The Cameron–Martin Theorem for (p-)Slepian Processes

Abstract

We show a Cameron–Martin theorem for Slepian processes $W_t:=\frac{1}{\sqrt{p}}(B_t-B_{t-p}), t\in [p,1]$, where $p\ge \frac{1}{2}$ and $B_s$ is Brownian motion. More exactly, we determine the class of functions $F$ for which a density of $F(t)+W_t$ with respect to $W_t$ exists. Moreover, we prove an explicit formula for this density. p-Slepian processes are closely related to Slepian processes. p-Slepian processes play a prominent role among others in scan statistics and in testing for parameter constancy when data are taken from a moving window.

Appendix

Proof of Lemma 3.1

Let $\frac{1}{\sqrt{p}}\nabla _p s_{f}\in \mathcal {H}_{p-\text {Sl}}, f\in L^2[0,1]$, be arbitrary and let $1/2\le p\le 1$. There are several places where the case $p=1/2$ must be considered separately. We do not do this since the specifications for $p=1/2$ are simpler than the following considerations for $1/2<p\le 1$. By proposition 4.1 of [12], we get

$$\begin{aligned} \Vert \frac{1}{\sqrt{p}}\nabla _p s_{f}\Vert _{p-Sl}&= \frac{1}{\sqrt{p}}\inf _{g\in L^2[0,1]:\nabla _p s_{f}=\nabla _p s_{g}}\Vert s_g\Vert _{\mathcal {H}_{B_{[0,1]}}}\\&= \frac{1}{\sqrt{p}}\inf _{g\in L^2[0,1]:\nabla _p s_{f}=\nabla _p s_{g}}\Vert g\Vert _{L^2[0,1]}, \end{aligned}$$

where the second equality follows by Theorem 1.3. Each function $f\in L^2[0,1]$ can be written $\lambda -a.s.$ in the form

$$\begin{aligned} f(t)=\alpha \mathbf{1}_{[0,1-p]}(t)+a(t)+\beta \mathbf{1}_{(1-p,p)}(t) +b(t)+\gamma \mathbf{1}_{[p,1]}(t)+c(t), t\in [0,1], \end{aligned}$$

(7)

where

$$\begin{aligned} \alpha&= \frac{1}{1-p}s_f(1-p), a(t)=\mathbf{1}_{[0,1-p]}(t)(f(t)-\alpha ),\\ \beta&= \frac{1}{2p-1}(s_f(p)-s_f(1-p)), b(t)=\mathbf{1}_{(1-p,p)}(t)(f(t)-\beta ),\\ \gamma&= \frac{1}{1-p}(s_f(1)-s_f(p)), c(t)=\mathbf{1}_{[p,1]}(t)(f(t)-\gamma ). \end{aligned}$$

Note that the six summands in (7) build orthogonal functions in $L^2[0,1]$. In the following, we use the consequence of this representation at several places without citing it explicitly. Furthermore, we have

$$\begin{aligned} s_f(p)=\alpha (1-p)+ \beta (2p-1), f(t)-f(t-p)=\gamma -\alpha +c(t)-a(t), t\in [p,1]. \end{aligned}$$

Next, we consider $g\in L^2[0,1]$ of the specific form

$$\begin{aligned} g(t)=\alpha \mathbf{1}_{[0,1-p]}(t)+\beta \mathbf{1}_{(1-p,p)}(t)+(\alpha +\delta )\mathbf{1}_{[p,1]}(t), t\in [0,1], \alpha ,\beta ,\delta \in \mathbb {R}. \end{aligned}$$

Hence,

$$\begin{aligned} s_g(p)=\alpha (1-p)+\beta (2p-1) \Leftrightarrow \beta =\frac{1}{2p-1}(s_g(p)-\alpha (1-p)) \end{aligned}$$

and

$$\begin{aligned} g(t)-g(t-p)=\delta , t\in [p,1]. \end{aligned}$$

Thus, $\nabla _p(g)(t)=s_g(p)+\delta t, t\in [p,1]$. Let $g_0(t)=\alpha _0 \mathbf{1}_{[0,1-p]}(t)+\beta _0\mathbf{1}_{(1-p,p)}(t)+(\alpha _0+\delta _0)\mathbf{1}_{[p,1]}(t)\in L^2[0,1]$ be fixed and let $g\in L^2[0,1]$ with $\nabla _p(g)(t)=\nabla _p(g_0)(t)=s_{g_0}(p)+\delta _0 t, t\in [p,1]$. Then,

$$\begin{aligned} g(t)=\alpha \mathbf{1}_{[0,1-p]}(t)+\frac{1}{2p-1}(s_{g_0}(p)-\alpha (1-p))\mathbf{1}_{(1-p,p)}(t)+(\alpha +\delta _0)\mathbf{1}_{[p,1]}(t) \end{aligned}$$

and the square of its norm is given by

$$\begin{aligned} \Vert g\Vert ^2= \frac{1-p}{2p-1} (\alpha ^2 (3p-1)-2\alpha (s_{g_0}(p)-(2p-1)\delta _0))+\frac{s_{g_0}(p)^2}{2p-1}+\delta _0^2(1-p). \end{aligned}$$

This norm is minimal if and only if

$$\begin{aligned} \alpha =\frac{1}{3p-1}(s_{g_0}(p)-(2p-1)\delta _0). \end{aligned}$$

Hence, we obtain after some calculation

$$\begin{aligned} \Vert \nabla _p s_{g_0}\Vert ^2_{p-Sl}= \frac{1}{3p-1}(2s_{g_0}(p)^2+2(1-p)s_{g_0}(p)\delta _0+\delta _0^2(1-p)p). \end{aligned}$$

Next, we consider a general function

$$\begin{aligned} f=\alpha \mathbf{1}_{[0,1-p]}(t)+a(t)+\beta \mathbf{1}_{(1-p,p)}(t)+b(t)+(\alpha +\delta )\mathbf{1}_{[p,1]}(t)+c(t)\in L^2[0,1]. \end{aligned}$$

By the above considerations, we obtain

$$\begin{aligned} \Vert \nabla _p s_f\Vert _{p-Sl}^2&= \frac{1}{3p-1}(2s_{f}(p)^2+2(1-p)s_{f}(p)\delta +\delta ^2(1-p)p)\\&\quad +\,\min (\Vert a(t)\Vert ^2+\Vert b(t)\Vert ^2+\Vert c(t)\Vert ^2)\nonumber \end{aligned}$$

where the minimum is taken over all $a:[0,1-p]\rightarrow \mathbb {R}, b:(1-p,p)\rightarrow \mathbb {R}, c:[p,1]\rightarrow \mathbb {R}$ with $s_a(1-p)=s_b(2p-1)=s_c(1)=0$ and $c(t)-a(t-p)=f(t)-f(t-p)-\delta$. It holds true for $t\in [p,1]$ fixed

$$\begin{aligned} \min |a(t-p)|^2+|c(t)|^2&= |\frac{1}{2}(-c(t)+a(t-p))|^2+|\frac{1}{2}(c(t)-a(t-p))|^2\\&= \frac{1}{2}|(c(t)-a(t-p))|^2 =\frac{1}{2}|f(t)-f(t-p)-\delta |^2, \end{aligned}$$

where the minimum is taken over all $a(t-p), c(t)\in \mathbb {R}$ with $c(t)-a(t-p)=f(t)-f(t-p)-\delta$. Hence,

$$\begin{aligned} \Vert \nabla _p s_f\Vert _{p-Sl}^2&= \frac{1}{3p-1}(2s_{f}(p)^2+2(1-p)s_{f}(p)\delta +\delta ^2(1-p)p)\\&\quad +\,\min \Vert a(t)\Vert _{L^2[0,1]}^2+\Vert c(t)\Vert _{L^2[0,1]}^2.\\&= \frac{1}{3p-1}(2s_{f}(p)^2+2(1-p)s_{f}(p)\delta +\delta ^2(1-p)p)\\&\quad +\,\frac{1}{2}\Vert (f(t)-f(t-p)-\delta )_{t\in [p,1]}\Vert _{L^2[p,1]}^2\\&= \frac{1}{3p-1}(2s_{f}(p)^2+2(1-p)s_{f}(p)\delta +\delta ^2(1-p)p)\\&\quad +\,\frac{1}{2}(\Vert (f(t)-f(t-p))_{t\in [p,1]}\Vert _{L^2[p,1]}^2 -(1-p)\delta ^2)\\&= \frac{1}{2(3p-1)}(2s_{f}(p)+\delta (1-p))^2\\&\quad +\, \frac{1}{2}\Vert (f(t)-f(t-p))_{t\in [p,1]}\Vert _{L^2[p,1]}^2. \end{aligned}$$

This minimum is obtained at the function

$$\begin{aligned} f^*(t)&= \left( \frac{1}{3p-1}(s_{f}(p)+\frac{1-p}{2}\delta )+\frac{1}{2}(-f(t+p)+f(t))\right) \mathbf{1}_{[0,1-p]}(t)\\&\quad +\,\frac{1}{2p-1}(s_f(p)-\frac{1-p}{3p-1}(s_{f}(p)-(2p-1)\delta ))\mathbf{1}_{(1-p,p)}(t)\\&\quad +\,\left( \frac{1}{3p-1}(s_{f}(p)+\frac{1-p}{2}\delta )+\frac{1}{2}(f(t+p)-f(t))\right) \mathbf{1}_{[p,1]}(t), t\in [0,1]. \end{aligned}$$

Proof of Lemma 3.2

We have to prove Eq. (1) for the p-Slepian-process. To this end, let $p\le t\le 1, f\in L^2[0,1]$ and $(\nabla _p s_f)(\cdot )=s_f(p)+s_{\nabla _p f}(\cdot )\in \mathcal {H}_{p-\text {Sl}}.$ It holds true for $t\in [p,1]$:

$$\begin{aligned}&\mathbf {E}\,\left( \frac{1}{\sqrt{p}}(\nabla _pB)_{[p,1]}(t)\cdot \frac{3s_f(p)+s_{\nabla _p f}(1)}{2(3p-1)} \left( (\nabla _pB)_{[p,1]}(p)+(\nabla _pB)_{[p,1]}(1)\right) \right) \\&\qquad +\,\mathbf {E}\,\left( \frac{1}{\sqrt{p}}(\nabla _pB)_{[p,1]}(t)\cdot \frac{1}{2}\int _p^{1} \nabla _p f(s) \; d(\nabla _pB)_{[p,1]}(s))\right) \\&\quad = \frac{3s_f(p)+s_{\nabla _p f}(1)}{2\sqrt{p}(3p-1)} \mathbf {E}\,\left( (B_{[0,1]}(t)-B_{[0,1]}(t-p))(B_{[0,1]}(p) +B_{[0,1]}(1)-B_{[0,1]}(1-p)) \right) \\&\qquad +\,\frac{1}{2\sqrt{p}}\mathbf {E}\,\left( (B_{[0,1]}(t)-B_{[0,1]}(t-p))\int _p^{1} \nabla _p f(s) \; d (B_{[0,1]}(s)-B_{[0,1]}(s-p))\right) \\&\quad = \frac{3s_f(p)+s_{\nabla _p f}(1)}{2\sqrt{p}}-\frac{1}{2\sqrt{p}}\left( s_{\nabla _p f}(1)-s_{\nabla _p f}(p)\right) = \frac{1}{\sqrt{p}}s_f(p)+s_{\frac{1}{\sqrt{p}}\nabla _p f}(t). \end{aligned}$$