Chapter VI

Markov Processes. Discrete and Continuous Time

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Problem VI.1

Prove that definition (1) is equivalent to the following: for each s∈Ts \in T and any A∈Fs,B∈F⩾sA \in \mathscr {F}_{s}, B \in \mathscr {F}_{ \geqslant s} the following equality holds

P(AC∣Xt)=P(A∣Xt)P(C∣Xt).(83) \mathbb {P}\left(A C \mid X_{t}\right) = \mathbb {P}\left(A \mid X_{t}\right) \mathbb {P}\left(C \mid X_{t}\right) . \tag {83}

Definition 1: a process XX adapted to the filtration (Ft)t∈T\left(\mathscr {F}_{t}\right)_{t \in T} is called Markov (with respect to this filtration and the measure P\mathbb {P}) if for each s∈Ts \in T and any event C∈F⩾s=σ{Xu,u⩾s,u∈T}C \in \mathscr {F}_{ \geqslant s} = \sigma \left\{ X_{u}, u \geqslant s, u \in T\right\}

P(C∣Fs)=P(C∣Xs) a.s.(1) \mathbb {P}\left(C \mid \mathscr {F}_{s}\right) = \mathbb {P}\left(C \mid X_{s}\right) \text{ a.s.} \tag {1}
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Problem VI.2

Let X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\} be a Markov process, T⊂RT \subset \mathbb {R}. Explain why {Xt,t∈U}\left\{ X_{t}, t \in U\right\}, where U⊂TU \subset T, is also a Markov process. In particular, if {Xt,t⩾0}\left\{ X_{t}, t \geqslant 0\right\} is a Markov process, then for any Δ>0\Delta > 0 the process XΔ={XkΔ,k⩾0}X^{\Delta } = \left\{ X_{k \Delta }, k \geqslant 0\right\} is Markov. Is the converse true?

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Problem VI.3

Let X={Xt,t⩾0}X = \left\{ X_{t}, t \geqslant 0\right\} be a Markov process taking values, for each tt, in a measurable space (S,B)(\mathrm{S}, \mathscr {B}). Let (V,E)(\mathrm{V}, \mathscr {E}) be a measurable space and ht:S→V,ht∈B∣E,t⩾0h_{t}: \mathrm{S} \rightarrow \mathrm{V}, h_{t} \in \mathscr {B} \mid \mathscr {E}, t \geqslant 0. Prove that if hth_{t} is a one-to-one mapping for each t⩾0t \geqslant 0, then Y={Yt=ht(Xt),t⩾0}Y = \left\{ Y_{t} = h_{t}\left(X_{t}\right), t \geqslant 0\right\} is also a Markov process. Give an example showing that if the condition that the mapping hth_{t} be one-to-one is dropped, this assertion may fail.

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Problem VI.4

Let X={Xt,t⩾0}X = \left\{ X_{t}, t \geqslant 0\right\} be a Markov process with values in S⊂RS \subset \mathbb {R}. Set Yt=[Xt]Y_{t} = \left[X_{t}\right], where [⋅]\left[\cdot \right] is the integer part of a number. Can one assert that the process Y={Yt,t⩾0}Y = \left\{ Y_{t}, t \geqslant 0\right\} is Markov?

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Problem VI.5

(compare with Exercise 3). Let W={W(t)=(W1(t),…,Wm(t)),t⩾0}W = \left\{ W(t) = \left(W_{1}(t), \ldots , W_{m}(t)\right), t \geqslant 0\right\} be an mm-dimensional Brownian motion and Xm={Xm(t)=(∑k=1mWk2(t))1/2,t⩾0}X_{m} = \left\{ X_{m}(t) = \left(\sum_{k = 1}^{m} W_{k}^{2}(t)\right)^{1 / 2}, t \geqslant 0\right\} the Bessel process, i.e., the radial part of the process WW. Is the process XmX_{m} Markov? Note that X1(t)=∣W1(t)∣X_{1}(t) = \left|W_{1}(t)\right| for t⩾0,m=1t \geqslant 0, m = 1.

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Problem VI.6

Let X={Xt,t⩾0}X = \left\{ X_{t}, t \geqslant 0\right\} and Y={Yt,t⩾0}Y = \left\{ Y_{t}, t \geqslant 0\right\} be real Markov processes. Are the processes X+Y={Xt+Yt,t⩾0}X+Y = \left\{ X_{t}+Y_{t}, t \geqslant 0\right\} and XY={XtYt,t⩾0}X Y = \left\{ X_{t} Y_{t}, t \geqslant 0\right\} Markov? What can be said about the special case where Yt=c(t)Y_{t} = c(t), with c(t)c(t) a deterministic function?

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Problem VI.7

Let {Xk,k=0,1,…}\left\{ X_{k}, k = 0,1, \ldots \right\} be a real Markov process. Set Xt=(t−k)Xk+(k+1−t)Xk+1X_{t} = (t-k) X_{k}+(k+1-t) X_{k+1} for t∈[k,k+1),k=0,1,…t \in [k, k+1), k = 0,1, \ldots, i.e., consider the continuous polygonal line on [0,∞)[0, \infty ) with vertices (k,Xk)\left(k, X_{k}\right). Is the process X={Xt,t⩾0}X = \left\{ X_{t}, t \geqslant 0\right\} Markov? Is the process Y={Yt=X[t],t⩾0}Y = \left\{ Y_{t} = X_{\left[t\right]}, t \geqslant 0\right\}, where [⋅]\left[\cdot \right] is the integer part of a number, Markov?

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Problem VI.8

Let ξ1,ξ2,…\xi_{1}, \xi_{2}, \ldots be independent identically distributed random variables taking the values 1 and -1 with probability 1/21 / 2 each. Set S0=0S_{0} = 0, Sn=ξ1+⋯+ξn,n⩾1S_{n} = \xi_{1}+\cdots +\xi_{n}, n \geqslant 1, and Xn=max⁡0⩽k⩽nSkX_{n} = \max_{0 \leqslant k \leqslant n} S_{k}. Prove that the process X={Xn,n⩾0}X = \left\{ X_{n}, n \geqslant 0\right\} is not Markov.

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Problem VI.9

Give an example of a martingale that is not a Markov process. Construct an example of a real Markov process X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\} with E[∣Xt∣]<∞\mathbb {E}\left[\left|X_{t}\right|\right] < \infty for all t∈Tt \in T that is not a martingale.

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Problem VI.10

Let η,ξ1,ξ2,…\eta , \xi_{1}, \xi_{2}, \ldots be independent variables, where ξ1,ξ2,…\xi_{1}, \xi_{2}, \ldots are uniformly distributed on [0,1][0,1] and η\eta is a real random variable with distribution function Fη(x)F_{\eta }(x). Set Sn=1S_{n} = 1 if ξn⩽η\xi_{n} \leqslant \eta, and Sn=−1S_{n} = -1 if ξn>η(n=1,2,…)\xi_{n} > \eta (n = 1,2, \ldots ). Is {Sn,n⩾1}\left\{ S_{n}, n \geqslant 1\right\} a homogeneous Markov process?

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Problem VI.11

Is it true that a real Gaussian process X={Xt,t⩾0}X = \left\{ X_{t}, t \geqslant 0\right\} is Markov if condition (12) holds only for the function f(x)≡x,x∈Rf(x) \equiv x, x \in \mathbb {R} ?

Theorem 2. A process X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\} taking, for each t∈Tt \in T, values in a Borel space (St,Bt)\left(\mathrm{S}_{t}, \mathscr {B}_{t}\right) is Markov if and only if it has the following property: for any m⩾1m \geqslant 1 and all s1<⋯<sm<s⩽ts_{1} < \cdots < s_{m} < s \leqslant t (s,ts, t and all sis_{i} in TT), and for an arbitrary bounded Bt\mathscr {B}_{t}-measurable function ff

E[f(Xt)∣Xs1,…,Xsm,Xs]=E[f(Xt)∣Xs].(12) \mathbb {E}\left[f\left(X_{t}\right) \mid X_{s_{1}}, \ldots , X_{s_{m}}, X_{s}\right] = \mathbb {E}\left[f\left(X_{t}\right) \mid X_{s}\right] . \tag {12}
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Problem VI.12

Prove that a real Gaussian process X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\}, T⊂RT \subset \mathbb {R}, is Markov if and only if for any t1<t2<t3(t1,t2,t3∈T)t_{1} < t_{2} < t_{3} \left(t_{1}, t_{2}, t_{3} \in T\right) the following equality holds

r(t1,t3)r(t2,t2)=r(t1,t2)r(t2,t3) r\left(t_{1}, t_{3}\right) r\left(t_{2}, t_{2}\right) = r\left(t_{1}, t_{2}\right) r\left(t_{2}, t_{3}\right)

where r(s,t)=cov⁡(Xs,Xt),s,t∈Tr(s, t) = \operatorname {cov}\left(X_{s}, X_{t}\right), s, t \in T. Is fractional Brownian motion a Markov process?

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Problem VI.13

Is the Ornstein-Uhlenbeck process (Definition 13 of Chapter III) a homogeneous Markov process?

Definition 13 of Chapter III. The Ornstein-Uhlenbeck process with parameters α,β>0\alpha , \beta > 0 is the process given by the formula

Vt=e−βtW(αe2βt),t∈R(III.88) V_{t} = e^{-\beta t} W\left(\alpha e^{2 \beta t}\right), \quad t \in \mathbb {R} \tag {III.88}

where W={W(t),t⩾0}W = \left\{ W(t), t \geqslant 0\right\} is a standard Brownian motion.

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Problem VI.14

Let P(X0=1)=P(X0=−1)=1/2\mathbb {P}\left(X_{0} = 1\right) = \mathbb {P}\left(X_{0} = -1\right) = 1 / 2, and let X0X_{0} be independent of the Poisson process N={Nt,t⩾0}N = \left\{ N_{t}, t \geqslant 0\right\} with intensity λ>0\lambda > 0. Let the process X={Xt,t⩾0}X = \left\{ X_{t}, t \geqslant 0\right\} change its value to the value of the opposite sign at the jump points of the Poisson process. Is the process XX, called the telegraph signal, Markov? Draw the graph of a sample path of this process.

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Problem VI.15

For x⩾0x \geqslant 0 let τx=inf⁡{t⩾0:W(t)=x}\tau_{x} = \inf \left\{ t \geqslant 0: W(t) = x\right\}, where W={W(t),t⩾0}W = \left\{ W(t), t \geqslant 0\right\} is a Wiener process. Prove that the process {τx,x⩾0}\left\{ \tau_{x}, x \geqslant 0\right\} is a process with independent increments (hence Markov, by Theorem 3).

Theorem 3. Let X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\} be a process with independent increments taking values in Rd\mathbb {R}^{d}, d⩾1d \geqslant 1. Then XX is a Markov process (with respect to the natural filtration).

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Problem VI.16

Let X={Xt,t⩾0}X = \left\{ X_{t}, t \geqslant 0\right\} be a Markov process. For s>0s > 0 define the process Y={Yt=Xs−t,t∈[0,s]}Y = \left\{ Y_{t} = X_{s-t}, t \in [0, s]\right\}. Will YY be a Markov process? Must the process YY be a homogeneous process if XX is one?

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Problem VI.17

Prove that the Brownian bridge (see the definition in §9\S 9 of Chapter V) is an inhomogeneous Markov process.

Definition from §9\S 9 of Chapter V: W0W^{0} is the Brownian bridge, defined as the process W0(t)={W(t)−tW(1),t∈[0,1]}W^{0}(t) = \left\{ W(t)-t W(1), t \in [0,1]\right\}, where W={W(t),t∈[0,1]}W = \left\{ W(t), t \in [0,1]\right\} is a Wiener process on the interval [0,1][0,1].

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Problem VI.18

Let X={Xt,t⩾0}X = \left\{ X_{t}, t \geqslant 0\right\} be a stochastic process on [0,∞)[0, \infty ), and

Y={Yt=(Xt,t),t⩾0} Y = \left\{ Y_{t} = \left(X_{t}, t\right), t \geqslant 0\right\}

Prove that the processes XX and YY can only be Markov simultaneously. Prove that if YY is a Markov process, then this process is necessarily homogeneous. How are the transition functions of the Markov processes XX and YY related?

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Problem VI.19

Let X={Xt,t⩾0}X = \left\{ X_{t}, t \geqslant 0\right\} be a homogeneous Markov process with values (for each tt) in some Polish space S\mathrm{S}. Prove that there exist measurable functions hs:S×[0,1]→S,s⩾0h_{s}: \mathrm{S} \times [0,1] \rightarrow \mathrm{S}, s \geqslant 0 (i.e. hs∈B(S)⊗B([0,1])∣B(S),s⩾0h_{s} \in \mathscr {B}(\mathrm{S}) \otimes \mathscr {B}([0,1]) \mid \mathscr {B}(\mathrm{S}), s \geqslant 0) and uniformly distributed random variables θt,s\theta_{t, s} independent of {Xu,0⩽u⩽t}\left\{ X_{u}, 0 \leqslant u \leqslant t\right\} for t,s⩾0t, s \geqslant 0, such that Xt+s=hs(Xt,θt,s)X_{t+s} = h_{s}\left(X_{t}, \theta_{t, s}\right) a.s. for all t,s⩾0t, s \geqslant 0.

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Problem VI.20

Let a homogeneous Markov process take values in Rm\mathbb {R}^{m}, and let its transition function be such that P(x,t,B)=P(x+y,t,B+y)P(x, t, B) = P(x+y, t, B+y) for all x,y∈Rmx, y \in \mathbb {R}^{m}, t⩾0,B∈B(Rm)t \geqslant 0, B \in \mathscr {B}\left(\mathbb {R}^{m}\right). Prove that then this process has independent increments.

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Problem VI.21

Prove that for the standard Poisson process with intensity λ>0\lambda > 0 the "strong law of large numbers" holds: N(t)/t→λN(t) / t \rightarrow \lambda a.s. as t→∞t \rightarrow \infty.

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Problem VI.22

("waiting time paradox") Let the Poisson process N={Nt,t⩾0}N = \left\{ N_{t}, t \geqslant 0\right\} be constructed according to Theorem 2 of Chapter II from a sequence {ξj}\left\{ \xi_{j}\right\} of independent exponential (with parameter λ\lambda) random variables.

Theorem 2 of Chapter II (explicit construction of the Poisson process). Let ξ1,ξ2,…\xi_{1}, \xi_{2}, \ldots be independent random variables having the exponential distribution with density

pξi(x)={λe−λx,x⩾0,0,x<0,(II.4) p_{\xi _{i}}(x) = \begin{cases} \lambda e^{-\lambda x}, & x \geqslant 0, \\ 0, & x < 0, \end{cases} \tag {II.4}

where λ\lambda is a positive parameter, i∈Ni \in \mathbb {N}. Then the corresponding renewal process (I.10) is a Poisson process with intensity λ\lambda.

Formula (I.10) (Example 2 of Chapter I). For a sequence ξ1,ξ2,…\xi_{1}, \xi_{2}, \ldots of independent identically distributed nondegenerate nonnegative random variables, the renewal process is defined by the formula

X0(ω)=0,Xt(ω)=sup⁡{n:∑j⩽nξj(ω)⩽t},t>0.(I.10) X_{0}(\omega ) = 0, \quad X_{t}(\omega ) = \sup \left\{ n: \sum _{j \leqslant n} \xi _{j}(\omega ) \leqslant t\right\} , \quad t > 0 . \tag {I.10}

For fixed t>0t > 0 find the distribution of the random variable ζt=∑j=1Nt+1ξj−t\zeta_{t} = \sum_{j = 1}^{N_{t}+1} \xi_{j}-t, i.e. of the "overshoot" over the level tt of the accumulating partial sums Sk=∑j=1kξj,k⩾1S_{k} = \sum_{j = 1}^{k} \xi_{j}, k \geqslant 1. The following amusing question is related to this problem. Let ξj\xi_{j} be the random intervals between the arrival times of buses at a given stop. The question is: how long will you wait for the nearest bus to appear if you happen to be at the stop at time tt?

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Problem VI.23

(continuation of Exercise 22) Prove that for each fixed t>0t > 0 the random variables ζt,ξNt+2,ξNt+3,…\zeta_{t}, \xi_{N_{t}+2}, \xi_{N_{t}+3}, \ldots introduced above form a sequence of independent exponentially distributed random variables with parameter λ\lambda.

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Problem VI.24

Define m(B)=E[μ(B,⋅)]m(B) = \mathbb {E}\left[\mu (B, \cdot )\right] for all B∈B(R+)B \in \mathscr {B}\left(\mathbb {R}_{+}\right), where μ(B,⋅)\mu (B, \cdot ) appears in (85).

Suppose that on some probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) we are given a sequence of random variables {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} such that for ω∈Ω0\omega \in \Omega_{0}, where P(Ω0)=1\mathbb {P}\left(\Omega_{0}\right) = 1,

0<τ1<τ2<… and τn→∞ as n→∞. (84) 0 < \tau _{1} < \tau _{2} < \ldots \text{ and } \tau _{n} \rightarrow \infty \text{ as } n \rightarrow \infty \text{. } \tag {84}

Definition 11. The random point measure μ:B(R+)×Ω→Z‾+=Z+∪{∞}\mu : \mathscr {B}\left(\mathbb {R}_{+}\right) \times \Omega \rightarrow \overline{\mathbb {Z}}_{+} = \mathbb {Z}_{+} \cup \left\{ \infty \right\} is given by the formula

μ(B,ω)=∑k=1∞δτk(ω)(B),B∈B(R+),ω∈Ω(85) \mu (B, \omega ) = \sum _{k = 1}^{\infty } \delta _{\tau _{k}(\omega )}(B), \quad B \in \mathscr {B}\left(\mathbb {R}_{+}\right), \quad \omega \in \Omega \tag {85}

where the sequence {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} is described above, and δx\delta_{x} is the Dirac measure.

Will the function m(⋅)m(\cdot ) be a measure on B(R+)\mathscr {B}\left(\mathbb {R}_{+}\right)? Can a Poisson process with a locally finite intensity measure mm be represented in the form (86)?

Definition 12. The random point process Y={Y(t),t⩾0}Y = \left\{ Y(t), t \geqslant 0\right\}, where Y(0)=0Y(0) = 0, is introduced for t>0t > 0 as follows:

Y(t,ω)=μ((0,t],ω)=∑k=1∞1(0,t](τk(ω))(86) Y(t, \omega ) = \mu ((0, t], \omega ) = \sum _{k = 1}^{\infty } \mathbf{1}_{(0, t]}\left(\tau _{k}(\omega )\right) \tag {86}

where the random point measure μ\mu is given by formula (85).

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Problem VI.25

(compare with Theorem 14 of Chapter V). Prove that the random point process (86) is a Poisson process if YY has independent increments and YY is stochastically continuous on R+\mathbb {R}_{+} (see (I.56)).

Theorem 14 of Chapter V (Doob). Let a process X={Xt,t⩾0}X = \left\{ X_{t}, t \geqslant 0\right\} take values in Rm\mathbb {R}^{m} for each tt and have independent increments. If the sample paths of the process are continuous a.s., then the process Y={Yt=Xt−X0,t⩾0}Y = \left\{ Y_{t} = X_{t}-X_{0}, t \geqslant 0\right\} is Gaussian. Moreover, for 0⩽s⩽t<∞0 \leqslant s \leqslant t < \infty

E[Xt−Xs]=Mt−Ms,Var⁡[Xt−Xs]=Gt−Gs(V.42) \mathbb {E}\left[X_{t}-X_{s}\right] = M_{t}-M_{s}, \quad \operatorname {Var}\left[X_{t}-X_{s}\right] = G_{t}-G_{s} \tag {V.42}

where M:R+→RmM: \mathbb {R}_{+} \rightarrow \mathbb {R}^{m} and G:R+→Rm2G: \mathbb {R}_{+} \rightarrow \mathbb {R}^{m^2} are continuous functions.

Suppose that on some probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) we are given a sequence of random variables {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} such that for ω∈Ω0\omega \in \Omega_{0}, where P(Ω0)=1\mathbb {P}\left(\Omega_{0}\right) = 1,

0<τ1<τ2<… and τn→∞ as n→∞. (84) 0 < \tau _{1} < \tau _{2} < \ldots \text{ and } \tau _{n} \rightarrow \infty \text{ as } n \rightarrow \infty \text{. } \tag {84}

Definition 11. The random point measure μ:B(R+)×Ω→Z‾+=Z+∪{∞}\mu : \mathscr {B}\left(\mathbb {R}_{+}\right) \times \Omega \rightarrow \overline{\mathbb {Z}}_{+} = \mathbb {Z}_{+} \cup \left\{ \infty \right\} is given by the formula

μ(B,ω)=∑k=1∞δτk(ω)(B),B∈B(R+),ω∈Ω(85) \mu (B, \omega ) = \sum _{k = 1}^{\infty } \delta _{\tau _{k}(\omega )}(B), \quad B \in \mathscr {B}\left(\mathbb {R}_{+}\right), \quad \omega \in \Omega \tag {85}

where the sequence {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} is described above, and δx\delta_{x} is the Dirac measure.

Definition 12. The random point process Y={Y(t),t⩾0}Y = \left\{ Y(t), t \geqslant 0\right\}, where Y(0)=0Y(0) = 0, is introduced for t>0t > 0 as follows:

Y(t,ω)=μ((0,t],ω)=∑k=1∞1(0,t](τk(ω))(86) Y(t, \omega ) = \mu ((0, t], \omega ) = \sum _{k = 1}^{\infty } \mathbf{1}_{(0, t]}\left(\tau _{k}(\omega )\right) \tag {86}

where the random point measure μ\mu is given by formula (85).

Formula (I.56): a process X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\} with values in a metric space (S,ρ)(\mathrm{S}, \rho ) is stochastically continuous on TT if for each t∈Tt \in T and any ε>0\varepsilon > 0

lim⁡s→tP(ρ(Xs,Xt)>ε)=0.(I.56) \lim _{s \rightarrow t} \mathbb {P}\left(\rho \left(X_{s}, X_{t}\right) > \varepsilon \right) = 0 . \tag {I.56}
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Problem VI.26

By Theorem 3, the random variables τk=ξ1+⋯+ξk,k∈N\tau_{k} = \xi_{1}+\cdots +\xi_{k}, k \in \mathbb {N}, form a Markov chain if the summands are independent real random variables.

Theorem 3. Let X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\} be a process with independent increments taking values in Rd\mathbb {R}^{d}, d⩾1d \geqslant 1. Then XX is a Markov process (with respect to the natural filtration).

Will the process (86) be Markov if {τk,k∈N}\left\{ \tau_{k}, k \in \mathbb {N}\right\} is an arbitrary Markov chain satisfying condition (84)?

Suppose that on some probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) we are given a sequence of random variables {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} such that for ω∈Ω0\omega \in \Omega_{0}, where P(Ω0)=1\mathbb {P}\left(\Omega_{0}\right) = 1,

0<τ1<τ2<… and τn→∞ as n→∞. (84) 0 < \tau _{1} < \tau _{2} < \ldots \text{ and } \tau _{n} \rightarrow \infty \text{ as } n \rightarrow \infty \text{. } \tag {84}

Definition 11. The random point measure μ:B(R+)×Ω→Z‾+=Z+∪{∞}\mu : \mathscr {B}\left(\mathbb {R}_{+}\right) \times \Omega \rightarrow \overline{\mathbb {Z}}_{+} = \mathbb {Z}_{+} \cup \left\{ \infty \right\} is given by the formula

μ(B,ω)=∑k=1∞δτk(ω)(B),B∈B(R+),ω∈Ω(85) \mu (B, \omega ) = \sum _{k = 1}^{\infty } \delta _{\tau _{k}(\omega )}(B), \quad B \in \mathscr {B}\left(\mathbb {R}_{+}\right), \quad \omega \in \Omega \tag {85}

where the sequence {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} is described above, and δx\delta_{x} is the Dirac measure.

Definition 12. The random point process Y={Y(t),t⩾0}Y = \left\{ Y(t), t \geqslant 0\right\}, where Y(0)=0Y(0) = 0, is introduced for t>0t > 0 as follows:

Y(t,ω)=μ((0,t],ω)=∑k=1∞1(0,t](τk(ω))(86) Y(t, \omega ) = \mu ((0, t], \omega ) = \sum _{k = 1}^{\infty } \mathbf{1}_{(0, t]}\left(\tau _{k}(\omega )\right) \tag {86}

where the random point measure μ\mu is given by formula (85).

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Problem VI.27

Definition 13. Suppose that on some probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) we are given a Poisson process N={N(t),t⩾0}N = \left\{ N(t), t \geqslant 0\right\} with intensity λ=1\lambda = 1 and a process Λ={Λ(t),t⩾0}\Lambda = \left\{ \Lambda (t), t \geqslant 0\right\} independent of it, with a.s. nondecreasing, finite and right-continuous sample paths starting from zero. The Cox process is the process Z={Z(t),t⩾0}Z = \left\{ Z(t), t \geqslant 0\right\}, where Z(t,ω)=N(Λ(t,ω),ω),t⩾0,ω∈ΩZ(t, \omega ) = N(\Lambda (t, \omega ), \omega ), t \geqslant 0, \omega \in \Omega.

Is the Cox process Markov?

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Problem VI.28

Find the covariance function of the process (87) if the sequence {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} defines a Poisson point process with intensity λ\lambda.

Suppose that on some probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) we are given a sequence of random variables {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} such that for ω∈Ω0\omega \in \Omega_{0}, where P(Ω0)=1\mathbb {P}\left(\Omega_{0}\right) = 1,

0<τ1<τ2<… and τn→∞ as n→∞. (84) 0 < \tau _{1} < \tau _{2} < \ldots \text{ and } \tau _{n} \rightarrow \infty \text{ as } n \rightarrow \infty \text{. } \tag {84}

Definition 14. We say that a stochastic process X={X(t),t⩾0}X = \left\{ X(t), t \geqslant 0\right\} is "shot noise" if

X(t,ω)=∑k=1∞ϕ(t−τk(ω)),t⩾0(87) X(t, \omega ) = \sum _{k = 1}^{\infty } \phi \left(t-\tau _{k}(\omega )\right), \quad t \geqslant 0 \tag {87}

where the bounded Borel function ϕ:R→R\phi : \mathbb {R} \rightarrow \mathbb {R} has compact support, and the random variables τk\tau_{k} (k∈Nk \in \mathbb {N}) satisfy condition (84) (which ensures that the series (87) converges for each t⩾0t \geqslant 0 on the set Ω0\Omega_{0} of probability one).

Is it sufficient, for the convergence of the series (87) with probability one for each t⩾0t \geqslant 0, to require that the bounded function ϕ\phi be Lebesgue integrable?

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Problem VI.29

Suppose that on some probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) we are given a sequence of random variables {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} such that for ω∈Ω0\omega \in \Omega_{0}, where P(Ω0)=1\mathbb {P}\left(\Omega_{0}\right) = 1,

0<τ1<τ2<… and τn→∞ as n→∞. (84) 0 < \tau _{1} < \tau _{2} < \ldots \text{ and } \tau _{n} \rightarrow \infty \text{ as } n \rightarrow \infty \text{. } \tag {84}

Suppose that, besides the sequence {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} of random variables satisfying condition (84), on the same probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) there is some sequence {ηk}k⩾1\left\{ \eta_{k}\right\}_{k \geqslant 1} of real random variables "attached" to the times τk,k⩾1\tau_{k}, k \geqslant 1.

Definition 15. A marked random point process is a process (indexed by sets CC) of the following form

μ~(C;ω)=∑k=1∞δ(τk(ω),ηk(ω))(C),C∈B(R+)×B(R),ω∈Ω \widetilde{\mu }(C ; \omega ) = \sum _{k = 1}^{\infty } \delta _{\left(\tau _{k}(\omega ), \eta _{k}(\omega )\right)}(C), \quad C \in \mathscr {B}\left(\mathbb {R}_{+}\right) \times \mathscr {B}(\mathbb {R}), \quad \omega \in \Omega

The function

Y(t,D;ω)=μ~((0,t]×D;ω),t⩾0,D∈B(R) Y(t, D ; \omega ) = \widetilde{\mu }((0, t] \times D ; \omega ), \quad t \geqslant 0, \quad D \in \mathscr {B}(\mathbb {R})

is called the counting function.

Draw the graph of the sample path Y(t,D),t⩾0Y(t, D), t \geqslant 0, for fixed D∈B(R)D \in \mathscr {B}(\mathbb {R}).

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Problem VI.30

Let N={N(t),t⩾0}N = \left\{ N(t), t \geqslant 0\right\} be a Poisson process with mean measure m=m(⋅)m = m(\cdot ), determined by the sequence {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} that specifies the jump points of NN. Let {ηk}k⩾1\left\{ \eta_{k}\right\}_{k \geqslant 1} be a sequence of independent identically distributed real random variables with Law⁡(ηk)=P~\operatorname {Law}\left(\eta_{k}\right) = \widetilde{\mathbb {P}}. Take D∈B(R)D \in \mathscr {B}(\mathbb {R}) such that P~(D)>0\widetilde{\mathbb {P}}(D) > 0. Prove that the "thinned process" Y(t,D),t⩾0Y(t, D), t \geqslant 0, defined by formula (88), is a Poisson process with mean measure m(⋅)P~(D)m(\cdot ) \widetilde{\mathbb {P}}(D).

Definition 11. The random point measure μ:B(R+)×Ω→Z‾+=Z+∪{∞}\mu : \mathscr {B}\left(\mathbb {R}_{+}\right) \times \Omega \rightarrow \overline{\mathbb {Z}}_{+} = \mathbb {Z}_{+} \cup \left\{ \infty \right\} is given by the formula

μ(B,ω)=∑k=1∞δτk(ω)(B),B∈B(R+),ω∈Ω(85) \mu (B, \omega ) = \sum _{k = 1}^{\infty } \delta _{\tau _{k}(\omega )}(B), \quad B \in \mathscr {B}\left(\mathbb {R}_{+}\right), \quad \omega \in \Omega \tag {85}

where the sequence {τk}k⩾1\left\{ \tau_{k}\right\}_{k \geqslant 1} is described above, and δx\delta_{x} is the Dirac measure.

Definition 12. The random point process Y={Y(t),t⩾0}Y = \left\{ Y(t), t \geqslant 0\right\}, where Y(0)=0Y(0) = 0, is introduced for t>0t > 0 as follows:

Y(t,ω)=μ((0,t],ω)=∑k=1∞1(0,t](τk(ω))(86) Y(t, \omega ) = \mu ((0, t], \omega ) = \sum _{k = 1}^{\infty } \mathbf{1}_{(0, t]}\left(\tau _{k}(\omega )\right) \tag {86}

where the random point measure μ\mu is given by formula (85).

Formula (88):

Y(t,D;ω)={0 if Y(t,ω)=0∑k=1Y(t)1D(ηk) if Y(t)⩾1(88) Y(t, D ; \omega ) = \begin{cases} 0 & \text{ if } Y(t, \omega ) = 0 \\ \sum _{k = 1}^{Y(t)} \mathbf{1}_{D}\left(\eta _{k}\right) & \text{ if } Y(t) \geqslant 1 \end{cases} \tag {88}

where the process YY is given by formula (86).

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Problem VI.31

Let X={X(t),t⩾0}X = \left\{ X(t), t \geqslant 0\right\} be a homogeneous standard (i.e., pij(t)→δijp_{i j}(t) \rightarrow \delta_{i j} as t→0+t \rightarrow 0+) Markov chain with a finite or countable state space S\mathrm{S} and transition probabilities pij(t)p_{i j}(t), considered in its separable modification, and let Q=(qij)i,j∈SQ = \left(q_{i j}\right)_{i, j \in \mathrm{S}} be its infinitesimal matrix:

qij=d+pij(t)dt∣t=0,qi=−qii∈[0,∞]. q_{i j} = \left.\frac{d^{+} p_{i j}(t)}{d t}\right|_{t = 0}, \quad q_{i} = -q_{i i} \in [0, \infty ] .

Let qi=∞q_{i} = \infty for a state ii. Then the probability that the process XX stays in this state for a positive time is zero, i.e., having entered this state, the process leaves it instantly.

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Problem VI.32

Let Q=(qij)i,j∈SQ = \left(q_{i j}\right)_{i, j \in \mathrm{S}} be the infinitesimal matrix of a homogeneous Markov chain, and qi=−qiiq_{i} = -q_{i i}.

Definition 10. A chain is called conservative if all entries qijq_{i j} of the infinitesimal matrix QQ are finite and, moreover, for every ii

∑j≠iqij=qi.(63) \sum _{j \neq i} q_{i j} = q_{i} . \tag {63}

Let S\mathrm{S} be a finite or countable set, and let X={X(t),t⩾0}X = \left\{ X(t), t \geqslant 0\right\} be a conservative Markov chain with infinitesimal matrix QQ, where qi∈(0,∞)q_{i} \in (0, \infty ), i∈Si \in \mathrm{S}. Let us construct recursively sequences of random variables {ηn}n⩾1\left\{ \eta_{n}\right\}_{n \geqslant 1} and {τn}n⩾1\left\{ \tau_{n}\right\}_{n \geqslant 1} with values in S\mathrm{S} and R+=[0,∞)\mathbb {R}_{+} = [0, \infty ) respectively, and such that η1\eta_{1} has an arbitrary distribution,

P(τ1>t∣η1=i)=e−qit,t>0,i∈S(92) \mathbb {P}\left(\tau _{1} > t \mid \eta _{1} = i\right) = e^{-q_{i} t}, \quad t > 0, \quad i \in \mathrm{S} \tag {92}

and for n⩾1,i,j,i1,…,in∈S,x1,…,xn∈R+n \geqslant 1, i, j, i_{1}, \ldots , i_{n} \in \mathrm{S}, x_{1}, \ldots , x_{n} \in \mathbb {R}_{+} let

P(ηn+1=j∣τ1=x1,…,τn=xn,η1=i1,…,ηn−1=in−1,ηn=i)=qij/qi,i≠jP(τn+1−τn>t∣τ1=x1,…,τn=xn,η1=i1,…,ηn=in,ηn+1=j)=e−qjt(93) \begin{gathered} \mathbb {P}\left(\eta _{n+1} = j \mid \tau _{1} = x_{1}, \ldots , \tau _{n} = x_{n}, \eta _{1} = i_{1}, \ldots , \eta _{n-1} = i_{n-1}, \eta _{n} = i\right) = q_{i j} / q_{i}, \quad i \neq j \\ \mathbb {P}\left(\tau _{n+1}-\tau _{n} > t \mid \tau _{1} = x_{1}, \ldots , \tau _{n} = x_{n}, \eta _{1} = i_{1}, \ldots , \eta _{n} = i_{n}, \eta _{n+1} = j\right) = e^{-q_{j} t} \end{gathered} \tag {93}

Prove that on a suitable probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) there exist the sequences of random variables {ηn}n⩾1\left\{ \eta_{n}\right\}_{n \geqslant 1} and {τn}n⩾1\left\{ \tau_{n}\right\}_{n \geqslant 1} described above, satisfying conditions (92) and (93). Moreover, 0=τ0<τ1(ω)<τ2(ω)<⋯0 = \tau_{0} < \tau_{1}(\omega ) < \tau_{2}(\omega ) < \cdots a.s. Can one assert that always lim⁡n→∞τn=∞\lim_{n \rightarrow \infty } \tau_{n} = \infty a.s.? Prove that the latter certainly holds if sup⁡qi<∞\sup q_{i} < \infty.

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Problem VI.33

Let τ0(ω)=0\tau_{0}(\omega ) = 0 a.s. Using the random variables ηn,τn(n⩾1)\eta_{n}, \tau_{n}(n \geqslant 1) appearing in Exercise 32 , and assuming that condition (84) is satisfied, let us introduce the process Y={Y(t),t⩾0}Y = \left\{ Y(t), t \geqslant 0\right\} by setting

Y(t,ω)=ηn(ω) for τn−1(ω)⩽t<τn(ω),n⩾1 Y(t, \omega ) = \eta _{n}(\omega ) \text{ for } \tau _{n-1}(\omega ) \leqslant t < \tau _{n}(\omega ), \quad n \geqslant 1

(for ω\omega such that τn−1(ω)=τn(ω)\tau_{n-1}(\omega ) = \tau_{n}(\omega ) for some n⩾1n \geqslant 1, let Y(t,ω)=0Y(t, \omega ) = 0, t⩾0)t \geqslant 0).

Condition (84): for ω∈Ω0\omega \in \Omega_{0}, where P(Ω0)=1\mathbb {P}\left(\Omega_{0}\right) = 1,

0<τ1<τ2<… and τn→∞ as n→∞. (84) 0 < \tau _{1} < \tau _{2} < \ldots \text{ and } \tau _{n} \rightarrow \infty \text{ as } n \rightarrow \infty \text{. } \tag {84}

Prove that YY is a Markov process having the same finite-dimensional distributions as the process XX that determined the matrix QQ. How should the above construction be modified if some of the quantities qiq_{i} are equal to zero?

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Problem VI.34

Let X={X(t),t⩾0}X = \left\{ X(t), t \geqslant 0\right\} be a homogeneous standard (i.e., pij(t)→δijp_{i j}(t) \rightarrow \delta_{i j} as t→0+t \rightarrow 0+) Markov chain with a finite or countable state space S\mathrm{S} and transition probabilities pij(t)p_{i j}(t), and let Q=(qij)i,j∈SQ = \left(q_{i j}\right)_{i, j \in \mathrm{S}} be its infinitesimal matrix:

qij=d+pij(t)dt∣t=0,qi=−qii∈[0,∞]. q_{i j} = \left.\frac{d^{+} p_{i j}(t)}{d t}\right|_{t = 0}, \quad q_{i} = -q_{i i} \in [0, \infty ] .

Prove that sup⁡iqi<∞\sup_{i} q_{i} < \infty if and only if pii(t)→1p_{i i}(t) \rightarrow 1 as t→0+t \rightarrow 0+ uniformly in i∈Si \in \mathrm{S} (or, equivalently, pij(t)→δijp_{i j}(t) \rightarrow \delta_{i j} as t→0+t \rightarrow 0+ uniformly in i,j∈Si, j \in \mathrm{S} ).

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Problem VI.35

Give an example of transition probabilities for which the standardness condition (49) does not hold.

Definition 9. A homogeneous Markov chain X={Xt,t⩾0}X = \left\{ X_{t}, t \geqslant 0\right\} (or the semigroup P(t)=(pij(t))P(t) = \left(p_{i j}(t)\right) generated by it) is called standard if P(t)→IP(t) \rightarrow I as t→0+t \rightarrow 0+, i.e., for all i,j∈Si, j \in \mathrm{S}

lim⁡t→0+pij(t)=δij,(49) \lim _{t \rightarrow 0+} p_{i j}(t) = \delta _{i j}, \tag {49}

where, as usual, δij\delta_{i j} is the Kronecker symbol.

Give an example where property (49) holds but (89) does not.

Condition (89), where qij=d+pij(t)dt∣t=0q_{i j} = \left.\frac{d^{+} p_{i j}(t)}{d t}\right|_{t = 0} are the entries of the infinitesimal matrix QQ:

qi:=−qii<∞ for all i∈S.(89) q_{i}: = -q_{i i} < \infty \text{ for all } i \in \mathrm{S} . \tag {89}
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Problem VI.36

Definition 9. A homogeneous Markov chain X={Xt,t⩾0}X = \left\{ X_{t}, t \geqslant 0\right\} (or the semigroup P(t)=(pij(t))P(t) = \left(p_{i j}(t)\right) generated by it) is called standard if P(t)→IP(t) \rightarrow I as t→0+t \rightarrow 0+, i.e., for all i,j∈Si, j \in \mathrm{S}

lim⁡t→0+pij(t)=δij,(49) \lim _{t \rightarrow 0+} p_{i j}(t) = \delta _{i j}, \tag {49}

where, as usual, δij\delta_{i j} is the Kronecker symbol.

Let the state space S\mathrm{S} consist of a finite number NN of elements, and let the transition matrix P(t)P(t) be standard. Prove that then this matrix admits the following representation:

P(t)=exp(tQ),t⩾0,(94) P(t) = \mathrm{exp}\left(t Q\right), \quad t \geqslant 0, \tag {94}

where the N×NN \times N matrix QQ satisfies the condition

qij⩾0 for i≠j and ∑jqij=0 for each i.(95) q_{i j} \geqslant 0 \text{ for } i \neq j \text{ and } \sum _{j} q_{i j} = 0 \text{ for each } i . \tag {95}

Conversely, if conditions (95) hold for a matrix QQ, then formula (94) defines a standard matrix of transition probabilities.

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Problem VI.37

The backward system of Kolmogorov equations for the transition probabilities pij(t)p_{i j}(t) of a homogeneous Markov chain with infinitesimal matrix Q=(qij)i,j∈SQ = \left(q_{i j}\right)_{i, j \in \mathrm{S}} has the form

pij′(t)=∑kqikpkj(t),(65) p_{i j}^{\prime }(t) = \sum _{k} q_{i k} p_{k j}(t), \tag {65}

and the forward system has the form

pij′(t)=∑kpik(t)qkj for all i,j∈S.(72) p_{i j}^{\prime }(t) = \sum _{k} p_{i k}(t) q_{k j} \text{ for all } i, j \in \mathrm{S} . \tag {72}

Equations (65) and (72) can be understood either in the narrow sense (i.e., that all pij′(t)p_{i j}^{\prime }(t) are continuous for t⩾0t \geqslant 0 and the equations under study hold for all t⩾0t \geqslant 0 ), or in the broad sense, i.e., that pij(t)p_{i j}(t) are absolutely continuous and the equations under consideration hold for almost all (with respect to Lebesgue measure) t⩾0t \geqslant 0.

Prove that for the systems of Kolmogorov equations both interpretations given above (in the narrow and in the broad sense) are equivalent. For the forward equations this exercise should be classed among the very difficult ones.

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Problem VI.38

Let Q=(qij)i,j∈SQ = \left(q_{i j}\right)_{i, j \in \mathrm{S}} be a matrix with real entries, where S\mathrm{S} is countable, such that

qij⩾0 for i≠j,qi=−qii⩾0, and ∑jqij⩽0 for i∈S.(96) q_{i j} \geqslant 0 \text{ for } i \neq j, \quad q_{i} = -q_{i i} \geqslant 0, \text{ and } \sum _{j} q_{i j} \leqslant 0 \text{ for } i \in \mathrm{S} . \tag {96}

The backward and forward systems of Kolmogorov equations for the matrices P(t)=(pij(t))i,j∈SP(t) = \left(p_{i j}(t)\right)_{i, j \in \mathrm{S}}, t⩾0t \geqslant 0, have, respectively, the form

pij′(t)=∑kqikpkj(t)(65) p_{i j}^{\prime }(t) = \sum _{k} q_{i k} p_{k j}(t) \tag {65}

and

pij′(t)=∑kpik(t)qkj for all i,j∈S.(72) p_{i j}^{\prime }(t) = \sum _{k} p_{i k}(t) q_{k j} \quad \text{ for all } \quad i, j \in \mathrm{S} . \tag {72}

(see also Exercise 34). Prove that if sup⁡iqi<∞\sup_{i} q_{i} < \infty, then both systems of Kolmogorov equations have a unique solution.

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Problem VI.39

Definition 17. A homogeneous Markov process X={X(t),t⩾0}X = \left\{ X(t), t \geqslant 0\right\} with state space S={0,1,…}S = \left\{ 0,1, \ldots \right\} is called a birth-and-death process if its infinitesimal matrix QQ has the following form

qi,i−1=μi,qii=−(λi+μi),qi,i+1=λi,qij=0 for ∣i−j∣>1, q_{i, i-1} = \mu _{i}, \quad q_{i i} = -\left(\lambda _{i}+\mu _{i}\right), \quad q_{i, i+1} = \lambda _{i}, \quad q_{i j} = 0 \quad \text{ for } \quad \left|i-j\right| > 1,

where μ0=0,μi>0,i⩾1\mu_{0} = 0, \mu_{i} > 0, i \geqslant 1 and λi>0,i⩾0\lambda_{i} > 0, i \geqslant 0. If μi=0\mu_{i} = 0 for all i⩾0i \geqslant 0, one speaks of a birth process, and if λi=0,i⩾0\lambda_{i} = 0, i \geqslant 0, of a death process.

The backward and forward systems of Kolmogorov equations for the matrices P(t)=(pij(t))i,j∈SP(t) = \left(p_{i j}(t)\right)_{i, j \in \mathrm{S}}, t⩾0t \geqslant 0, have, respectively, the form

pij′(t)=∑kqikpkj(t)(65) p_{i j}^{\prime }(t) = \sum _{k} q_{i k} p_{k j}(t) \tag {65}

and

pij′(t)=∑kpik(t)qkj for all i,j∈S.(72) p_{i j}^{\prime }(t) = \sum _{k} p_{i k}(t) q_{k j} \quad \text{ for all } \quad i, j \in \mathrm{S} . \tag {72}

Solutions are sought in the class of semistochastic matrices (P(t))t⩾0(P(t))_{t \geqslant 0}, for which ∑jpij(t)⩽1,t⩾0,i∈S\sum_{j} p_{i j}(t) \leqslant 1, t \geqslant 0, i \in \mathrm{S}. If it turns out that ∑jpij(t)=1\sum_{j} p_{i j}(t) = 1 for all t⩾0,i∈St \geqslant 0, i \in \mathrm{S}, then one speaks of a proper solution.

Suppose a birth process is such that ∑iλi−1<∞\sum_{i} \lambda_{i}^{-1} < \infty. Prove that then the system of forward Kolmogorov equations has a unique solution, while the system of backward equations has infinitely many. Prove that if ∑iμi−1<∞\sum_{i} \mu_{i}^{-1} < \infty for a death process, then the system of backward equations has a unique solution, while the system of forward equations has infinitely many, exactly one of which is proper.

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Problem VI.40

Definition 19. Let a probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) and σ\sigma-algebras E,A1,A2⊂F\mathscr {E}, \mathscr {A}_{1}, \mathscr {A}_{2} \subset \mathscr {F} be given. The σ\sigma-algebra E\mathscr {E} is said to split A1\mathscr {A}_{1} and A2\mathscr {A}_{2} if

P(A1A2∣E)=P(A1∣E)P(A2∣E) for any Ak∈Ak,k=1,2(99) \mathbb {P}\left(A_{1} A_{2} \mid \mathscr {E}\right) = \mathbb {P}\left(A_{1} \mid \mathscr {E}\right) \mathbb {P}\left(A_{2} \mid \mathscr {E}\right) \quad \text{ for any } \quad A_{k} \in \mathscr {A}_{k}, \quad k = 1,2 \tag {99}

(compare with (1)). Show that if E\mathscr {E} splits A1\mathscr {A}_{1} and A2\mathscr {A}_{2}, then E\mathscr {E} splits A1∨E\mathscr {A}_{1} \vee \mathscr {E} and A2∨E\mathscr {A}_{2} \vee \mathscr {E} (as before, A∨E=σ{A,E}\mathscr {A} \vee \mathscr {E} = \sigma \left\{ \mathscr {A}, \mathscr {E}\right\} ). Check that condition (99) is equivalent to

P(A∣A1∨E)=P(A∣E) for all A∈A2 \mathbb {P}\left(A \mid \mathscr {A}_{1} \vee \mathscr {E}\right) = \mathbb {P}\left(A \mid \mathscr {E}\right) \text{ for all } \quad A \in \mathscr {A}_{2}

Definition 1: a process XX adapted to the filtration (Ft)t∈T\left(\mathscr {F}_{t}\right)_{t \in T} is called Markov (with respect to this filtration and the measure P\mathbb {P}) if for each s∈Ts \in T and any event C∈F⩾s=σ{Xu,u⩾s,u∈T}C \in \mathscr {F}_{ \geqslant s} = \sigma \left\{ X_{u}, u \geqslant s, u \in T\right\}

P(C∣Fs)=P(C∣Xs) a.s.(1) \mathbb {P}\left(C \mid \mathscr {F}_{s}\right) = \mathbb {P}\left(C \mid X_{s}\right) \text{ a.s.} \tag {1}

Let σ\sigma-algebras Eα⊂A1\mathscr {E}_{\alpha } \subset \mathscr {A}_{1} be given, and suppose that for each α∈I\alpha \in I the σ\sigma-algebra Eα\mathscr {E}_{\alpha } splits A1\mathscr {A}_{1} and A2\mathscr {A}_{2}. Verify that then ⋂α∈IEα\bigcap_{\alpha \in I} \mathscr {E}_{\alpha } splits A1\mathscr {A}_{1} and A2\mathscr {A}_{2}, and that if property (99) holds, then A1∩A2⊂E\mathscr {A}_{1} \cap \mathscr {A}_{2} \subset \mathscr {E}. In particular, (83) involves the minimal splitting algebra, since σ{Xt}⊂Ft∩F⩾t\sigma \left\{ X_{t}\right\} \subset \mathscr {F}_{t} \cap \mathscr {F}_{ \geqslant t}.

Relation (83): for each s∈Ts \in T and any A∈Fs,B∈F⩾sA \in \mathscr {F}_{s}, B \in \mathscr {F}_{ \geqslant s} the following equality holds

P(AC∣Xt)=P(A∣Xt)P(C∣Xt).(83) \mathbb {P}\left(A C \mid X_{t}\right) = \mathbb {P}\left(A \mid X_{t}\right) \mathbb {P}\left(C \mid X_{t}\right) . \tag {83}
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