samples/texts/3179895/page_23.md

$$B^0(t_k) := \sum_{i=1}^{k-1} a(i)b(i), \quad M^0(t_k) := \sum_{i=1}^{k-1} a(i)\xi(i),$$

$${\tau }^0(t_k):=\sum _{i=1}^{k-1}\tau (i),{\mathrm{\Phi }}^0(t_k):=\sum _{i=1}^{k-1}\varphi (i)\backslash tau^0(t\_k)\; :=\; \backslash sum\_\{i=1\}^\{k-1\}\; \backslash tau(i),\; \backslash qquad\; \backslash Phi^0(t\_k)\; :=\; \backslash sum\_\{i=1\}^\{k-1\}\; \backslash phi(i)$$

and the corresponding piecewise linear interpolations for $t \in (t_k, t_{k+1})$ are defined as

$$B^0(t):=\frac{t_{k+1}-t}{\widehat{a}(k)}{B}^0(t_k)+\frac{t-t_k}{\widehat{a}(k)}{B}^0(t_{k+1}),B^0(t)\; :=\; \backslash frac\{t\_\{k+1\}-t\}\{\backslash hat\{a\}(k)\}\; B^0(t\_k)\; +\; \backslash frac\{t-t\_k\}\{\backslash hat\{a\}(k)\}\; B^0(t\_\{k+1\}),$$

$$M^0(t):=\frac{t_{k+1}-t}{\widehat{a}(k)}{M}^0(t_k)+\frac{t-t_k}{\widehat{a}(k)}{M}^0(t_{k+1}),M^0(t)\; :=\; \backslash frac\{t\_\{k+1\}-t\}\{\backslash hat\{a\}(k)\}\; M^0(t\_k)\; +\; \backslash frac\{t-t\_k\}\{\backslash hat\{a\}(k)\}\; M^0(t\_\{k+1\}),$$

$${\tau }^0(t):=\frac{t_{k+1}-t}{\widehat{a}(k)}{\tau }^0(t_k)+\frac{t-t_k}{\widehat{a}(k)}{\tau }^0(t_{k+1}),\backslash tau^0(t)\; :=\; \backslash frac\{t\_\{k+1\}-t\}\{\backslash hat\{a\}(k)\}\; \backslash tau^0(t\_k)\; +\; \backslash frac\{t-t\_k\}\{\backslash hat\{a\}(k)\}\; \backslash tau^0(t\_\{k+1\}),$$

$${\mathrm{\Phi }}^0(t):=\frac{t_{k+1}-t}{\widehat{a}(k)}{\varphi }^0(t_k)+\frac{t-t_k}{\widehat{a}(k)}{\varphi }^0(t_{k+1})\mathrm{.}\backslash Phi^0(t)\; :=\; \backslash frac\{t\_\{k+1\}-t\}\{\backslash hat\{a\}(k)\}\; \backslash phi^0(t\_k)\; +\; \backslash frac\{t-t\_k\}\{\backslash hat\{a\}(k)\}\; \backslash phi^0(t\_\{k+1\}).$$

Using (14), $X^0(t)$ can be written as

where

$$Z^0(t)=\sum _{i=0}^{k-1}(\widehat{a}(i)-a(i))sg(x(i))+\frac{t-t_k}{\widehat{a}(k)}(\widehat{a}(i)-a(i))sg(x(k)),Z^0(t)\; =\; \backslash sum\_\{i=0\}^\{k-1\}\; (\backslash hat\{a\}(i)\; -\; a(i))sg(x(i))\; +\; \backslash frac\{t-t\_k\}\{\backslash hat\{a\}(k)\}\; (\backslash hat\{a\}(i)\; -\; a(i))sg(x(k)),$$

$$H^0(t)=-{\int }_0^{t}sg(\bar{x}(s))ds,H^0(t)\; =\; -\backslash int\_0^t\; sg(\backslash bar\{x\}(s))ds,$$

$$\bar{x}(s)=x(k),s\in [t_k,t_{k+1})\mathrm{.}\backslash bar\{x\}(s)\; =\; x(k),\; s\; \backslash in\; [t\_k,\; t\_\{k+1\}).$$

Since we are interested in the tail properties of the interpolated process, let us define the time shifted and centered process $X^k(t)$:

$$X^k(t):=x(k)+B^k(t)+M^k(t)+H^k(t)+Z^k(t)+{\tau }^k(t)+{\mathrm{\Phi }}^k(t),X^k(t)\; :=\; x(k)\; +\; B^k(t)\; +\; M^k(t)\; +\; H^k(t)\; +\; Z^k(t)\; +\; \backslash tau^k(t)\; +\; \backslash Phi^k(t),$$

where

$$B^k(t)=B^0(t_k+t)-B^0(t_k),B^k(t)\; =\; B^0(t\_k\; +\; t)\; -\; B^0(t\_k),$$