CIR paths stay positive and do not explode under the Feller condition

Put $z_0=\sqrt{x_0}$ and $q(z)=a\mathrm{/}z-\kappa z\mathrm{/}2$ on $U=(0,\mathrm{\infty })$, with constant diffusion $c=\sigma \mathrm{/}2$. The function $q$ is $C^1$, hence bounded and Lipschitz on every compact subset of $U$; $c$ is constant, and $z_0>0$ is deterministic with finite second moment. Apply [F2] directly to these coefficients and initial value to obtain the unique maximal local solution $Z$ of $dZ=q(Z)dt+cdW$. This construction uses the local theorem for the displayed transformed coefficients; it does not reverse [F1], which only transforms a pre-existing CIR solution. Put $\delta =\frac{8a}{{\sigma }^2}=\frac{4\kappa \theta }{{\sigma }^2}-1.$ The Feller inequality gives $\delta \ge 1$.

Define $s(z)={\int }_1^z{u}^{-\delta }\exp \text{\hspace{0.17em}⁣}\left(\frac{2\kappa u^2}{{\sigma }^2}\right)du,z>0.$ Then $s^{\mathrm{\prime }\mathrm{\prime }}(z)\mathrm{/}{s}^{\mathrm{\prime }}(z)=-\delta \mathrm{/}z+4\kappa z\mathrm{/}{\sigma }^2$, and direct substitution gives $\mathcal{L}s(z)=s^{\mathrm{\prime }}(z)(\frac{a}{z}-\frac{\kappa }{2}z+\frac{{\sigma }^2}{8}(-\frac{\delta }{z}+\frac{4\kappa z}{{\sigma }^2}))=0.$ Fix $0<\epsilon <z_0<R$, define ${\tau }_{\epsilon }=\inf \{t<\zeta :Z_t\le \epsilon \}$ and ${\tau }_R=\inf \{t<\zeta :Z_t\ge R\}$ (with $\inf \mathrm{\varnothing }=\mathrm{\infty }$), and set ${\tau }_{\epsilon ,R}={\tau }_{\epsilon }\wedge {\tau }_R\wedge T$. By [F5], this interval-exit time is a stopping time. Extend $s$ globally by [F7] and apply [F3] to the stopped process. The stopped stochastic integrand is bounded and has finite energy, so [F4] gives $\mathbb{E}s(Z_{T\wedge {\tau }_{\epsilon ,R}})=s(z_0)$. On $A_{\epsilon ,R}=\{{\tau }_{\epsilon }\le T,{\tau }_{\epsilon }<{\tau }_R\}$ the stopped value is $s(\epsilon )$; on its complement it is at most $s(R)$. Therefore $\mathbb{P}(A_{\epsilon ,R})\le \frac{s(R)-s(z_0)}{s(R)-s(\epsilon )}\mathrm{.}$ Since $s^{\mathrm{\prime }}(z)=z^{-\delta }\exp (2\kappa z^2\mathrm{/}{\sigma }^2)$ and $\delta \ge 1$, $s(\epsilon )\to -\mathrm{\infty }$ as $\epsilon \downarrow 0$. Thus, for each fixed upper level $R$, the probability of reaching arbitrarily small positive levels before $R$ by time $T$ is zero. Taking a countable union over integer $R>z_0$ excludes lower-boundary termination on paths bounded above; lower-boundary termination can therefore remain only on paths with unbounded upper excursions by time $T$.

For $V(z)=z^2$, the global Itô formula gives $\mathcal{L}V(z)=2zq(z)+c^2=\kappa \theta -\kappa z^2\le \kappa \theta \mathrm{.}$ For $0<\epsilon <z_0<R$, stop at ${\tau }_{\epsilon ,R}$ as above. The stopped stochastic integral has mean zero by [F4], so the pathwise generator bound gives $\mathbb{E}{Z}_{T\wedge {\tau }_{\epsilon ,R}}^2\le z_0^2+\kappa \theta T\mathrm{.}$ On the event of hitting $R$ before $\epsilon$ by time $T$, the stopped value is $R$, hence $R^2\mathbb{P}({\tau }_R\le T,{\tau }_R<{\tau }_{\epsilon })\le z_0^2+\kappa \theta T\mathrm{.}$ For fixed $R$, as $\epsilon \downarrow 0$ these upper-exit events increase to $\{{\tau }_R\le T\}$: on every path that reaches $R$ by $T$, the positive continuous path up to its first hit has a positive minimum. Thus $\mathbb{P}({\tau }_R\le T)\le (z_0^2+\kappa \theta T)\mathrm{/}{R}^2$, and letting $R\to \mathrm{\infty }$ rules out upper escape by time $T$. A local solution cannot terminate while remaining in a compact subset of $(0,\mathrm{\infty })$, by [F2]. Step 1.2 excludes lower-boundary termination on bounded paths, and this estimate excludes paths with unbounded upper excursions; hence $Z$ is global and strictly positive on $[0,T]$.

Set $X=Z^2$. Applying [F3] to the global $C^2$ function $z\mapsto z^2$ gives $d{X}_t=(2Z_{t}q(Z_t)+c^2)dt+2c{Z}_{t}d{W}_t=\kappa (\theta -X_t)dt+\sigma \sqrt{X_t}d{W}_t,$ since $2a+c^2=\kappa \theta$ and $2c=\sigma$. Thus the CIR equation has a global positive strong solution. If $\tilde{X}$ is any other positive local strong CIR solution with initial value $x_0$, [F1] makes $\sqrt{\tilde{X}}$ a local solution of the same $q,c$ equation; local uniqueness in [F2] makes it equal to $Z$ up to every compact exit, so $\tilde{X}=X$.

For any $p\ge 1$, let $V_p(z)=z^{2p}$ on $(0,\mathrm{\infty })$. For each $0<\epsilon <z_0<R$, use [F7] to extend $V_p$ to a global $C^2$ function agreeing on $[\epsilon ,R]$, and apply [F3] to the stopped process. Its generator on the stopped path is $\mathcal{L}{V}_p(z)=(2pa+\frac{p(2p-1){\sigma }^2}{4})z^{2p-2}-p\kappa z^{2p}\le C_p(1+z^{2p})\mathrm{.}$ The stopped stochastic integral has mean zero by [F4]. For fixed $\epsilon ,R$, dominated convergence [F8] makes $u(t)=\mathbb{E}{V}_p(Z_{t\wedge {\tau }_{\epsilon ,R}})$ continuous, and Tonelli [F9] justifies the expectation of its time-integrated generator. Gronwall [F6] gives ${\sup }_{0<\epsilon <z_0<R}{\sup }_{t\le T}\mathbb{E}{V}_p(Z_{t\wedge {\tau }_{\epsilon ,R}})\le C_{p,T}(1+z_0^{2p})\mathrm{.}$ Since $Z$ is global, continuous and strictly positive on $[0,T]$, these stopping times equal $T$ eventually along any sequence $\epsilon \downarrow 0$, $R\uparrow \mathrm{\infty }$ on almost every path. Fatou's lemma [F10] then yields ${\sup }_{t\le T}\mathbb{E}{Z}_t^{2p}<\mathrm{\infty }$, and $X=Z^2$ gives the claimed CIR moment bound. AC is exactly [A1].