Let H∈(0,1) be an arbitrary Hurst index, {Yt,t≥0} be the process that satisfies the equation (5) and τ be the first moment of reaching zero by the latter. The fractional Cox–Ingersoll–Ross process is the process {Xt,t≥0} such that for all t≥0,ω∈Ω : (6) Xt(ω)=Yt2(ω)1{t<τ(ω)}.

Let τ:=inf{s>0:Ys=0} 0:Y_{s}=0\}$]]>. For 0≤t≤τ the fractional CIR process from Definition 1 satisfies the following SDE: (7) dXt=(k−aXt)dt+σXt∘dBtH, where X0=Y02>0 0$]]> and the integral with respect to the fractional Brownian motion is defined as the pathwise Stratonovich integral.

Let us fix an ω∈Ω and consider an arbitrary t<τ(ω) .

According to (5) and (6), (8) Xt=Yt2=(X0+12∫0t(kYs−aYs)ds+σ2BtH)2.

Consider an arbitrary partition of the interval [0,t] : 0=t0<t1<t2<⋯<tn−1<tn=t.

Using (8), we get Xt= ∑i=1n(Xti−Xti−1)+X0= ∑i=1n([X0+12∫0ti(kYs−aYs)ds+σ2BtiH]2−[X0+12∫0ti−1(kYs−aYs)ds+σ2Bti−1H]2)+X0.

Factoring each summand as the difference of squares, we get: Xt=X0+ ∑i=1n[2X0+12(∫0ti(kYs−aYs)ds+∫0ti−1(kYs−aYs)ds)+σ2(BtiH+Bti−1H)]×[12∫ti−1ti(kYs−aYs)ds+σ2(BtiH−Bti−1H)].

Expanding the brackets in the last expression, we obtain: (9) Xt=X0+ ∑i=1nX0∫ti−1ti(kYs−aYs)ds+14 ∑i=1n(∫0ti(kYs−aYs)ds+∫0ti−1(kYs−aYs)ds)×∫ti−1ti(kYs−aYs)ds+σ4 ∑i=1n(BtiH+Bti−1H)∫ti−1ti(kYs−aYs)ds+σX0 ∑i=1n(BtiH−Bti−1H)+σ24 ∑i=1n(BtiH−Bti−1H)(BtiH+Bti−1H)+σ4 ∑i=1n(∫0ti(kYs−aYs)ds+∫0ti−1(kYs−aYs)ds)(BtiH−Bti−1H).

Let the mesh Δt of the partition tend to zero. The first three summands (10) ∑i=1nX0∫ti−1ti(kYs−aYs)ds+14 ∑i=1n(∫0ti(kYs−aYs)ds+∫0ti−1(kYs−aYs)ds)×∫ti−1ti(kYs−aYs)ds+σ4 ∑i=1n(BtiH+Bti−1H)∫ti−1ti(kYs−aYs)ds→∫0t(kYs−aYs)(X0+12∫0s(kYu−aYu)du+σ2BsH)ds=∫0t(k−aYs2)ds=∫0t(k−aXs)ds,Δt→0, and the last three summands (11) σX0 ∑i=1n(BtiH−Bti−1H)+σ24 ∑i=1n(BtiH−Bti−1H)(BtiH+Bti−1H)+σ4 ∑i=1n(∫0ti(kYs−aYs)ds+∫0ti−1(kYs−aYs)ds)(BtiH−Bti−1H)→σ∫0t(X0+12∫0s(kYu−aYu)du+σ2BsH)∘dBsH=σ∫0tYs∘dBsH=σ∫0tXs∘dBsH,Δt→0.

Note that the left-hand side of (9) does not depend on the partition and the limit in (10) exists as the pathwise Riemann integral, therefore the corresponding pathwise Stratonovich integral exists and the passage to the limit in (11) is correct.

Thus, the fractional Cox–Ingersoll–Ross process, introduced in Definition 1, satisfies the SDE of the form (12) Xt=X0+∫0t(k−aXs)ds+σ∫0tXs∘dBsH, where ∫0tXs∘dBsH is the pathwise Stratonovich integral.  □

In the case of k=0 , the process (5) is the fractional Ornstein–Uhlenbeck process and the definition coincides with the one given in [16].

Let k>0,H>1/2 0,H>1/2$]]>. Then the process {Yt,t≥0} , defined by the equation (5), is strictly positive a.s.

The proof is by contradiction.

Let Ω′ be the same as in Proposition 1. First, assume that a>0 0$]]> and let for some ω∈Ω′ , τ(ω)=inf{t>0:Xt=0}=inf{t>0:Yt=0}<∞ 0:X_{t}=0\}=\inf \{t>0:Y_{t}=0\}<\infty $]]>.

For all ε∈(0,min(Y0,ka)) (the condition ε<ka provides the inequality kε−aε>0 0$]]>) let us introduce the last moment of hitting the level of ε before the first zero reaching: τε:=sup{t∈(0,τ):Yt=ε}.

Consider δ>0 0$]]> such that the inequality H−δ>1/2⇔1+δ−H<1/2 1/2\Leftrightarrow 1+\delta -H<1/2$]]> holds. According to the definitions of τ,τε and Y, the following equality is true: −ε=Yτ−Yτε=12∫τετ(kYs−aYs)ds+σ2(BτH−BτεH).

The process Ys∈(0,ε) on the interval (τε,τ) , hence ∀s∈(τε,τ) : (13) kYs−aYs≥kε−aε.

From this and Proposition 1, it follows that ∃C=C(τ(ω),ω,δ) : σ2C|τ−τε|H−δ≥σ2|BτH−BτεH|≥12(kε−aε)(τ−τε)+ε.

It is clear that there exists ε˜>0 0$]]> such that ∀ε<ε˜ : kε−aε>k2ε \frac{k}{2\varepsilon }$]]>. Then, by choosing an arbitrary ε<ε˜ , we have: (14) σ2C|τ−τε|H−δ≥k4ε(τ−τε)+ε.

For x≥0 consider the function (15) Fε(x)=k4εx−σ2CxH−δ+ε.

Let us show that there exists ε∗∈(0,ε˜) such that, for all ε<ε∗ and for all x≥0 , Fε(x)>0 0$]]>. It is easy to check that Fε(0)=ε>0 0$]]> and Fε is convex on R+∖{0} (its second derivative is strictly positive on this set), so it is enough to examine the sign of the function in its critical points. F′(x˜)=k4ε−σ(H−δ)2Cx˜H−δ−1=0⟹x˜=(k2σεC(H−δ))1/(H−δ−1)=(2σC(H−δ)k)1/(1+δ−H)ε1/(1+δ−H). After some calculations we get F(x˜)=12(2(H−δ)k)H−δ1+δ−H(σC)11+δ−H(H−δ−1)εH−δ1+δ−H+ε.

From the choice of δ it follows that H−δ1+δ−H>1 1$]]>, hence ∀K∈R∃ε∗>0 0$]]>: (16) ε−KεH−δ1+δ−H>0,∀ε<ε∗. 0,\hspace{1em}\forall \varepsilon <{\varepsilon }^{\ast }.\]]]>

Choosing the corresponding ε∗ for K:=−12(2(H−δ)k)H−δ1+δ−H(σC)11+δ−H(H−δ−1), and choosing an arbitrary ε<min{ε˜,ε∗} we obtain that Fε(x)>0∀x>0. 0\hspace{1em}\forall x>0.\]]]> However, from (14) it follows that Fε(τ−τε)≤0.

The contradiction obtained proves the theorem for a>0 0$]]>. If a≤0 , instead of (13) the following bound can be used: (17) kYs−aYs≥kε.  □

Let k1<k2 . Then ∀ω∈Ω,∀t≥0 : (i)

τ(k1)(ω)≤τ(k2)(ω) ;

This lemma holds for an arbitrary Hurst index H∈(0,1) .

Let ω∈Ω be fixed (we will omit ω in brackets in further formulas). Consider the function δ on the interval [0,min{τ(k1),τ(k2)}) , such that δ(t)=Yt(k2)−Yt(k1) . It is obvious that δ is differentiable, δ(0)=0 and δ+′(0)=12(k2Y0−aY0)−12(k1Y0−aY0)=k2−k12Y0>0. 0.\]]]>

As δ(t)=δ+′(0)t+o(t) , t→0+ , it is easy to see that there exists the maximal interval (0,t∗)⊂(0,min{τ(k1),τ(k2)}) such that δ(t)>0 0$]]> for all t∈(0,t∗) . It is also clear that t∗=sup{t∈(0,min{τ(k1),τ(k2)})∣∀s∈(0,t):δ(s)>0}. 0\big\}.\]]]>

Assume that t∗<min{τ(k1),τ(k2)} . According to the definition of t∗ , δ(t∗)=0 . Hence Yt∗(k2)=Yt∗(k1)=Y∗>0 0$]]> and δ′(t∗)=k2−k12Y∗>0. 0.\]]]>

As δ(t)=δ′(t∗)(t−t∗)+o(t−t∗) , t→t∗ , there exists ε>0 0$]]> such that δ(t)<0 for all t∈(t∗−ε,t∗) , that contradicts the definition of t∗ .

Therefore, ∀t∈(0,min{τ(k1),τ(k2)}) : (19) Yt(k2)>Yt(k1). {Y_{t}^{(k_{1})}}.\]]]>

Now it is easy to show that (i) holds: indeed, if τ(k1)>τ(k2) {\tau }^{(k_{2})}$]]>, then 0=Yτ(k2)(k2)<Yτ(k2)(k1).

This means that ∃t∗<τ(k2) such that Yt(k2)<Yt(k1) for all t∈(t∗,τ(k2)) , which contradicts (19).

Finally, as Yt(k2)>Yt(k1) {Y_{t}^{(k_{1})}}$]]> for all t∈(0,τ(k1)) , Yt(k2)>Yt(k1)=0 {Y_{t}^{(k_{1})}}=0$]]> for all t∈[τ(k1),τ(k2)) and Yt(k2)=Yt(k1)=0 for all t≥τ(k2) , (ii) also holds.  □

The proof is by contradiction.

Assume that ∃T∗>0 0$]]>, ∃{kn,n≥1},kn↑∞ as n→∞ such that: P(τ(kn)≤T∗)→α>0,n→∞. 0,\hspace{1em}n\to \infty .\]]]>

Let us consider the case of a>0 0$]]>. Let Ω′ be from Proposition 1, and for all ε∈(0,min(Y0,1,k12a)) denote τε(kn):=sup{t∈(0,τ):Yt(kn)=ε} and DT∗(kn):={ω∈Ω′∣τ(kn)≤T∗}.

According to Lemma 1, ∀n≥1:DT∗(kn+1)⊂DT∗(kn) , so P(⋂n≥1DT∗(kn))=limn→∞P(DT∗(kn))=α>0. 0.\]]]>

Just like in Theorem 2, ∀n≥1 , ∀ω∈DT∗(kn) : −ε=Yτ(kn)(kn)−Yτε(kn)(kn)=12∫τε(kn)τ(kn)(knYs(kn)−aYs(kn))ds+σ2(Bτ(kn)H−Bτε(kn)H).

The process Ys(kn)∈(0,ε) on the interval (τε(kn),τ(kn)) , hence (21) knYs(kn)−aYs(kn)≥knε−aε,∀s∈(τε(kn),τ(kn)).

Let δ>0 0$]]> satisfy the condition 0<H−δ<1/2 .

According to Proposition 1, ∃Cω=C(T∗,ω,δ) , ∀0<s<t<T∗ : |BtH−BsH|≤Cω|t−s|H−δ.

As ε<k12a , the following inequality is true: knε−aε>kn2ε∀n≥0, \frac{k_{n}}{2\varepsilon }\hspace{1em}\forall n\ge 0,\]]]> so just like in the proof of Theorem 2 we can obtain that ∀n≥1 , ∀ω∈DT∗(kn) : (22) σ2Cω(τ(kn)−τε(kn))H−δ≥kn4ε(τ(kn)−τε(kn))+ε. According to (22), ∀n≥1 , ∀ω∈⋂n≥0DT∗(kn) : (23) σ2Cω(τ(kn)−τε(kn))H−δ>ε. \varepsilon .\]]]> However, it is easy to see from (22) that (24) σ2Cω(τ(kn)−τε(kn))H−δ>kn4ε(τ(kn)−τε(kn)). \frac{k_{n}}{4\varepsilon }\big({\tau }^{(k_{n})}-{\tau _{\varepsilon }^{(k_{n})}}\big).\]]]>

Let us transform (24): σ2Cω>kn4ε(τ(kn)−τε(kn))1−H+δ,2σεknCω>(τ(kn)−τε(kn))1−H+δ,(2σεknCω)11−H+δ>τ(kn)−τε(kn), \frac{k_{n}}{4\varepsilon }{\big({\tau }^{(k_{n})}-{\tau _{\varepsilon }^{(k_{n})}}\big)}^{1-H+\delta },\\{} \displaystyle \frac{2\sigma \varepsilon }{k_{n}}C_{\omega }& \displaystyle >{\big({\tau }^{(k_{n})}-{\tau _{\varepsilon }^{(k_{n})}}\big)}^{1-H+\delta },\\{} \displaystyle {\bigg(\frac{2\sigma \varepsilon }{k_{n}}C_{\omega }\bigg)}^{\frac{1}{1-H+\delta }}& \displaystyle >{\tau }^{(k_{n})}-{\tau _{\varepsilon }^{(k_{n})}},\end{array}\]]]> hence σ2Cω(τ(kn)−τε(kn))H−δ<σ2Cω(2σεknCω)H−δ1−H+δ=(22H−2δ−11−H+δσ11−H+δ)kn−H−δ1−H+δCω11−H+δεH−δ1−H+δ=C˜kn−H−δ1−H+δCω11−H+δεH−δ1−H+δ.