} {.}$$

The absolute value of twice the third term of $\mathcal{G}$ can be bounded by

\begin{align*} & \|{Z_{\Delta,s}}\|^2 \|{M_{\Delta}}(s)\|^2 \|{L_{\Delta}}(s) (a(s)-\tilde{a}(s)) {L_{\Delta}}(s)^{\prime}\| \\[3pt] & \quad \le k^2 (T-s) \log(1/(T-s)) \bar{c}^2 (T-s)^{-2} c_3 (T-s)^\alpha\\[3pt] & \quad \le k^2 \bar{c}^2 c_3 (T-s)^{\alpha-1} \log(1/(T-s)).\end{align*}

We conclude that all three terms in $\mathcal{G}$ are integrable on [0, T].

Lemma 6.2. For all bounded, continuous $f\colon [0,T] \times {\mathbb{R}}^d \to {\mathbb{R}}$ ,

$$\lim_{t\uparrow T} \int f(t,x) \tilde{p}(t,x;\ T,v) \,{\mathrm{d}} x = f(T,v).$$

Proof.The proof is just as in Lemma 7 of [Reference Schauer, van der Meulen and van Zanten35].

Lemma 6.3. If Assumption 2.5 holds true, $0<t_1 < t_2 < \dots < t_N< t < T$ , and $g \in C_b({\mathbb{R}}^{Nd})$ , then

$$\lim_{t \uparrow T} {{\mathbb{E}}}\bigg[{g(X^\star_{t_1}, \ldots, X^\star_{t_N}) \frac{\tilde \rho(t, X^\star_{t})}{ \rho(t, X^\star_{t})}}\bigg] = {{\mathbb{E}}}[{g(X^\star_{t_1}, \ldots, X^\star_{t_N})}].$$

Proof.The joint density q of ${(X_{t_1}, \ldots, X_{t_N})}$ , conditional on $X_{t_0}=x_0$ , is given by

$$q(x_1,\ldots, x_N) = \prod_{i=1}^N p(t_{i-1}, x_{i-1};\ t_i, x_i) .$$

Hence

(6.3) \begin{align}& {{\mathbb{E}}}\bigg[{g(X^\star_{t_1}, \ldots, X^\star_{t_N}) \frac{\tilde \rho(t, X^\star_{t})}{ \rho(t, X^\star_{t})}}\bigg] \notag \\* & \quad = \int g(x_1,\ldots,x_N) \frac{\tilde{ \rho}(t,x)}{ \rho(t,x)} q(x_1,\ldots,x_N) \frac{p(t_N,x_N;\ t,x) \rho(t,x)}{ \rho(0,x_0)} \,{\mathrm{d}} x_1\ldots {\mathrm{d}} x_N \,{\mathrm{d}} x \notag \\* & \quad = \frac{1}{\rho(0,x_0)} \int g(x_1,\ldots,x_N) q(x_1,\ldots, x_N) F(t;\ t_N, x_N) \,{\mathrm{d}} x_1\ldots {\mathrm{d}} x_N,\end{align}

where for $t_N<t<T$

$$F(t;\ t_N, x_N)=\int p(t_N, x_N;\ t,x) \tilde{\rho}(t,x) \,{\mathrm{d}} x.$$

We can assume $t\ge (T+t_N)/2$ . For fixed $t_N$ and $x_N$ , the mapping ${(t,x) \mapsto p(t_N, x_N;\ t,x)}$ is continuous and bounded, for t bounded away from $t_N$ . By Lemma 6.2 it follows that $F(t;\ t_N, x_N) \to \rho(t_N,x_N)$ when $t\uparrow T$ . The argument is finished by taking the limit $t\uparrow T$ on both sides of equation (6.3), interchanging limit and integral on the right-hand side and noting that the limit on the right-hand side coincides with ${{\mathbb{E}}}[{g(X^\star_{t_1}, \ldots, X^\star_{t_N})}]$ .

The interchange is permitted by dominated convergence. To see this, first note that g is assumed to be bounded. Next,

$$\int \Bigg(\prod_{i=1}^n p(t_{i-1}, x_{i-1};\ t_i, x_i)\Bigg) p(t_N,x_N;\ t,x) \tilde\rho(t,x) \,{\mathrm{d}} x \,{\mathrm{d}} x_1\ldots {\mathrm{d}} x_N \le C^{N+1} \tilde\rho(t_0,x_0),$$

which follows from repeated application of Assumption 2.5.

Lemma 6.4. Assume that there exists a positive $\delta$ such that $|\Delta(t)| \lesssim (T-t)^{-\delta}$ . If Assumption 2.5 holds true, then

$$\lim_{k\to \infty} \lim_{t \uparrow T}{{\mathbb{E}}}[{ \tilde\rho(t,X_t) {\textbf{1}}\{t> {\sigma}_k\}}] .$$

Proof.As in the proof of [Reference Schauer, van der Meulen and van Zanten35, Lemma 5], it suffices to show that

\begin{equation*} \lim_{k\to \infty} \lim_{t \uparrow T} {{\mathbb{E}}}\bigg[{ {\textbf{1}}_{\{t >\sigma_k\}} \int p(\sigma_k,X_{\sigma_k};\ t,z) \tilde \rho(t ,z) \,{\mathrm{d}} z }\bigg] =0 .\end{equation*}

Applying Assumption 2.5 and using the Chapman–Kolmogorov relations, we obtain

$$\int p(\sigma_k,X_{\sigma_k};\ t,z) \tilde \rho(t ,z) \,{\mathrm{d}} z \le C \tilde\rho(\sigma_k,X_{\sigma_k}).$$

Define $\tilde{Z}_t=v-\mu(t) - L(t) \tilde{X}_t$ . If we denote its transition density by $\tilde{q}$ , then

$$\tilde\rho(t,y) = \tilde{q}(t,v-\mu(t)-L(t) y;\ T, 0), \quad y \in {\mathbb{R}}^d \quad \text{and}\quad t\in [0,T),$$

since $\tilde{r}(t,x)$ depends on x only via L(t)x. Define the set

$$\mathcal{A}_k = \{(t,y) \in [0,T) \times {\mathbb{R}}^d \colon \|\Delta(t)(v-\mu(t)-L(t)y)\| = k \eta(t)\},$$

where $\eta(t) = \sqrt{(T-t)\log(1/(T-t))}$ . Then

$${{\mathbb{E}}}\bigg[{ {\textbf{1}}_{\{t >\sigma_k\}} \int p(\sigma_k,X_{\sigma_k};\ t,z) \tilde \rho(t ,z) \,{\mathrm{d}} z }\bigg] \le {{\mathbb{E}}}\bigg[{\sup_{(t,y) \in \mathcal{A}_k}\tilde\rho(t,y)}\bigg],$$

since by definition of ${\sigma}_k$ , $\|\Delta({\sigma}_k)(v-\mu({\sigma}_k)-L({\sigma}_k)X_{{\sigma}_k})\| = k \eta({\sigma}_k)$ . The expectation on the right-hand side is now superfluous. It is easily derived that $\tilde{Z}_t$ satisfies the SDE

$${\mathrm{d}} \tilde{Z}_t = L(t)\tilde{\sigma}(t) \,{\mathrm{d}} W_t$$

and hence for $x\in {\mathbb{R}}^m$

$$\tilde{q}(t,x;\ T,0) = \phi_m(0;\ x, {M^\dagger}(t)),$$

where we denote the density of the multivariate normal distribution in ${\mathbb{R}}^m$ with mean vector $\nu$ and covariance matrix $\Upsilon$ , evaluated in u by $\phi_m(u;\ \nu,\Upsilon)$ . Hence, stitching the previous derivations together we obtain

$${{\mathbb{E}}}\bigg[{ {\textbf{1}}_{\{t >\sigma_k\}} \int p(\sigma_k,X_{\sigma_k};\ t,z) \tilde \rho(t ,z) \,{\mathrm{d}} z }\bigg] \le \sup_{(t,y) \in \mathcal{A}_k} \phi_m(0;\ x, {M^\dagger}(t)).$$

The right-hand side multiplied by ${(2\pi)^{m/2}}$ equals

$$|M(t)|^{1/2} \exp\bigg( -\frac12(v-\mu(t)-L(t)y)^{\prime} M(t)(v-\mu(t)-L(t)y)\bigg){,}$$

which can be further bounded by

\begin{align*}& \sup_{(t,y) \in \mathcal{A}_k} |\Delta(t)| |{M_{\Delta}}(t)|^{1/2}\exp\bigg( -\frac12(v-\mu(t)-L(t)y)^{\prime} \Delta(t) {M_{\Delta}}(t) \Delta(t) (v-\mu(t)-L(t)y)\bigg)\\[3pt] &\quad \le \sup_{(t,y) \in \mathcal{A}_k} |\Delta(t)| ( {\lambda_{\rm max}}({M_{\Delta}}(t))^{m/2} \exp\bigg(-\frac12\|\Delta(t)(v-\mu(t)-L(t)y))\|^2 {\lambda_{\rm min}}({M_{\Delta}}(t))\bigg)\\[3pt] & \quad\le \sup_{t\in [0,T)} |\Delta(t)| \bigg(\frac{\bar{c}}{T-t}\bigg)^{m/2} \exp\bigg(-\frac{\underline{c} k^2\eta(t)^2}{2(T-t)}\bigg)\\[3pt]& \quad \lesssim \sup_{t\in [0,T)} (T-t)^{-\delta-m/2} \exp\bigg(-\frac{\underline{c} k^2\eta(t)^2}{2(T-t)}\bigg).\end{align*}

Next, the maximum can be bounded, followed by taking the limit $k\to \infty$ , to see that this tends to zero. This is exactly as in the proof of [Reference Schauer, van der Meulen and van Zanten35, Lemma 5].

6.3. Proof of Lemma 2.3

By absolute continuity of the laws of $\tilde X$ and X and the abstract Bayes’ formula, for bounded $\mathcal{F}_T$ -measurable f we have

$${{\mathbb{E}}} [{f(X) \mid X_T=v}] = \frac{\tilde{p}(0,x_0;\ T,v)}{p(0,x_0;\ T,v)} {{\mathbb{E}}}\bigg[{f(\tilde{X}) \frac{{\mathrm{d}} {\mathbb{P}}_T}{{\mathrm{d}} \tilde{{\mathbb{P}}}_T}(\tilde{X} ) \biggm| \tilde{X}_T=v}\bigg].$$

Hence, upon taking $f\equiv 1$ and applying Girsanov’s theorem, we get

$$p(0,x_0;\ T,v) = \tilde{p}(0,x_0;\ T,v) {{\mathbb{E}}}\bigg[{\exp\bigg(\int_0^T \eta(\tilde{X}_s)^{\prime} \,{\mathrm{d}} W_s - \frac12 \int_0^T\|\eta(\tilde{X}_s)\|^2 \,{\mathrm{d}} s \bigg) \biggm| \tilde{X}_T=v}\bigg].$$

Since $\eta$ is bounded, this implies

$$p(0,x_0;\ T,v) \propto \tilde{p}(0,x_0;\ T,v) {{\mathbb{E}}}\bigg[{\exp\bigg(\int_0^T \eta(\tilde{X}_s)' \,{\mathrm{d}} W_s \bigg) \biggm| \tilde{X}_T=v}\bigg].$$

Upon defining $\tau_j =\int_0^T \eta_j(\tilde{X}_s)^2 \,{\mathrm{d}} s $ , the Dambis–Dubins–Schwarz theorem implies that the expectation on the right-hand side equals

$${{\mathbb{E}}}\big[{{\mathrm{e}}^{-\sum_{j=1}^{d'} \int_0^T \eta_j(\tilde{X}_s) \,{\mathrm{d}} W_s^j} \mid \tilde{X}_T=v}\big] ={{\mathbb{E}}}\big[{{\mathrm{e}}^{\sum_{j=1}^{d'} {W}_{\tau_j}^j} \mid \tilde{X}_T=v}\big] .$$

By boundedness of $\eta$ there exist constants $\{K_j\}_{j=1}^{d'}$ such that $\tau_j\le K_j$ . Hence the right-hand side of the above display can be bounded by

$${{\mathbb{E}}}\big[{{\mathrm{e}}^{\sum_{j=1}^{d'} \sup_{0\le s \le K_j}{W}_s^j } \mid \tilde{X}_T=v}\big] = {{\mathbb{E}}}\big[{{\mathrm{e}}^{\sum_{j=1}^{d'} \sup_{0\le s \le K_j} {W}_s^j}}\big] = \prod_{j=1}^{d'} {{\mathbb{E}}}[{{\mathrm{e}}^{\sup_{0\le s \le K_j} {W}_s^j}}],$$

where the final equality follows from the components of $\bar{W}$ being independent. The expectation on the right-hand side is finite, the constant only depending on T. To see this, if $B_t$ is a one-dimensional Brownian motion, then $\bar{B}_t =\sup_{0\le s\le t} B_s$ has density $f_{\bar{B}_t}(x)=\sqrt{2/(\pi t)} {\mathrm{e}}^{-x^2/(2t)}{\textbf{1}}_{[0,\infty)}(x)$ , which implies that ${{\mathbb{E}}}[{\exp(\bar{B}_t)}]<\infty$ .

The statement of the theorem now follows by considering the processes X and $\tilde{X}$ started in x at time s and noting that the derived constant only depends on T.