Again, please note that the input data sample or, equivalently, the trajectory (2) has been observed and serves as input. Also, let \(\mathcal {F}_{t}\) be the sigma algebra generated by \(X(v), v \le t\) .

### Preliminaries

Conditionally upon this trajectory the inter-event time distribution

$$\begin{aligned} P\{T_{i} \ge u| \mathcal {F}_{t}\} = E\left[\exp \left(-\left(\lambda _{i} u+ \int _{t}^{t+u}\sum _{j\in E}\alpha _{i,j}X_{j}(v)dv\right)\right)|\mathcal {F}_{t}\right] \end{aligned}$$

(9)

holds. Assuming stochastic independence conditionally upon \(X(v)_{v\le t}\) (9) immediately yields

$$\begin{aligned} R(t|X) := P\{\min _{i \in E} T_i \ge u| \mathcal {F}_{t}\}= E\left[ \exp \left( -\sum _{i \in E}(\lambda _{i}u+ \int _{t}^{t+u}\sum _{j \in E}\alpha _{i,j}X_{j}(v)dv)\right) |\mathcal {F}_{t}\right] \end{aligned}$$

(10)

Define further

$$\begin{aligned} F_i(t|X)& := P\{\min _{i\in E}T_i < u| \mathcal {F}_{t}\} = 1-P\{\min _{i \in E}T_i \ge u| \mathcal {F}_{t}\} =\int _0^tf_i(u|X)du\nonumber \\ f_i(t|X)& := R(t|X)\times \left( \lambda _{i}+\sum _{j \in E}\alpha _{i,j}X_{j}(t)\right) \nonumber \\ \end{aligned}$$

(11)

Accordingly, \(f_i(t|X)*dt + o(dt)\) is the probability for an event of type \(i \in E\) to occur in the time interval \((t,t+dt)\) conditionally upon the trajectory \(X(v)_{v\le t}\). Also let \(\hat{C}(t|X)\) be the vector of expected numbers of renewals at time *t*, if the process starts in state *X*. The following then holds:

###
**Theorem 2**

*Let*

$$\begin{aligned} X^{(i)}(X,u):=X+\sum _{j \ne i} e_j*u -X_i*e_i, i\in E \end{aligned}$$

(12)

*Then*

$$\begin{aligned} \hat{C}(t|X)=\sum _{i\in E}F_i(t|X)e_i + \sum _{i\in E}\int _0^tf_i(u|X)\hat{C}(t-u|X^{(i)}(X,u))du \end{aligned}$$

(13)

###
*Proof*

Equation (13), representing the expected number of renewals at time *t* under the condition that the process started in state *X*, can be conditioned upon the first occurrence of the event. If this occurs after time *t* (probability *R*(*t*|*X*)), then the expected numbers are \((0,\ldots ,0)^T\). If it occurs during the time interval \([u,u+du)\) somewhere in the interval [0, *t*) and is of type \(i \in E\) (probability \(f_i(u|X)du)\), then state *X* transforms into state \(X^{(i)}(X,u)\) and—therefore—the expected number, seen as a vector, is equal to \(e_i +\hat{C}(t-u|X^{(i)}(X,u))\). \(\square\)

### Iterative approximation

In this section the individual components of \(\hat{C}(t|X)\) will be considered one by one and the arguments \(\lambda\) and \(\alpha\) will be suppressed. Also, \(\hat{C}_i(0|X)=0\) will be assumed. Then the individual components of equation (13) can be written as

$$\begin{aligned} \hat{C}_i(t|X)=F_i(t|X)+\sum _{j \in E}\int _0^t f_j(u|X)\hat{C}_i(t-u|X^{(j)}(X,u))du \end{aligned}$$

(14)

Assume (14) has an iterative solution such that, for \(k=0,1,2,\ldots\)

$$\begin{aligned} \hat{C}_i^{(0)}(t|X)&=\,F_i(t|X) \nonumber \\ \hat{C}_i^{(k+1)}(t|X)&=\,{} F_i(t|X)+\sum _{j \in E}\int _0^t f_j(u|X)\hat{C}_i^{(k)}(t-u|X^{(j)}(X,u))du \end{aligned}$$

(15)

Then stages (1) and (2) of the iterative approximation can be written as

$$\begin{aligned} \hat{C}_i^{(1)}(t|X)&= {} F_i(t|X)+\sum _{j \in E}\int _0^t f_j(u|X) F_i(t-u|X^{(j)}(X,u))du \nonumber \\ \hat{C}_i^{(2)}(t|X)&= {} F_i(t|X)+\sum _{j \in E}\int _0^t f_j(u|X) F_i(t-u|X^{(j)}(X,u))du \nonumber \\&\quad + \sum _{j \in E}\int _0^t f_j(u|X)\sum _{k \in E}\int _0^{t-u} f_k(v|X^{(j)}(X,u))\nonumber \\&\quad \times F_i(t-u-v|X^{(k)}(X^{(j)}(X,u),v))dv du \end{aligned}$$

(16)

(16) can be used to approximate the cumulated number of events for rare failure codes, and only in the near term environment. The advantage is, however, that those numbers take into consideration the initial condition *X* and therefore are in agreement with the requirements of “Condition Based Maintenance”. It will be shown now, that the iteration given in (15) converges.

###
**Lemma 1**

*For any *
\(T \in {\mathrm {I\!R\!}}_+\)
* such that*

$$\begin{aligned} \rho := \sum _{j \in E}\int _0^Tf_j(u|X)du < 1,\quad{X \in {\mathrm {I\!R\!}}}_+^n \end{aligned}$$

(17)

*the following holds:*

$$\begin{aligned} \lim _{k \rightarrow \infty } ||\hat{C}^{(k+1)}(t|X)-\hat{C}^{(k)}(t|X)|| = 0,\quad0\le t \le T,\quad{X \in {\mathrm {I\!R\!}}}_+^n \end{aligned}$$

(18)

###
*Proof*

Let

$$\begin{aligned} D^{(k+1)}(t|X):=\hat{C}^{(k+1)}(t|X)-\hat{C}^{(k)}(t|X),\quad{k=0,1,2,3}\ldots \nonumber \\ \gamma _k := \sup _{t \in [0,T]} \sup _{X \in {\mathrm {I\!R\!}}_+^n}\times\,\max _{l \in E}D^{(k)}_l(t|X) \end{aligned}$$

(19)

Then from (15) one obtains

$$\begin{aligned} D^{(k+1)}(t|X)=\int _0^tdu\sum _{i \in E} f_i(u|X)D^{(k)}(t|X) \end{aligned}$$

(20)

and therefore

$$\begin{aligned} ||D^{(k+1)}(t|X)|| = \rho *||D^{(k)}(t|X)|| \end{aligned}$$

(21)

as shown in Appendix 1, proving the lemma in the limit for \(k \rightarrow \infty.\)
\(\square\)

Please observe that (17) expresses the condition that the process will not “explode” at any time in a finite time interval.

### A special case: stochastic independence

Assume (7) holds. Let \(\tilde{C}_i(t)\) be the solution of (14) under (7). The expected cumulated numbers of events then become independent. For each \(i \in E\) the following result can be proven, whereby \(\lambda := \lambda _i\) and \(\alpha := \alpha _{ii}\) has been set.

###
**Corollary 2**

$$\begin{aligned} \lim _{t\rightarrow \infty }\tilde{C}_i(t)= \frac{t}{\exp \left( \frac{\lambda ^2}{2\alpha }\right) * \sqrt{\frac{2\pi }{\alpha }}*\left( 1-\Phi \left( \frac{\lambda }{\sqrt{\alpha }}\right) \right) }-1+o(1) \end{aligned}$$

(22)

*whereby *
\(\Phi (x)\)
*denotes the cumulative distribution function of the standard normal distribution.*

###
*Proof*

Note that

$$\begin{aligned} \tilde{C}_i(t)&= \int _{0}^t(\lambda +\alpha u)\exp \left( {-\left( \lambda u + \alpha \frac{u^2}{2}\right) }\right) du \nonumber \\&+ \int _{0}^t(\lambda +\alpha u)\exp \left( -\left( {\lambda u + \alpha \frac{u^2}{2}}\right) \right) \times\hat{C}_i(t-u)du \end{aligned}$$

(23)

Let

$$\begin{aligned} \phi (t)&:= {} \int _{0}^t(\lambda +\alpha u)\exp {\left( -\left( \lambda u + \alpha \frac{u^2}{2}\right) \right) }du \nonumber \\&=\,{} 1-\exp \left( -\left( \lambda t+\alpha \frac{t^2}{2}\right) \right) \end{aligned}$$

(24)

whereby the second equation above is proven in Appendix 2.

Furthermore, defining

$$\begin{aligned} f(u)&:=(\lambda +\alpha u)\exp {\left( -\left( \lambda u + \alpha \frac{u^2}{2}\right) \right) } \nonumber \\ L_{C}(s)&:=\int _{0}^\infty \exp (-st)\tilde{C}_i(t)dt\nonumber \\ L_{\phi }(s)&:=\,\int _{0}^\infty \exp (-st)\phi (t)dt \nonumber \\ L_{f}(s)&:=\int _{0}^\infty \exp (-st)f(t)dt \end{aligned}$$

(25)

and making use of the Laplace transformations introduced above yields

$$\begin{aligned} L_{C}(s)=\frac{L_{\phi }(s)}{1-L_{f}(s)} \end{aligned}$$

(26)

Next, we compute \(L_{\phi }(s)\) and \(L_{f}(s)\). It is easy to see that

$$\begin{aligned} L_{\phi }(s)= \frac{1}{s}- \exp \left( \frac{(\lambda +s)^2}{2\alpha }\right) \sqrt{\frac{2\pi }{\alpha }}\left( 1-\Phi \left( \frac{\lambda +s}{\sqrt{\alpha }}\right) \right) \end{aligned}$$

(27)

see Appendix 3. Also, \(L_{f}(s)\) is shown to be expressed as

$$\begin{aligned} L_f(s) = 1-\exp \left( \frac{(\lambda +s)^2}{2\alpha }\right) s \sqrt{\frac{2\pi }{\alpha }}\left( 1-\Phi \left( \frac{\lambda +s}{\sqrt{\alpha }}\right) \right) \end{aligned}$$

(28)

as shown in Appendix 4. Now, upon using (26), (27) and (28) the following is obtained:

$$\begin{aligned} L_C(s)= \frac{1-s\exp \left( \frac{(\lambda +s)^2}{2\alpha }\right) \sqrt{\frac{2\pi }{\alpha }}\left( 1-\Phi \left( \frac{\lambda +s}{\sqrt{\alpha }}\right) \right) }{s^2\exp \left( \frac{(\lambda +s)^2}{2\alpha }\right) \sqrt{\frac{2\pi }{\alpha }}\left( 1-\Phi \left( \frac{\lambda +s}{\sqrt{\alpha }}\right) \right) } \end{aligned}$$

(29)

which yields

$$\begin{aligned} \lim _{s\rightarrow 0}L_C(s)=\left( \frac{1}{s^2\exp \left( \frac{\lambda ^2}{2\alpha }\right) \sqrt{\frac{2\pi }{\alpha }}\left( 1-\Phi \left( \frac{\lambda }{\sqrt{\alpha }}\right) \right) }\right) -\frac{1}{s}+O(1) \end{aligned}$$

(30)

whereby—see Cox (1962)—*O*(1) is a function of *s* bounded as \(s\rightarrow 0\). According to Cox (1962), section 1.3, \(\tilde{C}_i(t)\) then satisfies

$$\begin{aligned} \lim _{t \rightarrow \infty }\tilde{C}_i(t)=\frac{t}{ \exp \left( \frac{\lambda ^2}{2\alpha }\right) * \sqrt{\frac{2\pi }{\alpha }} * \left( 1-\Phi \left( \frac{\lambda }{\sqrt{\alpha }}\right) \right) } - 1 + o(1) \end{aligned}$$

(31)

\(\square\)

An equivalent proof can be obtained from one of the central results in renewal theory which states that

$$\begin{aligned} \lim _{t \rightarrow \infty }\tilde{C}_i(t)=\frac{t}{\bar{T}} \end{aligned}$$

(32)

whereby \(\bar{T}\) is the expected renewal time, see chapter 4 in Cox (1962). By definition

$$\begin{aligned} \bar{T} = \int _0^\infty t(\lambda +\alpha *t)\exp \left( -\left( \lambda *t+\alpha *\frac{t^2}{2}\right) \right) dt \end{aligned}$$

(33)

Using substitutions in the style as shown above and properties of the incomplete Gamma function, as defined, for instance in Abramovitz and Stegun (1972), p. 262, one proves that

$$\begin{aligned} \bar{T} = \exp \left( \frac{\lambda ^2}{2\alpha }\right) *\sqrt{\frac{2\pi }{\alpha }}\left( 1-\Phi \left( \frac{\lambda }{\sqrt{\alpha }}\right) \right) \end{aligned}$$

(34)

Using (32) along with (34) proves the statement.

The following conclusions are now easy to draw:

###
**Corollary 3**

*If *
\(\lambda = 0\)
*then*

$$\begin{aligned} \lim _{t \rightarrow \infty } \hat{C}(t|\lambda =0)= 2\sqrt{\frac{\alpha }{2\pi }}*t - 1 + O(1) \end{aligned}$$

(35)

*If*
\(\,\alpha = 0\,\)
*then*

$$\begin{aligned} \lim _{t \rightarrow \infty } \hat{C}(t|\alpha =0)= \lambda t + O(1) \end{aligned}$$

(36)

###
*Proof*

(35) can immediately be derived from (22) by letting \(\lambda\) tend to zero. (36) must be concluded from equation (23), as (22) has been derived under the implicit assumption that \(\alpha \ne 0\) used in dividing the exponent by \(\alpha\), but the conclusion is straightforward. \(\square\)

Equation (35) has an interesting application. Assume (6) holds and, in addition \(\lambda =0\) can be safely assumed. In that case (35) allows to estimate \(\alpha\) by equating the slope of \(\hat{C}(t|\alpha =0)\) with the coefficient of *t*.

In the context of condition based maintenance residual lifetimes of components or residual forward times of critical events must be estimated conditionally upon the system state, which frequently is expressed by parameters such as age or backward time. Let

$$\begin{aligned} \bar{T}(\tau ) = E[T-\tau |T \ge \tau ] \end{aligned}$$

(37)

be the expected forward time conditionally upon the event that the forward time exceeds \(\tau\). Then, in close analogy with (34), the following can be proven:

$$\begin{aligned} \bar{T}(\tau ) = \exp \left( \frac{(\lambda +\alpha \tau )^2}{2\alpha }\right) *\sqrt{\frac{2\pi }{\alpha }}\left( 1-\Phi \left( \frac{\lambda +\alpha \tau }{\sqrt{\alpha }}\right) \right) \end{aligned}$$

(38)

### A first order correction

Let

$$\begin{aligned} \bar{\alpha }= \max _{i \in E}\sum _{j \in E, j \ne i}\alpha _{i,j}, \tilde{\alpha }_{i} = \sum _{j \in E}\alpha _{i,j} \end{aligned}$$

(39)

and assume, for the sake of simplicity, \(\alpha _{i,j} \ge 0, i \in E, j \in E, j \ne i\). With the definitions given in Appendix 5 one can show that, under (7)—i.e. stochastic independence between the event numbers—the renewal equation becomes

$$\begin{aligned} \tilde{C}(t|X)= \sum _{i \in E}e_i\tilde{F}_i(t|X)+ \sum _{i \in E}e_i \sum _{j \in E}\int _0^tdu \tilde{f}_j(u|X) \tilde{C}_i(t-u|X_{i}(u,X)) \end{aligned}$$

(40)

The following proposition holds:

###
**Proposition 1**

$$\begin{aligned} \hat{C}(t|X)&= \tilde{C}(t|X)+ D(t|X)\nonumber \\ D(t|X)&= D^{(1)}(t|X)+D^{(2)}(t|X) + \cdots \end{aligned}$$

(41)

*whereby*

$$\begin{aligned} D^{(1)}(t|X) = \sum _{i \in E}D^{(1)}_{i}(t|X)*e_{i} \end{aligned}$$

(42)

*and, approximately*

$$\begin{aligned} |D^{(1)}_{i}(t|X)| &\lessapprox \bar{\alpha }\times\left( \frac{t^2}{2}+n\lambda _i\frac{t^3}{6}+n\tilde{\alpha }_i\frac{t^4}{8}\right) \nonumber \\&+ {} \frac{\bar{\alpha }}{\tilde{T}_i}\sum _{ k \in E}\left( \frac{t^3}{6}+n\lambda _k\frac{t^4}{24}+n\tilde{\alpha }_k\frac{t^5}{40}\right) +o(\bar{\alpha }) \end{aligned}$$

(43)

*as well as*

$$\begin{aligned} D^{(2)}(t|X) := \sum _{ i \in E}e_i\sum _{ k \in E}\int _0^tdu(R(u|X)(\lambda _i + \sum _{j \in E}\alpha _{i,j}(u+X_j))\nonumber \\ \times\,{(\hat{C}_i(t-u|X^{(k)}(u,X))-\tilde{C}_i(t-u|X^{(k)}(u,X))} \end{aligned}$$

(44)

In (43) \(\tilde{T}_i\) is defined as in (31), however with explicit reference to a failure mode \(i \in E\). This statement is given as a a proposition rather than as a theorem, because the behavior of \(\tilde{C}_i(t|X)\) is used in its asymptotic approximation.

With (13), (40) and Appendix 6 one shows that

$$\begin{aligned} \hat{C}(t|X)-\tilde{C}(t)= &{} \sum _{i \in E}e_i(A_i+B_i)\nonumber \\&+ {} \sum _{ i \in E}e_i\sum _{ k \in E}\int _0^tdu(R(u|X)(\lambda _i + \sum _{j \in E}\alpha _{i,j}(u+X_j))\nonumber \\&\times(\hat{C}_i(t-u|X^{(k)}(u,X))-\tilde{C}_i(t-u|X^{(k)}(u,X))\nonumber \\ A_i: = &{} \int _0^tdu(R(u|X)(\lambda _i \nonumber \\&+ {} \sum _{j \in E}\alpha _{i,j}(u+X_j))- \tilde{R}(u|X)(\lambda _i+\alpha _{i,i}u))\nonumber \\ B_i := & {} \sum _{ k \in E}\int _0^t du(f_k(u|X)-\tilde{f}_k(u|X)) \nonumber \\&\times\tilde{C}_i(t-u|X^{(k)}(u,X))\nonumber \\ \end{aligned}$$

(45)

In Appendix 7 and Appendix 8 it is shown that

$$\begin{aligned} |A_i|&\le {} \bar{\alpha }*\left( \frac{t^2}{2} +n\lambda _i\frac{t^3}{3} +n\tilde{\alpha }_i\frac{t^4}{8}\right) +o(\bar{\alpha })\nonumber \\ |B_i|&\lessapprox\frac{\bar{\alpha }}{\tilde{T}_i}\sum _{ k \in E}\left( \frac{t^3}{6}+n\lambda _k\frac{t^4}{24} + n\tilde{\alpha }_k\frac{t^5}{40}\right) + o(\bar{\alpha }) \end{aligned}$$

(46)

and this proves (43).