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The CIR Lamperti step remains positive from the boundary value

Statement

Let $κ, θ, σ, h > 0$ satisfy $2 κ θ \geq σ^{2}$ . Starting the numerical Lamperti state at the boundary value $z_{n} = 0$ , even a negative Brownian increment $Δ W_{n} = - r$ with arbitrary $r \geq 0$ gives a strictly positive next state under the CIR scheme.

Facts & Assumptions

Given: AC, positive parameters satisfying the Feller condition, $z_{n} = 0$ , and $Δ W_{n} = - r$ for $r \geq 0$ .

[F1]

Put $a = (4 κ θ - σ^{2}) / 8$ , $A = 1 + κ h / 2$ , and $c = z_{n} + σ Δ W_{n} / 2$ . The scheme returns $(c + \sqrt{c^{2} + 4 A a h}) / (2 A)$ . Positive Lamperti drift-implicit scheme for the CIR process

[F2]

Under $κ, θ, σ > 0$ and $2 κ θ \geq σ^{2}$ , $a > 0$ and the displayed root is the unique positive solution of the drift-implicit Lamperti equation for every $z_{n} \geq 0$ and finite increment. CIR Lamperti equation and unique positive implicit root

[A1]

AC is inherited from the numerical Brownian and algebraic interfaces. The Axiom of Choice

Proof

Given: The hypotheses and supplier statements above.

1.1F1F2givenalgebra

From the parameter assumptions, $4 κ θ - σ^{2} \geq σ^{2} > 0$ , so $a > 0$ ; also $A = 1 + κ h / 2 > 0$ . With $z_{n} = 0$ and $Δ W_{n} = - r$ , $c = - σ r / 2 \leq 0$ .

2.1A1F1F2step 1.1givenalgebra∎

Since $4 A a h > 0$ , $\sqrt{c^{2} + 4 A a h} > ∣ c ∣ = σ r / 2.$ Therefore the numerator $c + \sqrt{c^{2} + 4 A a h}$ and denominator $2 A$ are both strictly positive, so the algorithm's next Lamperti state is positive. Its square is also strictly positive. The uniqueness and equation interpretation follow from [F2], including the case $r = 0$ . AC is exactly [A1].

Depends on

Used by

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Dependency tree · two levels

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Sources

Aurélien Alfonsi, Strong convergence of some drift implicit Euler scheme. Application to the CIR process, equation (3) and positive-root discussion (standard reference, not scraped)