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Robustness of movement models: can models bridge the gap between temporal scales of data sets and behavioural processes?

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Abstract

Discrete-time random walks and their extensions are common tools for analyzing animal movement data. In these analyses, resolution of temporal discretization is a critical feature. Ideally, a model both mirrors the relevant temporal scale of the biological process of interest and matches the data sampling rate. Challenges arise when resolution of data is too coarse due to technological constraints, or when we wish to extrapolate results or compare results obtained from data with different resolutions. Drawing loosely on the concept of robustness in statistics, we propose a rigorous mathematical framework for studying movement models’ robustness against changes in temporal resolution. In this framework, we define varying levels of robustness as formal model properties, focusing on random walk models with spatially-explicit component. With the new framework, we can investigate whether models can validly be applied to data across varying temporal resolutions and how we can account for these different resolutions in statistical inference results. We apply the new framework to movement-based resource selection models, demonstrating both analytical and numerical calculations, as well as a Monte Carlo simulation approach. While exact robustness is rare, the concept of approximate robustness provides a promising new direction for analyzing movement models.

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Acknowledgments

We thank members of the Lewis research group for fruitful discussions and an anonymous reviewer for helpful comments on the manuscript. UES was supported by a scholarship from iCORE, now part of Alberta Innovates-Technology Futures and funding from the University of Alberta. MAL gratefully acknowledges Natural Sciences and Engineering Research Council Discovery and Accelerator grants, a Canada Research Chair and a Killam Research Fellowship.

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Correspondence to Ulrike E. Schlägel.

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Appendices

Appendix 1: Proofs of results about exact robustness

Proof

(Theorem 1) First, note that for any standard deviation of the kernel, \(\sigma \), the integral \(\int _{\mathbb {R}} k_{\sigma }(y;x)w(y)\,\mathrm {d}y\) reduces to the weighting function evaluated at the kernel’s mean,

$$\begin{aligned}&\int _{\mathbb {R}} k_{\sigma }(y;x)w(y)\,\mathrm {d}y {=} \int _{\mathbb {R}} k_{\sigma }(y;x)(ay+b)\,\mathrm {d}y {=} \int _{\mathbb {R}} k_{\sigma }(y;x)(a(y-x+x)+b)\,\mathrm {d}y \nonumber \\&\quad = (ax+b)\int _{\mathbb {R}} k_{\sigma }(y;x)\,\mathrm {d}y + a\int _{\mathbb {R}} k_{\sigma }(y;x)(y-x)\,\mathrm {d}y = ax+b = w(x), \end{aligned}$$
(24)

because \(k_{\sigma }(\cdot | y)\) is a Gaussian density integrating to 1 and with vanishing first central moment. If we consider w as a linear transformation of a Normally distributed random variable with mean x, then Eq. (24) reflects a special case of Jensen’s inequality, in which equality holds.

We now show robustness of degree n with parameter transformation \(g_n(\sigma ,a,b) = (\sqrt{n}\sigma ,a,b)\) by induction. For \(n=1\), we have the trivial transformation \(g_1(\sigma , a, b) = (\sigma , a, b)\), and there is nothing to show for robustness of degree 1.

We assume that robustness or degree n holds, that is we have the relationship

$$\begin{aligned} p_{n}(x_n | x_0, \sigma , a, b) = p_1(x_n|x_0, \sqrt{n}\sigma , a, b). \end{aligned}$$
(25)

for all \(x_n, x_0 \in \mathbb {R}\). For \(n+1\), we use the Chapman–Kolmogorov equation and Markov property and obtain

$$\begin{aligned}&p_{n+1}(x_{n+1}|x_0, \sigma , a, b) = \int _{\mathbb {R}^n} \prod _{k=1}^{n+1} p_1(x_k|x_{k-1}, \sigma , a, b) \, \mathrm {d}x_1 \dots \mathrm {d}x_n \nonumber \\&\qquad = \int _{\mathbb {R}} p_1(x_{n+1}|x_n, \sigma , a, b) \left( \int _{\mathbb {R}^{n-1}} \prod _{k=1}^{n} p_1(x_k|x_{k-1}, \sigma , a, b) \, \mathrm {d}x_1 \dots \mathrm {d}x_{n-1} \right) \mathrm {d}x_n \nonumber \\&\qquad = \int _{\mathbb {R}} p_1(x_{n+1}|x_n, \sigma , a, b) \, p_{n}(x_n|x_0, \sigma , a, b) \, \mathrm {d}x_n \nonumber \\&\qquad = \int _{\mathbb {R}} p_1(x_{n+1}|x_n, \sigma , a, b) \, p_1(x_n|x_0, \sqrt{n}\sigma , a, b) \, \mathrm {d}x_n, \end{aligned}$$
(26)

where the last step follows by induction. We can now insert the model’s step probabilities and use Eq. (24) to further calculate,

$$\begin{aligned} p_{n+1}(x_{n+1}|x_0, \sigma , a, b)&= \int _{\mathbb {R}} \frac{k_{\sigma }(x_{n+1};x_n)\,w(x_{n+1})}{\int _{\mathbb {R}} k_{\sigma }(y;x_n) w(y) \, dy } \, \frac{k_{\sqrt{n}\,\sigma }(x_n;x_0)\,w(x_n)}{\int _{\mathbb {R}} k_{\sqrt{n}\,\sigma }(y;x_0) w(y)\, dy} \, \mathrm {d}x_n \nonumber \\&= \int _{\mathbb {R}} \frac{k_{\sigma }(x_{n+1};x_n)\,w(x_{n+1})}{w(x_n)} \, \frac{k_{\sqrt{n}\,\sigma }(x_n;x_0)\,w(x_n)}{w(x_0)} \, \mathrm {d}x_n \nonumber \\&= \frac{w(x_{n+1})}{w(x_0)} \int _{\mathbb {R}} k_{\sigma }(x_{n+1};x_n)\,k_{\sqrt{n}\sigma }(x_n;x_0) \, \mathrm {d}z. \end{aligned}$$
(27)

Note that we have assumed that all movement steps are within the domain \(\mathscr {I}\), where the weighting function is positive. Since \(k_{\sigma }(x_{n+1};x_n) = k_{\sigma }(x_{n+1}-x_n;0)\), the integral in the last expression is the convolution of two Gaussian densities with variances \(\sigma ^2\) and \(n\sigma ^2\) and with means 0 and \(x_0\), respectively. Because of the linearity of Gaussian random variables, this is again a Gaussian density with mean \(x_0\) and variance \((n+1)\sigma ^2\). Because Eq. (24) holds for the kernel with any standard deviation, we can rewrite the denominator as \(w(x_0) = \int _{\mathbb {R}} k_{\sqrt{n+1}\,\sigma }(y;x_0) w(y) \, dy\). Thus,

$$\begin{aligned} p_{n+1}(x_{n+1}|x_0, \sigma , a, b) = \frac{k_{\sqrt{n+1}\,\sigma }(x_{n+1};x_0) w(x_{n+1})}{\int _{\mathbb {R}} k_{\sqrt{n+1}\,\sigma }(y;x_0) w(y) \, dy} {=} p_{1}(x_{n+1}|x_0,\sqrt{n+1}\,\sigma , a, b). \end{aligned}$$
(28)

\(\square \)

Proof

(Theorem 2) We proceed analogously to the previous proof. The integral of weighting function and kernel with arbitrary standard deviation \(\sigma \) and mean x is here given by

$$\begin{aligned} \int _{\mathbb {R}} k_{\sigma }(y;x) \, w(y)\,\mathrm {d}y&= \int _{\mathbb {R}} k_{\sigma }(y;x)\,C e^{ay+b}\,\mathrm {d}y \\&= \frac{C}{\sqrt{2\pi }\sigma } \int _{\mathbb {R}} \exp \left( -\frac{(y-x)^2}{2\sigma ^2} + ay + b\right) \mathrm {d}y. \end{aligned}$$

By completing the square and using substitution \(u=\frac{1}{\sqrt{2} \sigma } (y-x-a\sigma ^2)\) we obtain

$$\begin{aligned} \int _{\mathbb {R}} k_{\sigma }(y;x) \, w(y)\,\mathrm {d}y&= \frac{C}{\sqrt{2\pi }\sigma } e^{ \frac{a^2\sigma ^2}{2} +ax +b } \int _{\mathbb {R}} \exp \left( - \left( \frac{y-x-a\sigma ^2}{\sqrt{2}\sigma } \right) ^2 \right) \mathrm {d}y \\&= \frac{C}{\sqrt{2\pi }\sigma } e^{ \frac{a^2\sigma ^2}{2} +ax +b } \int _{\mathbb {R}} \exp \left( - u^2 \right) \sqrt{2} \sigma \, \mathrm {d}u. \end{aligned}$$

The final integral reduces to \(\sqrt{2\pi }\sigma \), and therefore,

$$\begin{aligned} \int _{\mathbb {R}} k_{\sigma }(y;x) \, w(y)\,\mathrm {d}y&= C\, e^{\frac{a^2\sigma ^2}{2} + ax + b}. \end{aligned}$$
(29)

Again, we prove robustness of degree n by induction, using parameter transformation \(g_n(\sigma , C, a, b) = (\sqrt{n}\sigma , C,a,b)\). In the induction step, we obtain, with help of Eq. (29),

$$\begin{aligned} p_{n+1}(x_{n+1}|x_0, \sigma , a, b)&= \int _{\mathbb {R}} \frac{k_{\sigma }(x_{n+1};x_n)\,Ce^{ax_{n+1}+b}}{\int _{\mathbb {R}} k_{\sigma }(y;x_n) Ce^{ay+b} \, dy } \, \frac{k_{\sqrt{n}\,\sigma }(x_n;x_0)\,Ce^{ax_n+b}}{\int _{\mathbb {R}} k_{\sqrt{n}\,\sigma }(y;x_0) Ce^{ay+b}\, dy} \, \mathrm {d}x_n \nonumber \\&= \int _{\mathbb {R}} \frac{k_{\sigma }(x_{n+1};x_n)\,Ce^{ax_{n+1}+b}}{Ce^{\frac{a^2\sigma ^2}{2} + ax_n + b}} \, \frac{k_{\sqrt{n}\,\sigma }(x_n;x_0)\,Ce^{ax_n+b}}{Ce^{\frac{na^2\sigma ^2}{2} + ax_0 + b}} \, \mathrm {d}x_n \nonumber \\&= \frac{e^{x_{n+1}}}{e^{\frac{(n+1)a^2\sigma ^2}{2}+ax_0}} \int _{\mathbb {R}} k_{\sigma }(x_{n+1};x_n)\,k_{\sqrt{n}\,\sigma }(x_n;x_0) \, \mathrm {d}z \nonumber \\&= \frac{e^{x_{n+1}}}{e^{\frac{(n+1)a^2\sigma ^2}{2}+ax_0}} \, k_{\sqrt{n+1}\,\sigma }(x_{n+1}; x_0). \nonumber \\&= \frac{k_{\sqrt{n+1}\,\sigma }(x_{n+1}; x_0) \, C e^{ax_{n+1}+b}}{\int _{\mathbb {R}} k_{\sqrt{n+1}\,\sigma }(y; x_0) \, C e^{ay+b} \,\mathrm {d}y} \nonumber \\&= p_{1}(x_{n+1}|x_0, \sqrt{n+1}\,\sigma , a, b) \end{aligned}$$
(30)

\(\square \)

Appendix 2: Proof of result about asymptotic robustness

To highlight the main steps necessary to prove Theorem 3, we establish a series of intermediate results. As a first step, we show that the 2-step transition density can be broken up into a product of the form (5) in Definition 2.

Proposition 1

The 2-step transition density of model with transitions (14) can be written as

$$\begin{aligned} p_2(x_t|x_{t-2\tau }, \sigma , \varvec{\theta }) = p_1(x_t|x_{t-2\tau },\sqrt{2}\sigma ,\varvec{\theta }) \cdot v(x_t, x_{t-2\tau }; \tau ), \end{aligned}$$
(31)

where the function v is given by

$$\begin{aligned} v(x_t, x_{t-2\tau }; \tau )= & {} \frac{\int _{\mathbb {R}} k_{\sqrt{2}\sigma }(y;x) w_{\varvec{\theta }}(y)\, \mathrm {d}y}{\int _{\mathbb {R}} k_{\sigma }(y;x) w_{\varvec{\theta }}(y)\, \mathrm {d}y} \int _{\mathbb {R}} k_{\frac{\sigma }{\sqrt{2}}}\Bigl (z; \frac{1}{2}(x_t+x_{t-2\tau })\Bigr ) \nonumber \\&\times \, \frac{w_{\varvec{\theta }}(z)}{\int _{\mathbb {R}} k_{\sigma }(y;z) w_{\varvec{\theta }}(y) \, \mathrm {d}y} \, \mathrm {d}z. \end{aligned}$$
(32)

Note that v depends on \(\tau \) through \(\sigma \). For later convenience, we define

$$\begin{aligned} Q(x; \tau )&:= \frac{\int _{\mathbb {R}} k_{\sqrt{2}\sigma }(y;x) w_{\varvec{\theta }}(y)\, \mathrm {d}y}{\int _{\mathbb {R}} k_{\sigma }(y;x) w_{\varvec{\theta }}(y)\, \mathrm {d}y} \end{aligned}$$
(33)
$$\begin{aligned} I(x_1,x_2; \tau )&:= \int _{\mathbb {R}} k_{\frac{\sigma }{\sqrt{2}}}\Bigl (z; \frac{1}{2}(x_1+x_2)\Bigr ) \, \frac{w_{\varvec{\theta }}(z)}{\int _{\mathbb {R}} k_{\sigma }(y;z) w_{\varvec{\theta }}(y) \, \mathrm {d}y} \, \mathrm {d}z. \end{aligned}$$
(34)

Proof

The proposition can be shown with a straightforward calculation. The 2-step transition density is given by

$$\begin{aligned} p_2(x_t|x_{t-2\tau }&,\sigma ,\varvec{\theta }) \end{aligned}$$
(35)
$$\begin{aligned}&= \int _{\mathbb {R}} \frac{k_{\sigma }(x_t;z)\,w_{\varvec{\theta }}(x_t)}{\int _{\mathbb {R}} k_{\sigma }(y;z) w_{\varvec{\theta }}(y) \, \mathrm {d}y } \, \frac{k_{\sigma }(z;x_{t-2\tau })\,w_{\varvec{\theta }}(z)}{\int _{\mathbb {R}} k_{\sigma }(y;x_{t-2\tau }) w_{\varvec{\theta }}(y)\, \mathrm {d}y} \, \mathrm {d}z \end{aligned}$$
(36)
$$\begin{aligned}&= \frac{w_{\varvec{\theta }}(x_t)}{\int _{\mathbb {R}} k_{\sigma }(y;x_{t-2\tau }) w_{\varvec{\theta }}(y)\, \mathrm {d}y} \int _{\mathbb {R}} k_{\sigma }(x_t;z)\,k_{\sigma }(z;x_{t-2\tau }) \nonumber \\&\quad \times \,\frac{w_{\varvec{\theta }}(z)}{\int _{\mathbb {R}} k_{\sigma }(y;z) w_{\varvec{\theta }}(y) \, \mathrm {d}y} \, \mathrm {d}z. \end{aligned}$$
(37)

The product of the two Gaussian densities in the integrand can be transformed as follows

$$\begin{aligned} k_{\sigma }(x_t;z)\,k_{\sigma }(z;x_{t-2\tau }) = k_{\sqrt{2}\sigma }(x_t;x_{t-2\tau }) \, k_{\frac{\sigma }{\sqrt{2}}}\Bigl (z; \frac{1}{2}(x_t+x_{t-2\tau })\Bigr ). \end{aligned}$$
(38)

The two-step density therefore becomes

$$\begin{aligned}&p_2(x_t|x_{t-2\tau },\sigma ,\varvec{\theta }) \nonumber \\&\qquad = \frac{k_{\sqrt{2}\sigma }(x_t;x_{t-2\tau }) \,w_{\varvec{\theta }}(x_t)}{\int _{\mathbb {R}} k_{\sigma }(y;x_{t-2\tau }) w_{\varvec{\theta }}(y)\, \mathrm {d}y} \int _{\mathbb {R}} k_{\frac{\sigma }{\sqrt{2}}}\Bigl (z; \frac{1}{2}(x_t+x_{t-2\tau })\Bigr ) \nonumber \\&\qquad \quad \times \, \frac{w_{\varvec{\theta }}(z)}{\int _{\mathbb {R}} k_{\sigma }(y;z) w_{\varvec{\theta }}(y) \, \mathrm {d}y} \, \mathrm {d}z. \end{aligned}$$
(39)

The numerator of the first factor is the desired one-step density up to appropriate normalization. If we extend by the required normalization constant \(\int _{\mathbb {R}} k_{\sqrt{2}\sigma }(y;x_{t-2\tau }) w_{\varvec{\theta }}(y)\, \mathrm {d}y\), we obtain Eqs. (31) and (32). \(\square \)

We are now left to show that the function \(v-1\) is in the order of \(\tau \) on its entire domain \(\mathbb {R}^2\times \mathbb {R}^+\). In particular, this means that for any fixed \(\tau ^{*}\), the function \(v(x_1, x_2; \tau ^{*})-1\) is bounded on \(\mathbb {R}^2\) via \(c\tau ^{*}\) for a constant c. It turns out to be helpful to analyze v separately on \(\mathbb {R}^2\times (0,\tau _0)\) and \(\mathbb {R}^2\times [\tau _0,\infty )\) for some \(\tau _0\). Because the proof is simpler for large \(\tau \), we present this result first.

Lemma 1

Let w be continuous and bounded away from zero, that is there exist L and U such that \(0<L\le w_{\varvec{\theta }}(x) \le U\) for all \(x\in \mathbb {R}\). Let w further be twice differentiable on \(\mathbb {R}\) with \(|w''(x)|<M\) for some M and all \(x\in \mathbb {R}\). For any \(\tau _0 > 0\), we have \(v(x_1, x_2,;\tau ) -1 = \mathscr {O}(\tau )\) on \(\mathbb {R}^2\times [\tau _0,\infty )\).

Proof

Let \(\tau _0\) be a number away from zero and fixed. Our goal is to establish bounds on the functions Q and I, as defined in (33) and (34), and to use these to place a bound on \(v-1\). Because w is twice differentiable we can apply Taylor’s theorem to obtain a linear approximation for w using any point \(x\in \mathbb {R}\),

$$\begin{aligned} w_{\varvec{\theta }}(y) = w_{\varvec{\theta }}(x) + w'(x)(y-x)+R(y), \end{aligned}$$
(40)

where R(y) is the remainder term. This leads to

$$\begin{aligned} \int _{\mathbb {R}} k_{\sigma }(y;x)\, w_{\varvec{\theta }}(y)\, \mathrm {d}y= & {} w_{\varvec{\theta }}(x) \int _{\mathbb {R}} k_{\sigma }(y;x)\,\mathrm {d}y + w'(x) \int _{\mathbb {R}} k_{\sigma }(y;x)\,(y-x)\,\mathrm {d}y \nonumber \\&+ \int _{\mathbb {R}} k_{\sigma }(y;x)\,R(y)\,\mathrm {d}y, \end{aligned}$$
(41)

where the first term on the RHS becomes \(w_{\varvec{\theta }}(x)\), because the kernel integrates to 1, and the integral in the second term is the first central moment of the kernel, hence vanishes. The remainder R(y), using the Lagrange form, is given by \(R(y) = \frac{w''(\xi )}{2}(y-x)^2\), for some \(\xi \) between \(x_2\) and y. Since the second derivative of w is assumed to be globally bounded, we have \(|R(y)|\le \frac{M}{2}(y-x)^2\). We use this to place bounds on the third term, recognizing that the remaining integral \(\int _{\mathbb {R}} k_{\sigma }(y;x)\,(y-x)^2\,\mathrm {d}y\) is the second central moment of the Gaussian kernel \(k_{\sigma }\), which is given by its variance \(\sigma ^2 = \omega ^2\tau \). Therefore,

$$\begin{aligned} w_{\varvec{\theta }}(x) - \frac{M}{2}\omega ^2\tau \le \int _{\mathbb {R}} k_{\sigma }(y;x)\, w_{\varvec{\theta }}(y)\, \mathrm {d}y \le w_{\varvec{\theta }}(x) + \frac{M}{2}\omega ^2\tau . \end{aligned}$$
(42)

In general, the lower bound can be arbitrarily close to zero, therefore we cannot simply invert this inequality to obtain an estimate on the inverse of the integral. Instead, we use the bounds on w and again the fact \(\int _{\mathbb {R}} k_{\sigma }(y;x)\,\mathrm {d}y=1\) for any \(\sigma \) and any \(x\in \mathbb {R}\) to establish

$$\begin{aligned} 0 < L \le \int _{\mathbb {R}} k_{\sigma }(y;x) \, w_{\varvec{\theta }}(y) \,\mathrm {d}y \le U, \end{aligned}$$
(43)

which can be inverted. Since inequalities (42) and (43) hold for any \(\sigma \) and any \(x \in \mathbb {R}\), they allow us to place bounds on both Q and I. For Q, we obtain

$$\begin{aligned} \frac{1}{U}\bigl (w_{\varvec{\theta }}(x)-M\omega ^2\tau \bigr ) \le Q(x;\tau ) \le \frac{1}{L}\bigl ( w_{\varvec{\theta }}(x)+M\omega ^2\tau \bigr ) \end{aligned}$$
(44)

for all \(x\in \mathbb {R}, \tau \in \mathbb {R}^+\). We can avoid the dependency of the bounds on x by again invoking the bounds on w,

$$\begin{aligned} \frac{1}{U}\bigl (L-M\omega ^2\tau \bigr ) \le Q(x) \le \frac{1}{L}\bigl ( U+M\omega ^2\tau \bigr ). \end{aligned}$$
(45)

For the function I, we only make use of the bounds on w and inequality (43) and get

$$\begin{aligned} 0 < \frac{L}{U} \le I(x_1, x_2;\tau ) \le \frac{U}{L} \end{aligned}$$
(46)

for all \(x_1, x_2 \in \mathbb {R}, \tau \in \mathbb {R}^+\). We can now continue to calculate \(v-1\). An upper bound is immediately given by

$$\begin{aligned} v(x_1, x_2;\tau ) -1 = Q(x_1;\tau )\,I(x_1, x_2;\tau ) -1 \le \frac{U^2-L^2}{L^2} + \frac{MU}{L^2}\omega ^2\tau . \end{aligned}$$
(47)

With only few more additional steps, we obtain a lower bound by simply drawing upon \(L \le U\), its squared version and its inverse,

$$\begin{aligned} -\big (v(x_1, x_2;\tau )-1\big ) \le \frac{U^2-L^2}{U^2} + \frac{ML}{U^2}\omega ^2\tau \le \frac{U^2-L^2}{L^2} + \frac{MU}{L^2}\omega ^2\tau . \end{aligned}$$
(48)

Define \(C:= \frac{U^2-L^2}{L^2\tau _0} + \frac{MU}{L^2}\omega ^2\) for the \(\tau _0\) chosen up front. Then,

$$\begin{aligned} |v(x_1, x_2;\tau ) -1 |&\le \frac{U^2-L^2}{L^2} + \frac{MU}{L^2}\omega ^2\tau - C\tau + C\tau \end{aligned}$$
(49)
$$\begin{aligned}&= \frac{U^2-L^2}{L^2} - \frac{U^2-L^2}{L^2\tau _0}\tau + C\tau \end{aligned}$$
(50)
$$\begin{aligned}&= \left( 1-\frac{\tau }{\tau _0}\right) \frac{U^2-L^2}{L^2} +C\tau . \end{aligned}$$
(51)

The product on the RHS is non-positive for \(\tau \ge \tau _0\), and hence \(|v(x_1, x_2;\tau ) -1 |\le C\tau \) for all \(\mathbb {R}^2\times [\tau _0, \infty )\). \(\square \)

The bounds on Q and I, and thus \(v-1\), established in the preceding proof are not sufficient to conclude the result as \(\tau \rightarrow 0\), unless \(L=U\), which is the trivial case of a constant weighting function. More suitable bounds, however, can be found if inequality (42) can be inverted. This can be achieved by assuming \(\tau \) to be small enough.

Lemma 2

Let w be continuous and bounded away from zero, that is there exist L and U such that \(0<L\le w_{\varvec{\theta }}(x) \le U\) for all \(x\in \mathbb {R}\). Let w further be twice differentiable on \(\mathbb {R}\) with \(|w''(x)|<M\) for some M and all \(x\in \mathbb {R}\). Let \(\tau _0 = \frac{2L}{M\omega ^2}\). Then \(v(x_1, x_2,;\tau ) -1 = \mathscr {O}(\tau )\) on \(\mathbb {R}^2\times (0,\tau _0)\).

Proof

Here we develop bounds on Q and I such that both \(Q-1\) and \(I-1\) are in the order of \(\tau \). Let \(\tau \le \tau _{0}\) for \(\tau _0\) as defined in the lemma. Then the lower bound of Eq. (42) is bounded away from zero,

$$\begin{aligned} w_{\varvec{\theta }}(x) - \frac{M}{2}\omega ^2\tau \ge w_{\varvec{\theta }}(x) - \frac{M}{2}\omega ^2\tau _0 > w_{\varvec{\theta }}(x) - \frac{M}{2}\omega ^2 \frac{2L}{M\omega ^2} = w_{\varvec{\theta }}(x) - L\ge 0. \end{aligned}$$
(52)

Hence we can invert the inequality (42) and obtain

$$\begin{aligned} \frac{w_{\varvec{\theta }}(x)-M\omega ^2\tau }{w_{\varvec{\theta }}(x)+\frac{M}{2}\omega ^2\tau } \le Q(x;\tau ) \le \frac{w_{\varvec{\theta }}(x)+M\omega ^2\tau }{w_{\varvec{\theta }}(x)-\frac{M}{2}\omega ^2\tau }. \end{aligned}$$
(53)

Note that the values in the numerators and denominators differ slightly because the variances of the kernel k in the numerator and denominator of Q differ by a factor of 2.

Since \(2w_{\varvec{\theta }}(x)-M\omega ^2\tau \ge 2L-M\omega ^2\tau _0>0\), we can conclude

$$\begin{aligned}&Q(x;\tau ) - 1 \le \frac{w_{\varvec{\theta }}(x)+M\omega ^2\tau - w_{\varvec{\theta }}(x)-\frac{M}{2}\omega ^2\tau }{w_{\varvec{\theta }}(x)-\frac{M}{2}\omega ^2\tau } \nonumber \\&\quad = \frac{M\omega ^2\tau }{2w_{\varvec{\theta }}(x)-M\omega ^2\tau } \le \frac{M\omega ^2\tau }{2L-M\omega ^2\tau _0}, \end{aligned}$$
(54)

for all \(x\in \mathbb {R}\) and \(\tau <\tau _0\). Using \(2w_{\varvec{\theta }}(x)+M\omega ^2\tau \ge 2w_{\varvec{\theta }}(x)\ge 2L\), we similarly obtain,

$$\begin{aligned} -(Q(x;\tau )-1) \le \frac{3M\omega ^2\tau }{2w_{\varvec{\theta }}(x)+M\omega ^2\tau } \le \frac{3M}{2L}\omega ^2\tau \end{aligned}$$
(55)

for all \(x\in \mathbb {R}\) and \(\tau <\tau _0\). If we set \(C_1 := \max \left( \frac{M\omega ^2}{2L-2\omega ^2\tau _0},\frac{3M\omega ^2}{2L} \right) \), it follows that \(\vert Q(x;\tau ) - 1 \vert \le C_1\tau \) on \(\mathbb {R}^2\times (0,\tau _0)\).

Using analogous arguments as before, we can fine an find an upper bound on I,

$$\begin{aligned} I(x_1, x_2;\tau )&= \int _{\mathbb {R}} k_{\frac{\sigma }{\sqrt{2}}}\Bigl (z; \frac{1}{2}(x_1+x_2)\Bigr ) \, \frac{w_{\varvec{\theta }}(z)}{\int _{\mathbb {R}} k_{\sigma }(y;z) w_{\varvec{\theta }}(y) \, dy} \, \mathrm {d}z \end{aligned}$$
(56)
$$\begin{aligned}&\le \int _{\mathbb {R}} k_{\frac{\sigma }{\sqrt{2}}}\Bigl (z; \frac{1}{2}(x_1+x_2)\Bigr ) \, \frac{w_{\varvec{\theta }}(z)}{w_{\varvec{\theta }}(z) - \frac{M}{2}\omega ^2\tau } \, \mathrm {d}z \end{aligned}$$
(57)
$$\begin{aligned}&= \int _{\mathbb {R}} k_{\frac{\sigma }{\sqrt{2}}}\Bigl (z; \frac{1}{2}(x_1+x_2)\Bigr ) \, \frac{w_{\varvec{\theta }}(z)-\frac{M}{2}\omega ^2\tau +\frac{M}{2}\omega ^2\tau }{w_{\varvec{\theta }}(z) - \frac{M}{2}\omega ^2\tau } \, \mathrm {d}z \end{aligned}$$
(58)
$$\begin{aligned}&= \int _{\mathbb {R}} k_{\frac{\sigma }{\sqrt{2}}}\Bigl (z; \frac{1}{2}(x_1+x_2)\Bigr ) \,\mathrm {d}z + \int _{\mathbb {R}} k_{\frac{\sigma }{\sqrt{2}}}\Bigl (z; \frac{1}{2}(x_1+x_2)\Bigr )\nonumber \\&\quad \times \, \frac{\frac{M}{2}\omega ^2\tau }{w_{\varvec{\theta }}(z) - \frac{M}{2}\omega ^2\tau } \, \mathrm {d}z \end{aligned}$$
(59)
$$\begin{aligned}&\le 1 + \int _{\mathbb {R}} k_{\frac{\sigma }{\sqrt{2}}}\Bigl (z; \frac{1}{2}(x_1+x_2)\Bigr ) \, \frac{\frac{M}{2}\omega ^2\tau }{L - \frac{M}{2}\omega ^2\tau _0} \, \mathrm {d}z = 1+ \frac{M\omega ^2\tau }{2L - M\omega ^2\tau _0}. \end{aligned}$$
(60)

A lower bound is given by

$$\begin{aligned} I(x_1, x_2;\tau )&\ge \int _{\mathbb {R}} k_{\frac{\sigma }{\sqrt{2}}}\Bigl (z; \frac{1}{2}(x_1+x_2)\Bigr ) \, \frac{w_{\varvec{\theta }}(z)}{w_{\varvec{\theta }}(z) + \frac{M}{2}\omega ^2\tau } \, \mathrm {d}z \end{aligned}$$
(61)
$$\begin{aligned}&= 1 - \int _{\mathbb {R}} k_{\frac{\sigma }{\sqrt{2}}}\Bigl (z; \frac{1}{2}(x_1+x_2)\Bigr ) \, \frac{\frac{M}{2}\omega ^2\tau }{w_{\varvec{\theta }}(z) + \frac{M}{2}\omega ^2\tau } \, \mathrm {d}z \ge 1-\frac{M\omega ^2\tau }{2L}. \end{aligned}$$
(62)

Setting \(C_2 := \frac{M\omega ^2\tau }{2L - M\omega ^2\tau _0}\), we obtain \(|I(x_1,x_2;\tau )-1| \le C_2\tau \) on \(\mathbb {R}^2\times (0,\tau _0)\).

We can now estimate \(v-1\) as follows,

$$\begin{aligned} |v(x_1,x_2;\tau ) - 1|&= |Q_{\tau }\, I_{\tau } -1| \le |Q_{\tau }-1|\,|I_{\tau }-1| + |Q_{\tau }-1| + |I_{\tau }-1| \end{aligned}$$
(63)
$$\begin{aligned}&\le C_1\,C_2\tau ^2+(C_1+C_2)\,\tau \le \big (C_1\,C_2\tau _0+C_1+C_2\big )\,\tau , \end{aligned}$$
(64)

for all \(x_1, x_2\in \mathbb {R}\) and all \(\tau <\tau _0\). \(\square \)

Lemmata 1 and 2, together with proposition 1 prove Theorem 3.

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Schlägel, U.E., Lewis, M.A. Robustness of movement models: can models bridge the gap between temporal scales of data sets and behavioural processes?. J. Math. Biol. 73, 1691–1726 (2016). https://doi.org/10.1007/s00285-016-1005-5

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