Robot path planning in a dynamic environment with stochastic measurements

Abstract

We study the problem of trajectory planning for autonomous vehicles designed to minimize the travel distance while avoiding moving obstacles whose position and speed are not known. Because, in practice, observations from sensors have measurement errors, the stochasticity of the data is modeled using maximum likelihood estimators, which are shown to be consistent as the sample size increases. To comply with the kinematic constraints of the vehicle, we consider smooth trajectories that can be represented by a linear combination of B-spline basis functions, transforming the infinite-dimensional problem into a finite-dimensional one. Moreover, a smooth penalty function is added to the travel distance, transforming the constrained optimization problem into an unconstrained one. The planned stochastic trajectory, obtained from the minimization problem with stochastic confidence regions, is shown to converge almost surely to the deterministic one as the number of sensor observations increases. Finally, we present two simulation studies to demonstrate the proposed method.

Appendices

A Maximum likelihood estimation of ${\varvec{\eta }}_0^\ell , s^\ell , \alpha ^\ell$

Differentiating the likelihood $\varLambda ({\varvec{\eta }}_0^\ell , s^\ell , \alpha ^\ell |\mathbf{N}^\ell )$ in (1) with respect to ${\varvec{\eta }}_0^\ell , s^\ell , \mathbf{V}_\alpha ^\ell$ yields the equations

$$\begin{aligned} {\left\{ \begin{array}{ll} \mathbf 0 = \frac{\partial \log \varLambda ({\varvec{\eta }}_0^\ell ,s^\ell ,\alpha ^\ell )}{\partial {\varvec{\eta }}_0^\ell } =-\frac{1}{2}\sum \limits _{t=1}^T[-2(\varSigma ^\ell )^{-1}\mathbf{N}_t^\ell +2(\varSigma ^\ell )^{-1}({\varvec{\eta }}_0^\ell +ts^\ell \mathbf{V}_{\alpha }^\ell )] ,\\ 0 = \frac{\partial \log \varLambda ({\varvec{\eta }}_0^\ell ,s^\ell ,\alpha ^\ell )}{\partial s^\ell } = -\frac{1}{2}\sum \limits _{t=1}^T[-2t(\mathbf{V}_{\alpha }^\ell )^\top (\varSigma ^\ell )^{-1}\mathbf{N}_t^\ell +2(\mathbf{V}_{\alpha }^\ell )^\top (\varSigma ^\ell )^{-1}(t{\varvec{\eta }}_0^\ell +t^2s\mathbf{V}_{\alpha }^\ell )] ,\\ \mathbf 0 = \frac{\partial \log \varLambda ({\varvec{\eta }}_0^\ell ,s^\ell ,\alpha ^\ell )}{\partial \mathbf{V}_{\alpha }^\ell } =-\frac{1}{2}\sum \limits _{t=1}^T[-2ts^\ell (\varSigma ^\ell )^{-1}\mathbf{N}_t^\ell +2(\varSigma ^\ell )^{-1}(ts^\ell {\varvec{\eta }}_0^\ell +t^2 s^2 \mathbf{V}_{\alpha }^\ell )] . \end{array}\right. } \end{aligned}$$

Because the third equation is a vector equation with respect to $\alpha$ and s, the second equation is redundant, hence

$$\begin{aligned} {\left\{ \begin{array}{ll} \hat{{\varvec{\eta }}}_0^\ell = \frac{1}{T}\left[ \sum \limits _{t=1}^T\mathbf{N}_t^\ell - \hat{s}^\ell \hat{\mathbf{V}}_{\alpha }^\ell c_1\right] ,\\ \hat{s}^\ell \hat{\mathbf{V}}_{\alpha }^\ell c_2 = \sum \limits _{t=1}^Tt\mathbf{N}_t^\ell - \hat{{\varvec{\eta }}}_0^\ell c_1, \end{array}\right. } \end{aligned}$$

so that

$$\begin{aligned} \hat{{\varvec{\eta }}}_0^\ell&= \bar{\mathbf{N}}_T^\ell - \frac{c_1}{T}\left( \frac{\sum _{t=1}^T t\mathbf{N}_t^\ell - c_1\bar{\mathbf{N}}_T^\ell }{c_2 - c_1^2/T}\right) \text{ and } \hat{s}^\ell \hat{\mathbf{V}}_{\alpha }^\ell = \frac{\sum _{t=1}^Tt\mathbf{N}_t^\ell - c_1\bar{\mathbf{N}}_t}{c_2 - c_1^2/T}, \end{aligned}$$

which together with the fact that $\sin (x)/\cos (x) = \tan (x)$ yields

$$\begin{aligned} \hat{{\varvec{\eta }}}_0^\ell&= \bar{\mathbf{N}}_T^\ell - \frac{c_1}{T}\left( \frac{\sum _{t=1}^T t\mathbf{N}_t^\ell - c_1\bar{\mathbf{N}}_T^\ell }{c_2 - c_1^2/T}\right) , \\ \hat{\alpha }^\ell&= \arctan \left( \frac{\sum _{t=1}^T tN_{t2}^\ell - c_1\bar{N}_{T2}^\ell }{\sum _{t=1}^T tN_{t1}^\ell - c_1\bar{N}_{T1}^\ell }\right) \text{ and } \\ \hat{s}^\ell&= \frac{\sum _{t=1}^T tN_{t1}^\ell - c_1\bar{N}_{T1}^\ell }{(c_2 - c_1^2/T)\cos \left( \arctan \left( \frac{\sum _{t=1}^T tN_{t2}^\ell - c_1\bar{N}_{T2}^\ell }{\sum _{t=1}^T tN_{t1}^\ell - c_1\bar{N}_{T1}^\ell }\right) \right) }. \end{aligned}$$

B Variance of $\hat{{\varvec{\eta }}}^\ell _t$

The variance-covariance matrix of $\hat{{\varvec{\eta }}}_t^\ell$, the estimator of the position of obstacle $\ell$ at time t, is given by

$$\begin{aligned} \varSigma (\hat{{\varvec{\eta }}}^\ell _t)&= V(\hat{{\varvec{\eta }}}^\ell _t) = Var(\hat{{\varvec{\eta }}}_0^\ell + t\hat{s}^\ell \mathbf{V}_{\hat{\alpha }}^\ell )\\&= V\Big [\frac{1}{T}\sum _{k=1}^T\mathbf{N}_k^\ell - \frac{c_1}{T(c_2 - c_1^2/T)}\sum _{k=1}^Tk\mathbf{N}_k^\ell + \frac{c_1^2}{T^2(c_2 - c_1^2/T)}\sum _{k=1}^T\mathbf{N}_k^\ell \\&\qquad +\frac{t}{c_2 - c_1^2/T}\sum _{k=1}^Tk\mathbf{N}_k^\ell -\frac{c_1t}{T(c_2 - c_1^2/T)}\sum _{k=1}^T\mathbf{N}_k^\ell \Big ]\\&= V\Big [\sum _{k=1}^T \mathbf{N}_k^\ell \Big (\frac{1}{T} + \frac{c_1^2}{T^2(c_2 - c_1^2/T)} -\frac{c_1t}{T(c_2 - c_1^2/T)} \\&\qquad +\left( \frac{t}{c_2 - c_1^2/T} - \frac{c_1}{T(c_2 - c_1^2/T)}\right) k \Big )\Big ]\\&= \sum _{k=1}^TV\left( \mathbf{N}_k^\ell \right) \Big [ \Big (\frac{1}{T} + \frac{c_1^2}{T^2(c_2 - c_1^2/T)} -\frac{c_1t}{T(c_2 - c_1^2/T)} \\&\qquad +\left( \frac{t}{c_2 - c_1^2/T} - \frac{c_1}{T(c_2 - c_1^2/T)}\right) k \Big )\Big ]^2= d_{t}\varSigma ^\ell , \end{aligned}$$

where the fourth equality follows from the independence of ${{\varvec{\epsilon }}}_k$.

C Proofs of Lemmas and Theorems

Proof of Lemma 1

Part a) For each $\ell = 1, \ldots , L$,

$$\begin{aligned} E(\hat{{\varvec{\eta }}}_0^\ell )&= \left[ \frac{1}{T}+\frac{c_1^2}{T^2(c_2 - c_1^2/T)}\right] \sum _{t=1}^TE(\mathbf{N}_t^\ell ) -\frac{c_1}{T(c_2 - c_1^2/T)}\sum _{t=1}^TtE(\mathbf{N}_t^\ell ) \\&= \left[ \frac{1}{T}+\frac{c_1^2}{T^2(c_2 - c_1^2/T)}\right] \left[ T{\varvec{\eta }}_0^\ell + c_1s^\ell \mathbf{V}_\alpha ^\ell \right] - \frac{c_1}{T(c_2 - c_1^2/T)}\left[ c_1{\varvec{\eta }}_0^\ell + c_2s^\ell \mathbf{V}_\alpha ^\ell \right] \\&={\varvec{\eta }}_0^\ell \left[ 1+\frac{c_1^2}{T(c_2 - c_1^2/T)} - \frac{c_1^2}{T(c_2 - c_1^2/T)}\right] + s^\ell \mathbf{V}_\alpha ^\ell \left[ \frac{c_1T(c_2 - c_1^2/T) + c_1^3 - Tc_1c_2}{T^2(c_2 - c_1^2/T)} \right] \\&={\varvec{\eta }}_0^\ell . \end{aligned}$$

Writing $\hat{s}^\ell \mathbf{V}_{\hat{\alpha }}^\ell$ as $1/(c_2 - c_1^2/T)\sum _{t=1}^Tt\mathbf{N}_t^\ell - c_1/(T(c_2 - c_1^2/T))\sum _{t=1}^T\mathbf{N}_t^\ell$ we have

$$\begin{aligned} E(\hat{s}^\ell \mathbf{V}_{\hat{\alpha }}^\ell )&= \frac{1}{c_2 - c_1^2/T}\sum _{t=1}^TtE(\mathbf{N}_t^\ell )-\frac{c_1}{T(c_2 - c_1^2/T)}\sum _{t=1}^TE(\mathbf{N}_t^\ell ) \\&= \frac{1}{c_2 - c_1^2/T}\left[ c_1{\varvec{\eta }}_0^\ell + c_2s^\ell \mathbf{V}_\alpha ^\ell \right] -\frac{c_1}{T(c_2 - c_1^2/T)}\left[ T{\varvec{\eta }}_0^\ell + c_1s^\ell \mathbf{V}_\alpha ^\ell \right] \\&= s^\ell \mathbf{V}_{\alpha }^\ell \frac{Tc_2 - c_1^2}{T(c_2 - c_1^2/T)} = s^\ell \mathbf{V}_{\alpha }^\ell . \end{aligned}$$

Hence $E(\hat{{\varvec{\eta }}}_t^\ell ) = {\varvec{\eta }}_t^\ell$, completing the proof of unbiasedness.

Part b)

For each $\ell = 1, \ldots , L$,

$$\begin{aligned} \hat{{\varvec{\eta }}}_t^\ell&= \hat{{\varvec{\eta }}}_0^\ell + t \hat{s}^\ell \mathbf{V}_{\hat{\alpha }}^\ell = \left[ 1+\frac{c_1^2}{T(c_2 - c_1^2/T)}\right] \bar{\mathbf{N}}_T^\ell \\&- \frac{c_1}{(c_2 - c_1^2/T)}\frac{1}{T}\sum _{i=1}^T i \mathbf{N}_i^\ell + t\Big ( \frac{T}{c_2 - c_1^2/T}\frac{1}{T}\sum _{i=1}^T i \mathbf{N}_i^\ell - \frac{c_1}{(c_2 - c_1^2/T)}\bar{\mathbf{N}}_T^\ell \Big ). \end{aligned}$$

Since $\mathbf{N}_i^\ell , i = 1, \ldots , T$ are independent, with $E(\mathbf{N}_i^\ell ) = {\varvec{\eta }}_i^\ell$, $Var(\mathbf{N}_i^\ell ) = \varSigma ^\ell$, and $\sum _{i=1}^{\infty } \frac{1}{i^2} Var[N_{ij}^\ell ] < \infty , j = 1, 2$, it follows from Kolmogorov’s strong law ([21], Theorem 2.3.10) that $\bar{\mathbf{N}}_T^\ell - E(\bar{\mathbf{N}}_T^\ell ) {\mathop {\longrightarrow }\limits ^{a.s.}}0$.

Note that for $j = 1, 2$ and any $\epsilon > 0$, using the fact that $\sum _{i=1}^T i N_{ij}^\ell /T^2 - E(\sum _{i=1}^T i N_{ij}^\ell /T^2)$ has probability distribution $N(0,\sigma _{\ell ,j}^2(T+1)(2T+1)/(6T^3))$, one can compute the probability bound

$$\begin{aligned} P\left( \left| \frac{1}{T^2}\sum _{i=1}^T i N_{ij}^\ell - E\left( \frac{1}{T^2}\sum _{i=1}^T i N_{ij}^\ell \right) \right| > \epsilon \right)&\le E \left( \frac{1}{T^2}\sum _{i=1}^T i N_{ij}^\ell - E\left( \frac{1}{T^2}\sum _{i=1}^T i N_{ij}^\ell \right) \right) ^4\epsilon ^{-4}\\&= \frac{3\sigma _{\ell ,j}^4(T+1)^2(2T+1)^2}{36T^6}\epsilon ^{-4}. \end{aligned}$$

Since $\sum _{T = 1}^\infty 3\sigma _{\ell ,j}^4(T+1)^2(2T+1)^2/(36T^6) = \sigma _{\ell ,j}^4O(1) < \infty ,$ by the Borel–Cantelli Lemma we have

$$\begin{aligned} P\left( \left\{ \left| \frac{1}{T^2}\sum _{i=1}^T i N_{ij}^\ell - E\left( \frac{1}{T^2}\sum _{i=1}^T i N_{ij}^\ell \right) \right| > \epsilon \right\} i.o.\right) = 0, \end{aligned}$$

and hence $\sum _{i=1}^T i N_{ij}^\ell /T^2 - E(\sum _{i=1}^T i N_{ij}^\ell /T^2) {\mathop {\longrightarrow }\limits ^{a.s.}}0$. A similar argument shows that $\sum _{i=1}^T i N_{ij}^\ell /T^3 - E(\sum _{i=1}^T i N_{ij}^\ell /T^3) {\mathop {\longrightarrow }\limits ^{a.s.}}0$. Consequently $\hat{{\varvec{\eta }}}_0^\ell {\mathop {\longrightarrow }\limits ^{a.s.}}{\varvec{\eta }}_0^\ell$ and $\hat{s}^\ell \mathbf{V}_{\hat{\alpha }}^\ell {\mathop {\longrightarrow }\limits ^{a.s.}}s^\ell \mathbf{V}_{\alpha }^\ell$.

Note that

$$\begin{aligned}&\Big |\left( \hat{{\varvec{\eta }}}_0^\ell + t \hat{s}^\ell \mathbf{V}_{\hat{\alpha }}^\ell \right) - \left( {\varvec{\eta }}_0^\ell + ts\mathbf{V}_\alpha \right) \Big | \le \Big |\hat{{\varvec{\eta }}}_0^\ell - {\varvec{\eta }}_0^\ell \Big | + \Big |\hat{s}^\ell \mathbf{V}_{\hat{\alpha }}^\ell - s\mathbf{V}_\alpha \Big |T_{max} {\mathop {\longrightarrow }\limits ^{a.s.}}0, \end{aligned}$$

for any $T_{max} < \infty$, so that $\hat{{\varvec{\eta }}_t} = \hat{{\varvec{\eta }}}_0^\ell + t \hat{s}^\ell \mathbf{V}_{\hat{\alpha }}^\ell {\mathop {\longrightarrow }\limits ^{a.s.}}{\varvec{\eta }}_0^\ell + t s^\ell \mathbf{V}_{\alpha }^\ell = {\varvec{\eta }}_t$ uniformly in $t \in [0, T_{max}]$, completing the proof of Part b).

Part c)

Fix $t\in [0,T_{max}]$ and $\mathbf{x}\in \varvec{\varGamma }_t^{\gamma _T}$. By the definition of $\varvec{\varGamma }_t^{\gamma _T}$ there is some $\ell =1,\dots , L$ such that

$$\begin{aligned} \mathbf{x}\in \vartheta _t(\mathbf{N}^\ell ,\gamma _T)+B(\mathbf 0 ,r^\ell ). \end{aligned}$$

This means that there is a $\mathbf{z}\in \vartheta _t(\mathbf{N}^\ell ,\gamma _T)$ and $\mathbf{u}\in B(\mathbf 0 ,r^\ell )$ such that $\mathbf{x}=\mathbf{z}+\mathbf{u}$. It follows that

$$\begin{aligned} d(\mathbf{x},\varvec{\varGamma }_t)\le d(\mathbf{x},B({\varvec{\eta }}_t^\ell ,r^\ell ))\le d(\mathbf{x},{\varvec{\eta }}_t^\ell +\mathbf{u})&\le d(\mathbf{x},\hat{\varvec{\eta }}_t^\ell +\mathbf{u})+d(\hat{\varvec{\eta }}_t^\ell +\mathbf{u},{\varvec{\eta }}_t^\ell +\mathbf{u}))\\&\le d(\mathbf{z},\hat{\varvec{\eta }}_t^\ell )+d(\hat{\varvec{\eta }}_t^\ell ,{\varvec{\eta }}_t^\ell ). \end{aligned}$$

Since the radii of $\vartheta _t(\mathbf{N}^\ell ,\gamma _T)$ are $\sqrt{\chi ^2_2(\gamma _T)d_t\beta _{\ell ,j}/T}$, for $j=1,2$, and $d_t$ is an increasing function of t, we have

$$\begin{aligned} d(\mathbf{z},\hat{\varvec{\eta }}_t^\ell )\le \max _{\ell =1,\dots ,L}\max _{j=1,2}\sqrt{\chi ^2_2(\gamma _T)d_{T_{max}}\beta _{\ell ,j}/T}. \end{aligned}$$

Thus,

$$\begin{aligned}&\sup _{t\in [0,T_{max}]}\sup _{\mathbf{x}\in \varvec{\varGamma }_t^{\gamma _T}} d(\mathbf{x},\varvec{\varGamma }_t) \\&\quad \le \max _{\ell =1,\dots ,L}\max _{j=1,2}\sqrt{\chi ^2_2(\gamma _T)d_{T_{max}}\beta _{\ell ,j}/T}+ \sup _{t\in [0,T_{max}]} d(\hat{\varvec{\eta }}_t^\ell ,{\varvec{\eta }}_t^\ell ). \end{aligned}$$

The limit of the right-hand side of the previous inequality goes to zero as $T\rightarrow \infty$ by (C1) and Part b). This proves the claim.

Part d)

Note that by the Mean Value Theorem we can obtain

$$\begin{aligned} |Q_{\psi , \lambda , H, T}({\varvec{\theta }}) - Q_{\psi , \lambda , H}({\varvec{\theta }}) | =&\Big |\sup _{t\in [T_0,T_{{\varvec{\theta }}}]}w_{\varvec{\theta }}(t)\psi \varPhi (Z_\lambda + \sqrt{H}(r^v - d(\mathbf{v}_t({\varvec{\theta }}), \varvec{\varGamma }_t^{\gamma _T}))) \\&- \sup _{t\in [T_0,T_{{\varvec{\theta }}}]}w_{\varvec{\theta }}(t)\psi \varPhi (Z_\lambda + \sqrt{H}(r^v - d(\mathbf{v}_t({\varvec{\theta }}), \varvec{\varGamma }_t)))\Big |\\ \le&M \sup _{t\in [T_0,T_{{\varvec{\theta }}}]}|\varPhi '(c_t)\sqrt{H}(d(\mathbf{v}_t({\varvec{\theta }}), \varvec{\varGamma }_t^{\gamma _T}) - d(\mathbf{v}_t({\varvec{\theta }}), \varvec{\varGamma }_t))|\\ \le&M' \sup _{t\in [T_0,T_{{\varvec{\theta }}}]}|d(\mathbf{v}_t({\varvec{\theta }}), \varvec{\varGamma }_t^{\gamma _T}) - d(\mathbf{v}_t({\varvec{\theta }}), \varvec{\varGamma }_t)|\\ \le&M' \sup _{t\in [T_0,T_{{\varvec{\theta }}}]}\sup _{x\in \varvec{\varGamma }_t^{\gamma _T}}d(x, \varvec{\varGamma }_t) {\mathop {\longrightarrow }\limits ^{a.s.}}0, \end{aligned}$$

by Part c). $\square$

Proof of Lemma 2

We begin by showing that the family of functions $v_{t1}$, indexed by $t\in [T_0,\infty )$, is equicontinuous. Fix $t\in [T_0,\infty )$ and define the function F by

$$\begin{aligned} F({\varvec{\theta }},z)= {\left\{ \begin{array}{ll} \displaystyle \int _0^{z}\sqrt{1+\Big (\sum _{k=1}^K\theta _kB'_k(u)\Big )^2}\, du-(t-T_0)s^v &{}\text {if}\ z\le b_1\\ \displaystyle Q(f_{\varvec{\theta }})+s_{\varvec{\theta }}^f(e^{z-b_1}-1)-(t-T_0)s^v &{}\text {if}\ z>b_1, \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned} s_{\varvec{\theta }}^f=\sqrt{1+\Big (\sum _{k=1}^K\theta _kB'_k(b_1)\Big )^2}. \end{aligned}$$

It is readily checked that F is continuously differentiable. Define $w_{t}({\varvec{\theta }})$ implicitly by the equation

$$\begin{aligned} F({\varvec{\theta }},w_{t}({\varvec{\theta }}))=0 \end{aligned}$$

(7)

for $t\in [T_0,\infty )$. Notice that by (6), $w_t({\varvec{\theta }})=v_{t1}({\varvec{\theta }})$ for $t\in [T_0,T_{\varvec{\theta }}]$. Moreover, for $t>T_{\varvec{\theta }}$, we have $w_t({\varvec{\theta }})>b_1$.

By the Implicit Function Theorem, $w_{t}$ is a differentiable function of ${\varvec{\theta }}$ and for any $m=1,\dots ,K$ we have

$$\begin{aligned} \frac{\partial w_{t}}{\partial \theta _m}({\varvec{\theta }})=-\frac{\frac{\partial F}{\partial \theta _m}({\varvec{\theta }},w_t({\varvec{\theta }}))}{\frac{\partial F}{\partial z}({\varvec{\theta }},w_t({\varvec{\theta }}))}. \end{aligned}$$

When $w_t({\varvec{\theta }})\le b_1$ we have

$$\begin{aligned} \frac{\partial F}{\partial \theta _m}({\varvec{\theta }},w_t({\varvec{\theta }}))&=\int _0^{w_{t}({\varvec{\theta }})} \frac{\big (\sum _{k=1}^K\theta _kB_k'(u)\big )B'_m(u)}{\sqrt{\big (1+\big [\sum _{k=1}^K\theta _kB_k'(u)\big ]^2\big )}}\, du,\\ \frac{\partial F}{\partial z}({\varvec{\theta }},w_t({\varvec{\theta }}))&=\sqrt{1+\left[ \sum _{k=1}^K\theta _kB'_k(w_t({\varvec{\theta }}))\right] ^2} \end{aligned}$$

and hence

$$\begin{aligned} \Big | \frac{\partial w_{t}}{\partial \theta _m}({\varvec{\theta }})\Big |&\le \int _0^{w_{t}({\varvec{\theta }})} \frac{\left| \displaystyle \sum _{k=1}^K\theta _kB_k'(u)\Vert B'_m(u)\right| du}{\displaystyle \sqrt{\left( 1+\left[ \sum _{k=1}^K\theta _kB_k'(u)\right] ^2\right) \left( 1+\left[ \sum _{k=1}^K\theta _kB_k'(w_{t}({\varvec{\theta }}))\right] ^2\right) }},\\&\le \int _0^{b_1}|B_m'(u)|\, du \le M b_1, \end{aligned}$$

where $M = \max _{1\le m\le K}\sup _{0\le u\le b_1} |B_m'(u)|$ is independent of $t\in [T_0,\infty )$. On the other hand, when $w_t(\theta )>b_1$ we have

$$\begin{aligned} \frac{\partial F}{\partial \theta _m}({\varvec{\theta }},w_t({\varvec{\theta }}))&=\int _0^{b_1} \frac{\big (\sum _{k=1}^K\theta _kB_k'(u)\big )B'_m(u)}{\sqrt{\big (1+\big [\sum _{k=1}^K\theta _kB_k'(u)\big ]^2\big )}}\, du\\&\qquad +\frac{\big ( \sum _{k=1}^K \theta _kB'_k(b_1)\big ) B'_m(b_1)(e^{w_t({\varvec{\theta }})-b_1}-1)}{\sqrt{\big (1+\big [\sum _{k=1}^K\theta _kB_k'(b_1)\big ]^2\big )}}, \\ \frac{\partial F}{\partial z}({\varvec{\theta }},w_t({\varvec{\theta }}))&=s_{\varvec{\theta }}^fe^{w_t({\varvec{\theta }})-b_1} \end{aligned}$$

and hence

$$\begin{aligned} \Big | \frac{\partial w_{t}}{\partial \theta _m}({\varvec{\theta }})\Big |&\le \int _0^{b_1} \frac{|\sum _{k=1}^K\theta _kB_k'(u) ||B'_m(u)|}{s_{\varvec{\theta }}^fe^{w_t({\varvec{\theta }})-b_1}\sqrt{\big (1+\big [\sum _{k=1}^K\theta _kB_k'(u)\big ]^2\big )}}\, du\\&\qquad +\frac{| \sum _{k=1}^K \theta _kB'_k(b_1)| |B'_m(b_1)|(e^{w_t({\varvec{\theta }})-b_1}-1)}{s_{\varvec{\theta }}^fe^{w_t({\varvec{\theta }})-b_1}\sqrt{\big (1+\big [\sum _{k=1}^K\theta _kB_k'(b_1)\big ]^2\big )}}\\&\le \int _0^{b_1}|B_m'(u)|\, du+|B'_m(b_1)| \le M b_1 +M. \end{aligned}$$

It follows that regardless of the value of ${\varvec{\theta }}$ or t it is true that

$$\begin{aligned} \Big | \frac{\partial w_{t}}{\partial \theta _m}({\varvec{\theta }})\Big |\le Mb_1+M. \end{aligned}$$

This implies that $w_t$ is Lipschitz—that is, there is a $N>0$ such that

$$\begin{aligned} |w_{t}({\varvec{\theta }})-w_{t}({\varvec{\phi }})|\le N\Vert {\varvec{\theta }}-{\varvec{\phi }}\Vert \end{aligned}$$

for all ${\varvec{\theta }},{\varvec{\phi }}\in \mathbb {R}^K$. Notice that N is independent of t. To see that $v_{1t}$ is also Lipschitz, fix ${\varvec{\theta }}$ and ${\varvec{\phi }}$ and consider several cases. If

$$\begin{aligned} t\le \min \{T_{\varvec{\theta }},T_{\varvec{\phi }}\}, \end{aligned}$$

then since $v_{1t}({\varvec{\phi }})=w_t({\varvec{\phi }})$ for $t\in [T_0,T_{\varvec{\phi }}]$ for all ${\varvec{\phi }}\in \mathbb {R}^K$ we have

$$\begin{aligned} |v_{1t}({\varvec{\theta }})-v_{1t}({\varvec{\phi }})|=|w_{t}({\varvec{\theta }})-w_{t}({\varvec{\phi }})|\le N\Vert {\varvec{\theta }}-{\varvec{\phi }}\Vert . \end{aligned}$$

If

$$\begin{aligned} T_{\varvec{\theta }}\le t <T_{\varvec{\phi }}, \end{aligned}$$

then we know that $v_{1t}({\varvec{\phi }})< b_1=v_{1t}({\varvec{\theta }})\le w_t({\varvec{\theta }})$. Hence,

$$\begin{aligned} |v_{1t}({\varvec{\theta }})-v_{1t}({\varvec{\phi }})|=b_1-v_{1t}({\varvec{\phi }}) \le w_{t}({\varvec{\theta }})-w_{t}({\varvec{\phi }})\le N\Vert {\varvec{\theta }}-{\varvec{\phi }}\Vert . \end{aligned}$$

A similar argument works if

$$\begin{aligned} T_{\varvec{\phi }}\le t <T_{\varvec{\theta }}. \end{aligned}$$

Finally, if

$$\begin{aligned} t\ge \max \{T_{\varvec{\theta }},T_{\varvec{\phi }}\}, \end{aligned}$$

then since $v_{1t}({\varvec{\theta }})=v_{1t}({\varvec{\phi }})=b_1$,

$$\begin{aligned} |v_{t1}({\varvec{\theta }})-v_{t1}({\varvec{\phi }})|=0\le N\Vert {\varvec{\theta }}-{\varvec{\phi }}\Vert . \end{aligned}$$

Thus, $v_{1t}$ is also Lipschitz with Lipschitz constant N independent of t. It follows that $v_{t1}$ is equicontinuous.

Moreover, since the $B_k$ are independent of t and are continuous, it follows that the family of functions

$$\begin{aligned} {\varvec{\theta }}\mapsto \sum _{k=1}^K\theta _kB_k(v_{t1}({\varvec{\theta }})) \end{aligned}$$

is also equicontinuous. Hence, the family $\mathbf{v}_t$ defined in (5), indexed by $t\in [T_0,\infty )$, is equicontinuous. $\square$

Proof of Theorem 1

Define

$$\begin{aligned} g_t({\varvec{\theta }})={\left\{ \begin{array}{ll} w_{\varvec{\theta }}(t) \psi \varPhi (Z_\lambda + \sqrt{H}(r^v - d(\mathbf{v}_t({\varvec{\theta }}), \varvec{\varGamma }_t^{\gamma _T})))&{}\text {if}\ t\le T_{\varvec{\theta }}\\ w_{\varvec{\theta }}(T_{\varvec{\theta }}) \psi \varPhi (Z_\lambda + \sqrt{H}(r^v - d(\mathbf{v}_{T_{\varvec{\theta }}}({\varvec{\theta }}), \varvec{\varGamma }_{T_{\varvec{\theta }}}^{\gamma _T}))) &{} \text {if}\ t>T_{\varvec{\theta }}\end{array}\right. } \end{aligned}$$

for all ${\varvec{\theta }}\in \mathbb {R}^K$. To show the continuity of $Q_{\psi , \lambda , H, T}$ it suffices to prove the continuity of the function ${\varvec{\theta }}\mapsto \sup _{t\in [T_0,\infty )}g_t({\varvec{\theta }})$. Since the family of functions $\mathbf{v}_t$ and the family of functions ${\varvec{\theta }}\mapsto w_{\varvec{\theta }}(t)$ are equicontinuous, the family of functions $g_t$ for $t\in [T_0,\infty )$ is also equicontinuous. Moreover, because

$$\begin{aligned} |\sup _{t\in [T_0,\infty )} g_t({\varvec{\theta }})-\sup _{t\in [T_0,\infty )}g_t({\varvec{\phi }})|\le \sup _{t\in [T_0,\infty )}|g_t({\varvec{\theta }})-g_t({\varvec{\phi }})|, \end{aligned}$$

the function ${\varvec{\theta }}\mapsto \sup _{t\in [T_0,\infty )}g_t({\varvec{\theta }})$ is continuous. The proof for $Q_{\psi , \lambda , H}$ is similar. $\square$

Proof of Lemma 3

We prove the result for $Q_{\psi ,\lambda ,H}$. The proof for $Q_{\psi ,\lambda ,H,T}$ is similar. Let ${\varvec{\theta }}\in \mathbb {R}^K$ be given. If ${\varvec{\theta }}=\mathbf 0$, then the desired result holds for any $\mu >0$. Suppose ${\varvec{\theta }}\not = \mathbf 0$. Find m such that $\theta _m\not = 0$ and $|\theta _j|\le |\theta _m|$, and hence $-1\le \frac{\theta _j}{\theta _m}\le 1$, for all $j=1,\dots ,K$. We have

$$\begin{aligned} Q_{\psi ,\lambda ,H}({\varvec{\theta }})&\ge \int _0^{b_1}\sqrt{1 + \left( \sum _{k=1}^K\theta _kB_k'(u)\right) ^2}\,du \ge \int _0^{b_1}\Big |\sum _{j=1}^K\theta _jB'_j(u)\Big |\, du \end{aligned}$$

(8)

$$\begin{aligned}&= \Big (\int _0^{b_1}\Big |\sum _{j=1}^K\frac{\theta _j}{\theta _m}B'_j(u)\Big |\, du\Big ) |\theta _m| \ge M |\theta _m| \end{aligned}$$

(9)

where M is given by

$$\begin{aligned} M=\inf \Big \{\int _0^{b_1}\Big |\sum _{j=1}^K\xi _j B'_j(u)\Big |\, du\ \Big |\ {{\varvec{\xi }}}\in [-1,1]^K, \xi _k=1\ \text {for some}\ k=1,\dots ,K\Big \}. \end{aligned}$$

Notice that (8) implies that $Q_{\psi ,\lambda ,H}({\varvec{\theta }})\ge \frac{M}{\sqrt{K}}\Vert {\varvec{\theta }}\Vert$. Thus, if we show that $M>0$, the result holds with $\mu =\frac{M}{\sqrt{K}}$.

Suppose for the sake of contradiction that $M=0$. It follows that for each $n\in \mathbb {N}$ we can find a ${{\varvec{\xi }}}_n\in [-1,1]^K$ such that at least one of the components of ${{\varvec{\xi }}}_n$ is equal to one and

$$\begin{aligned} \frac{1}{n}\ge \int _0^{b_1}\Big |\sum _{j=1}^K\xi _{nj} B'_j(u)\Big |\, du. \end{aligned}$$

(10)

Since ${{\varvec{\xi }}}_n\in [-1,1]^K$ we have that the sequence ${{\varvec{\xi }}}_n$ is bounded and hence has a subsequence that converges to some ${{\varvec{\xi }}}_0\in [-1,1]^K$ such that $\xi _{0m}=1$ for some $m=1,\dots ,K$. Moreover, from (10) ${{\varvec{\xi }}}_0$ must satisfy

$$\begin{aligned} \int _0^{b_1}\Big |\sum _{j=1}^K\xi _{0j} B'_j(u)\Big |\, du=0 \end{aligned}$$

and hence

$$\begin{aligned} \sum _{j=1}^K\xi _{0j} B'_j(u)=0\quad \text {for all}\ u\in [0,b_1]. \end{aligned}$$

Since $\xi _{0m}=1$, this implies that $B'_m$ is a linear combination of $B'_j$ for $j\not = m$, which is not true. Thus, we must have $M>0$. $\square$

Proof of Corollary 1

We prove that $Q_{\psi ,\lambda ,H}$ has a minimizer. The proof for $Q_{\psi ,\lambda ,H,T}$ is similar. Let ${\varvec{\theta }}_n$, for $n\in \mathbb {N}$, be a minimizing sequence for $Q_{\psi ,\lambda ,H}$ so that

$$\begin{aligned} \inf _{{\varvec{\theta }}\in \mathbb {R}^K} Q_{\psi ,\lambda ,H}({\varvec{\theta }})+\frac{1}{n}\ge Q_{\psi ,\lambda ,H}({\varvec{\theta }}_n). \end{aligned}$$

(11)

It follows from (11) and Lemma 3 that the sequence ${\varvec{\theta }}_n$ is bounded and hence there is a subsequence that converges to some ${\varvec{\theta }}_0\in \mathbb {R}^K$. Taking the limit of (11) as n goes to infinity along this subsequence and utilizing the continuity of $Q_{\psi ,\lambda ,H}$ implies that

$$\begin{aligned} \inf _{{\varvec{\theta }}\in \mathbb {R}^k} Q_{\psi ,\lambda ,H}({\varvec{\theta }})\ge Q_{\psi ,\lambda ,H}({\varvec{\theta }}_0). \end{aligned}$$

Thus, ${\varvec{\theta }}_0$ is a minimizer. $\square$

Proof of Lemma 4

By the independence of the observations, for any $T_{max}$ we have

$$\begin{aligned} P(\varvec{\varGamma }_t \subset \varvec{\varGamma }_t^{\gamma _T} \text{ for } \text{ all } t \in [0, T_{max}])&= P(\varvec{\varGamma }_{T_{max}} \subset \varvec{\varGamma }_{T_{max}}^{\gamma _T}) \\&\ge \prod _{\ell = 1}^LP\Big (B({\varvec{\eta }}_{T_{max}}^\ell ,r^\ell ) \subset \vartheta _t(\mathbf{N}^\ell , \gamma _T) + B(\mathbf 0 ,r^\ell ) \Big ) \\&=\prod _{\ell = 1}^LP\Big ( \mathbf{z}\in \vartheta _{T_{max}}(\mathbf{N}^\ell , \gamma _T) + B(\mathbf 0 ,r^\ell ),\ \forall \mathbf{z}\in B({\varvec{\eta }}_{T_{max}}^\ell ,r^\ell ) \Big )\\&\ge \prod _{\ell = 1}^LP\big ({\varvec{\eta }}_{T_{max}}^\ell \in \vartheta _{T_{max}}(\mathbf{N}^\ell , \gamma _T) \big ) = (1 - \gamma _T)^L. \end{aligned}$$

follows from the Borel-Cantelli Lemma. $\square$

Proof of Lemma 5

By Lemma 4, for any $\omega \in F^c$, where $F = \big \{\{ \varvec{\varGamma }_t \not \subset \varvec{\varGamma }_t^{\gamma _T} \text{ for } \text{ some } t \in [0, T_{\varvec{\theta }}]\} \; i.o.\big \}$, there exists $n_0(\omega )$ such that for all $T \ge n_0(\omega )$ we have $\varvec{\varGamma }_t \subset \varvec{\varGamma }_t^{\gamma _T}(\omega )$ for all $t \in [0, T_{\varvec{\theta }}]$. Thus, for any ${\varvec{\theta }}\in \mathbb {R}^K$, any $\omega \in F^c$, and all $T \ge n_0(\omega )$

$$\begin{aligned} d\left( \mathbf{v}_t({\varvec{\theta }}), \varvec{\varGamma }_t^{\gamma _T}(\omega )\right) \le d\left( \mathbf{v}_t({\varvec{\theta }}), \varvec{\varGamma }_t\right) \text{ for } \text{ all } t \in [0, T_{\varvec{\theta }}]. \end{aligned}$$

Hence, for any ${\varvec{\theta }}$, $\lambda$, $\psi$, H, $\omega \in F^c$, and for all $T \ge n_o(\omega )$

$$\begin{aligned}&\sup _{t \in [T_0,T_{\varvec{\theta }}]} w_{\varvec{\theta }}(t) \psi \varPhi (Z_\lambda + \sqrt{H}(r^v - d(\mathbf{v}_t({\varvec{\theta }}), \varvec{\varGamma }_t^{\gamma _T}(\omega ))))\\& \ge \sup _{t \in [T_0,T_{\varvec{\theta }}]} w_{\varvec{\theta }}(t) \psi \varPhi (Z_\lambda + \sqrt{H}(r^v - d(\mathbf{v}_t({\varvec{\theta }}), \varvec{\varGamma }_t))) \end{aligned}$$

since $\varPhi$ is strictly increasing and H is positive.

Hence, for all $\omega \in F^c$ we have $Q_{\psi , \lambda , H, T}({\varvec{\theta }}, \omega ) \ge Q_{\psi , \lambda , H}({\varvec{\theta }})$ for all $T \ge n_0(\omega )$, so that $\{Q_{\psi , \lambda , H, T}({\varvec{\theta }}) < Q_{\psi , \lambda , H}({\varvec{\theta }})\;\; i.o.\}$ has probability 0. $\square$

Proof of Theorem 2

Utilizing Theorem 2.1 in [15] it suffices to show that $Q_{\psi , \lambda , H, T}$ epi-converges a.s. to $Q_{\psi , \lambda , H}$ and that there is a compact set $C\subset \mathbb {R}^K$ and an $\alpha \in \mathbb {R}$ such that

$$\begin{aligned} \emptyset \not =\{{\varvec{\theta }}\in \mathbb {R}^K\ |\ Q_{\psi , \lambda , H, T}({\varvec{\theta }})\le \alpha \}\subset C\quad \text {for all}\ T\in \mathbb {N}. \end{aligned}$$

(12)

The proof that $Q_{\psi , \lambda , H, T}$ epi-converges a.s. to $Q_{\psi , \lambda , H}$, which relies on Lemmas 1 and 5 and Theorem 1, is identical to the argument found in the proof of Theorem 3.2 in [6]. Thus, it will not be presented here.

To establish the existence of a compact C and $\alpha \in \mathbb {R}$ such that (12) holds begin by defining the set

$$\begin{aligned} R:=\{{\varvec{\theta }}\in \mathbb {R}^K\ |\ Q_{\psi , \lambda , H, T}({\varvec{\theta }})\le \inf _{{\varvec{\phi }}\in \mathbb {R}^K} Q_{\psi , \lambda , H}({\varvec{\phi }}) +2\psi \text { for all } T\in \mathbb {N}\}. \end{aligned}$$

This set is bounded because if ${\varvec{\theta }}\in R$, then by Lemma 3 we have for any $T\in \mathbb {N}$

$$\begin{aligned} \mu \Vert {\varvec{\theta }}\Vert \le Q_{\psi , \lambda , H, T}({\varvec{\theta }})\le \inf _{{\varvec{\phi }}\in \mathbb {R}^K} Q_{\psi , \lambda , H}({\varvec{\phi }})+2\psi . \end{aligned}$$

Thus, if we set C equal to the closure of R, C is compact. Now choose

$$\begin{aligned} \alpha :=\inf _{{\varvec{\phi }}\in \mathbb {R}^K} Q_{\psi , \lambda , H}({\varvec{\phi }})+\psi , \end{aligned}$$

and for any $T\in \mathbb {N}$ set

$$\begin{aligned} A_T:=\{{\varvec{\theta }}\in \mathbb {R}^K\ |\ Q_{\psi , \lambda , H, T}({\varvec{\theta }})\le \alpha \}. \end{aligned}$$

Notice that $A_T$ is not empty since $\mathbf 0 \in A_T$. Also, for any natural number $T'$ and ${\varvec{\theta }}\in A_T$ we have

$$\begin{aligned} Q_{\psi , \lambda , H, T'}({\varvec{\theta }})&=\int _0^{b_1}\sqrt{1+\Big (\sum _{k=1}^K\theta _kB'_k(u)\Big )^2} du +J_{\psi , \lambda , H, T'}(f_{\varvec{\theta }})\\&\le \int _0^{b_1}\sqrt{1+\Big (\sum _{k=1}^K\theta _kB'_k(u)\Big )^2} du +J_{\psi , \lambda , H, T}(f_{\varvec{\theta }})+\psi \\&= Q_{\psi , \lambda , H, T}({\varvec{\theta }})+\psi \le \inf _{{\varvec{\phi }}\in \mathbb {R}^K}Q_{\psi , \lambda , H}({\varvec{\phi }})+2\psi , \end{aligned}$$

and thus ${\varvec{\theta }}\in R$. This shows that $A_T\subset R\subset C$. $\square$