Consider a sequence of partitions of [0,T] : 0=t0n<t1n<⋯<tnn=T such that limn→∞maxk(tk+1n−tkn)=0 . Define ξkn by ξ0n=X(0) and ξk+1n=ξkn+a(tkn,ξkn)Δtkn+σ1(tkn,ξkn)σ2(tkn,Y(tkn))ΔWkn. It follows from Lemma 1, Lemma 2, and Proposition 1 in Section 5 that it is possible to choose a subsequence n′ and construct processes ξ˜n′ , W˜n′ , and Y˜n′ such that the finite-dimensional distributions of ξ˜n′ , W˜n′ , and Y˜n′ coincide with those of ξn′ , W, and Y and ξ˜n′→ξ˜ , W˜n′→W˜ and Y˜n′→Y˜ in probability, where ξ˜ , W˜ , and Y˜ are some stochastic processes (evidently, W˜ is a Wiener process). It suffices to prove that ξ˜ is a solution of Eq. (2) when W and Y are replaced by W˜ and Y˜ .

We have that ξ˜n′ satisfies the equation ξ˜n′(t)=ξ˜n′(0)+∑tk+1n′≤ta(tkn′,ξ˜n′(tkn′))Δtkn′+∑tk+1n′≤tσ1(tkn′,ξ˜n′(tkn′))σ2(tkn′,Y˜n′(tkn′))ΔW˜kn′. Since σ2 is bounded and σ1 is of linear growth, their product is of linear growth: |σ1(t,x)σ2(t,y)|≤K1(1+x), where K1>0 0$]]> is a constant. Therefore, P(sup0≤t≤Tσ1(t,ξ˜n′(t))σ2(t,Y˜n′(t))>C)≤P(sup0≤t≤TK1(1+ξ˜n′(t))>C)=P(sup0≤t≤Tξ˜n′(t)>CK1−1)=P(sup0≤t≤T|ξn′(t)|>CK1−1). C\Big)\\{} & \displaystyle \hspace{1em}\le \mathbb{P}\Big(\underset{0\le t\le T}{\sup }K_{1}\big(1+\left|\widetilde{\xi }_{{n^{\prime }}}(t)\right|\big)>C\Big)=\mathbb{P}\bigg(\underset{0\le t\le T}{\sup }\left|\widetilde{\xi }_{{n^{\prime }}}(t)\right|>\frac{C}{K_{1}}-1\bigg)\\{} & \displaystyle \hspace{1em}=\mathbb{P}\bigg(\underset{0\le t\le T}{\sup }\big|{\xi }^{{n^{\prime }}}(t)\big|>\frac{C}{K_{1}}-1\bigg).\end{array}\]]]> Using Lemma 1, we get that P(sup0≤t≤Tσ1(t,ξ˜n′(t))σ2(t,Y˜n′(t))>C)→0asC→∞. C\Big)\to 0\hspace{1em}\text{as}\hspace{2.5pt}C\to \infty .\]]]> Moreover, we have that σ1(t,x)σ2(t,y) is continuous w.r.t.  t∈[0,T] , x,y∈R . Then, for any ε>0 0$]]>, there exists δ>0 0$]]> such that |σ1(t1,x1)σ2(t1,y1)−σ1(t2,x2)σ2(t2,y2)|<ε whenever t1−t2<δ , x1−x2<δ , y1−y2<δ . Therefore, P(σ1(t1,ξ˜n′(t1))σ2(t1,Y˜n′(t1))−σ1(t2,ξ˜n′(t2))σ2(t2,Y˜n′(t2))>ε)≤P(ξ˜n′(t1)−ξ˜n′(t2)<δ,Y˜n′(t1)−Y˜n′(t2)<δ,σ1(t1,ξ˜n′(t1))σ2(t1,Y˜n′(t1))−σ1(t2,ξ˜n′(t2))σ2(t2,Y˜n′(t2))>ε)+P(ξ˜n′(t1)−ξ˜n′(t2)≥δ)+P(Y˜n′(t1)−Y˜n′(t2)≥δ)=P(ξn′(t1)−ξn′(t2)≥δ)+P(Y(t1)−Y(t2)≥δ), \varepsilon \big)\\{} & \displaystyle \le \mathbb{P}\big(\left|\widetilde{\xi }_{{n^{\prime }}}(t_{1})-\widetilde{\xi }_{{n^{\prime }}}(t_{2})\right|<\delta ,\left|\widetilde{Y}_{{n^{\prime }}}(t_{1})-\widetilde{Y}_{{n^{\prime }}}(t_{2})\right|<\delta ,\\{} & \displaystyle \hspace{1em}\left|\sigma _{1}\big(t_{1},\widetilde{\xi }_{{n^{\prime }}}(t_{1})\big)\sigma _{2}\big(t_{1},\widetilde{Y}_{{n^{\prime }}}(t_{1})\big)-\sigma _{1}\big(t_{2},\widetilde{\xi }_{{n^{\prime }}}(t_{2})\big)\sigma _{2}\big(t_{2},\widetilde{Y}_{{n^{\prime }}}(t_{2})\big)\right|>\varepsilon \big)\\{} & \displaystyle \hspace{1em}+\mathbb{P}\big(\left|\widetilde{\xi }_{{n^{\prime }}}(t_{1})-\widetilde{\xi }_{{n^{\prime }}}(t_{2})\right|\ge \delta \big)+\mathbb{P}\big(\left|\widetilde{Y}_{{n^{\prime }}}(t_{1})-\widetilde{Y}_{{n^{\prime }}}(t_{2})\right|\ge \delta \big)\\{} & \displaystyle =\mathbb{P}\big(\left|\xi _{{n^{\prime }}}(t_{1})-\xi _{{n^{\prime }}}(t_{2})\right|\ge \delta \big)+\mathbb{P}\big(\left|Y(t_{1})-Y(t_{2})\right|\ge \delta \big),\end{array}\]]]> and the last relation implies the following one: limh→0limn′→∞supt1−t2≤hP(|σ1(t1,ξ˜n′(t1))σ2(t1,Y˜n′(t1))−σ1(t2,ξ˜n′(t2))σ2(t2,Y˜n′(t2))|>ε)=0. \varepsilon \big)=0.\end{array}\]]]> Applying Lemma 1, we get that ∑tk+1n′≤tσ1(tkn′,ξ˜n′(tkn′))σ2(tkn′,Y˜n′(tkn′))ΔWkn′→∫0Tσ1(s,ξ˜(s))σ2(s,Y˜(s))dW˜(s) in probability as n′→∞ , and we also have that ∑tk+1n′≤ta(tkn′,ξ˜n′(tkn′))Δtkn′→∫0Ta(s,ξ˜(s))ds, whence the proof follows.  □

Equations (3) and (4) are equivalent to the two-dimensional stochastic differential equation dZ(t)=A(t,Zt)dt+B(t,Zt)dW(t), where Z(t)=(X(t)Y(t)) , W(t)=(W1(t)W3(t)) is a two dimensional Wiener process, A(t,x,y)=a(t,x)α(t,y),andB(t,x,y)=σ1(t,x)σ2(t,y)0ρβ(t,y)1−ρ2β(t,y). It follows from the measurability and boundedness of a and α and from the continuity and boundedness of σ1 , σ2 , and β that the coefficients of matrices A and B are nonrandom, measurable, and bounded, and additionally the coefficients of B are continuous in all arguments. Then we can apply Theorems 4.2 and 5.6 from [21] (see also Prop. 1.14 in [3]) and deduce that we have to prove the following relation: for any (t,x,y)∈R+×R2 , there exists ε(t,x,y)>0 0$]]> such that, for all λ∈R2 , (5) ‖B(t,x,y)λ‖≥ε(t,x,y)‖λ‖. Relation (5) is equivalent to the following one (we omit arguments): σ12σ22λ12+β2(ρλ1+1−ρ2λ2)2≥ε2(λ12+λ22) or (6) (σ12σ22+β2ρ2)λ12+β2(1−ρ2)λ22+2ρ1−ρ2β2λ1λ2≥ε2(λ12+λ22). The quadratic form Q(λ1,λ2)=(σ12σ22+β2ρ2)λ12+β2(1−ρ2)λ22+2ρ1−ρ2β2λ1λ2 in the left-hand side of (6) is positive definite since its discriminant D=ρ2(1−ρ2)β4−β2(1−ρ2)(σ12σ22+β2ρ2)=−β2(1−ρ2)σ12σ22<0. The continuity of Q(λ1,λ2) implies the existence of minλ12+λ22=1Q(λ1,λ2)>0 0$]]>. Then, putting ε=minλ12+λ22=1Q(λ1,λ2) and using homogeneity, we get (6).  □

Let 1>a1>a2>⋯>an>⋯>0 a_{1}>a_{2}>\cdots >a_{n}>\cdots >0$]]> be defined by ∫a11ρ−2(u)du=1,∫a2a1ρ−2(u)du=2,…,∫anan−1ρ−2(u)du=n,…. We have that an→0 as n→∞ . Let ϕn(u) , n=1,2,… , be a continuous function with support contained in (an,an−1) such that 0≤ϕn(u)≤2ρ−2(u)n and ∫anan−1ϕn(u)du=1 . Such a function obviously exists. Set φn(x)=∫0x∫0yϕn(u)dudy,x∈R. Clearly, φn∈C2(R) , φn′(x)≤1 , and φn(x)↗x as n→∞ .

Let X1 and X2 be two solutions of Eq. (2) on the same probability space with the same Wiener process and such that X1(0)=X2(0) . Then we can present their difference as X1(t)−X2(t)=∫0tσ2(s,Y(s))(σ1(s,X1(s))−σ1(s,X2(s)))dW(s)+∫0t(a(s,X1(s))−a(s,X2(s)))ds. By the Itô formula, φn(X1(t)−X2(t))=∫0tφn′(X1(s)−X2(s))σ2(s,Y(s))(σ1(s,X1(s))−σ1(s,X2(s)))dW(s)+∫0tφn′(X1(s)−X2(s))(a(s,X1(s))−a(s,X2(s)))ds+12∫0tφn″(X1(s)−X2(s))σ2(s,Y(s))2(σ1(s,X1(s))−σ1(s,X2(s)))2ds=J1+J2+J3. We have that E(J1)=0 , E(J2)≤∫0tEa(s,X1(s))−a(s,X2(s))ds≤∫0tE(k(X1(s)−X2(s))ds≤∫0tk(E(X1(s)−X2(s))ds by Jensen’s inequality, and E(J3)≤C22∫0tE(2nρ−2(|X1(s)−X2(s)|)ρ2(|X1(s)−X2(s)|))ds≤tn→0asn→∞. So by letting n→∞ we get E(|X1(s)−X2(s)|)≤∫0tk(E(|X1(s)−X2(s)|)ds. We have that ∫0∞k−1(u)du=+∞ . Then we get E(X1(s)−X2(s))=0 , and hence X1(s)=X2(s) a.s.  □

Let a, σ1 , and σ2 be nonrandom measurable functions, and let Y be an adapted continuous stochastic process. Consider the following assumptions: (i)

There exists K>0 0$]]> such that, for all t≥0 and x∈R , |σ1(t,x)|2+|a(t,x)|2≤K2(1+x2);

Note that, under condition (9) the process Mt=∫0tg(s,Xs,Ys)dWs is a square-integrable local martingale with quadratic variation ⟨M⟩t=∫0tg2(s,Xs,Ys)ds . According to the strong law of large numbers for martingales [14, Ch. 2, § 6, Thm. 10, Cor. 1], under the condition ⟨M⟩T→∞ a.s. as T→∞ , we have that MT⟨M⟩T→0 a.s. as T→∞ . Therefore, it follows from representation (11) that θˆT is strongly consistent.  □

Under assumptions (A1)–(A8), the estimator θˆT is strongly consistent as T→∞ .

We need to verify conditions (8)–(10) of Theorem 5. For model (12), they read as follows: (15) Xtσ2(Yt)≠0,t≥0,a.s., (16) ∫0tσ2−2(Ys)ds<∞,t>0,a.s., 0,\hspace{1em}\text{a.s.},\end{array}\]]]> (17) ∫0∞σ2−2(Ys)ds=∞a.s. Note that (15) is assumption (A8). By [17, Thm. 2.7] the local integrability condition (A6), together with (A2)–(A5), implies (16). Further, if assumption (A7)(i) holds, then (17) is satisfied by [17, Thm. 2.11]. In the remaining case s(r)<∞ or s(l)>−∞ -\infty $]]>, we have that Ω=limt↑∞Yt=r∪limt↑∞Yt=l; see [17]. Moreover, if s(r)=∞ , then P(limt↑∞Yt=r)=0 by [17, Prop. 2.4]. If s(r)<∞ and s(r)−sρβ2σ22∉Lloc1(r−) , then ∫0∞σ2−2(Ys)ds=∞ a.s. on {limt↑∞Yt=r} by [17, Thm. 2.12]. The similar statements hold for {limt↑∞Yt=l} . This implies that (17) is satisfied under each of conditions (ii)–(iv) of assumption (A7).  □