Stability Analysis of Riccati Differential Equations related to Affine Diffusion Processes —Online Supplement— Kyoung-Kuk Kim∗ KAIST, South Korea November 2009

In this online supplement, we show that a canonical affine diffusion process characterized by Filipovi´c and Mayerhofer [2] admits a limiting stationary distribution, say Y∞ , and that the region of finite exponential moments © ª n−m S = u ∈ Rn : E exp(2u · Y∞ ) < ∞, ∀Y0 ∈ Rm + ×R is the same as the stability region of the origin of the quadratic differential system x˙ = Ax + Q(x) (1) in Kim [4]. We suppose that all the parametric restrictions hold. Also, Λv À 0 is assumed. The proofs below are almost identical to the proofs of Lemma 5.1 and Theorem 2.2 in Glasserman and Kim [3] except for some obvious changes. This online supplement is only for a reader’s convenience. All the equation numbers correspond to the numbers in [4]. Lemma 1 Let S be the stability region of the origin for the system (1). Then, n o n−m S = u ∈ Rn : lim E exp(2u · Yt ) < ∞, ∀Y0 ∈ Rm . + ×R t→∞

Proof Suppose u ∈ S. Then, Φt (u) converges to the origin exponentially as t → ∞ and there exist positive constants C0 , µ0 , and δ such that |x(t)| ≤ C0 |x(0)|e−µ0 t whenever |x(0)| ≤ δ; we may therefore define tδ = inf{t : |Φt (u)| ≤ δ} < ∞. ∗

ISE dept., KAIST, Daejeon, South Korea, Email: [email protected], Tel.: +82 42 350 3128.

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Then, for t ≥ tδ , Z

Z

t

0

|Λ · Φs (u)|ds ≤

0

Z ≤

t

|Λ| · |Φs (u)|ds tδ

0

Z |Λ| · |Φs (u)|ds + C0 δ|Λ|

t

e−µ0 (s−tδ ) ds.

tδ

The last integral converges to a finite value as t → ∞. The integrability of Φds (u)> B0 Φds (u) as a function of t follows similarly from (3). Therefore, the validity of the Fourier transform in [2] implies that lim E exp(2u · Yt ) = exp(2Ψ∞ (u)) < ∞

t→∞

with Ψt (u) =

Rt 0

Λ · x(s)ds +

Rt 0

xd (s)> B0 xd (s)ds + x(t) · Y0 .

For the converse, suppose u ∈ / S. If Φt (u) blows up in finite time τ , then limt→τ exp(2Ψt (u)) = ∞ n−m and Lemma A.2 imply that the first term of Ψ (u) is bounded below, because, first, Λ ∈ Rm t + ×R

and second, Φi,t (u) for some i diverges to infinity. Hence, no further argument is required in this case. Now, assume that Φt (u) exists for all t ≥ 0. Since S open and it contains the origin, we can choose K large enough to make u/K ∈ S. Then Lemma A.1 implies KΦt (u/K) ≤ Φt (u) for all t. This gives us that

Z lim inf t→∞

Z

t

0

Φi,s (u)ds ≥ ci :=

∞

0

KΦi,s (u/K)ds,

for some real number ci , for each i ∈ {1, . . . , m}. We also have lim inf Φt (u) ≥ lim inf KΦt (u/K) = 0. t→∞

t→∞

But this liminf cannot be the zero vector; if it were, Φt (u) would reach S in finite time and then converge to 0, which would contradict the assumption that u ∈ / S. Thus, some component i of Φt (u) has a positive liminf, and i must be in {1, . . . , m} because Φdt (u) converges to zero. As a consequence, Z lim inf t→∞

0

t

Λv · Φvs (u)ds ≥

X

Z Λj cj + lim inf t→∞

j6=i

0

t

Λi Φi,s (u)ds = ∞.

It follows that lim inf t→∞ Ψt (u) = ∞ and thus lim inf t→∞ E exp(2u · Yt ) = ∞.

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Theorem 1 The process Y has a stationary distribution, which is also the limiting distribution of Yt , as t → ∞, for any Y0 . Moreover, if Y∞ is the limiting stationary distribution of Y , then we have n−m S = {u ∈ Rn : E exp(2u · Y∞ ) < ∞, ∀Y0 ∈ Rm }. + ×R

Proof We start by showing that the sequence {Yt } is tight (as defined, for example, in Chung [1], p.90). For this, we need to show limr→∞ supt P(|Yt | > r) = 0. But we have Ã

[© √ ª P (|Yt | > r) ≤ P |Yi,t | > r/ n

!

i

X ¡ √ ¢ P |Yi,t | > r/ n ≤ i

X© √ √ ª = P(Yi,t > r/ n) + P(−Yi,t > r/ n) i

Xn √ √ o = P(e2δYi,t > e2δr/ n ) + P(e−2δYi,t > e2δr/ n ) i

≤

X ½ Ee2δYi,t i

e2δr/

√ n

Ee−2δYi,t + 2δr/√n e

¾ ,

where δ is a positive constant such that Bδ (0) ⊂ S. From Lemma 1, we get supt E exp(±2δYt,i ) ≤ Mi < ∞, for some Mi , for each i. Therefore, sup P(|Yt | > r) ≤ 2 t

X

√ Mi exp(−2δr/ n)

i

which converges to zero as r → ∞. Because the sequence {Yt } is tight, it is relatively compact (Chung [1], p.90), so each subsequence {Yt0 } contains a further subsequence {Yt00 } converging weakly to some limiting random vector Y a . Since we have supt00 E exp(2u · Yt00 ) < ∞, for any u ∈ Bδ (0) (by Lemma 1) and since Yt00 ⇒ Y a , Theorem 4.5.2 in Chung [1] implies that lim E exp(2θu · Yt00 ) = E exp(2θu · Y a ),

t00 →∞

∀θ ∈ (0, 1).

Equality continues to hold if we replace θu by u because Bδ (0) is open: we can find u0 ∈ Bδ (0) such that u = θu0 for some θ ∈ (0, 1) and then apply the above equation at u0 . We know that the original sequence {Yt } satisfies limt→∞ E exp(2u · Yt ) = exp(2Ψ∞ (u)) for u ∈ Bδ (0), so the same 3

limit applies to {Yt00 }. Applying the same argument to any other weakly convergent subsequence of {Yt }, say with limit Y b , we find that E exp(2u · Y a ) = exp(2Ψ∞ (u)) = E exp(2u · Y b ),

∀u ∈ Bδ (0).

But the distribution of a random vector is uniquely determined by its moment generating function in a neighborhood of the origin, so Y a and Y b have the same distribution. Since every convergent subsequence has the same limiting distribution, the original sequence {Yt } also converges to Y a in distribution, so we now denote Y a by Y∞ . We have shown that E exp(2u · Yt ) → E exp(2u · Y∞ ) for all u ∈ Bδ (0). Our next step will be to show that this holds for all u ∈ S, and to show that E exp(2u · Y∞ ) = ∞ if u ∈ / S. For any u ∈ S, we can find u0 ∈ S and θ ∈ (0, 1) with u = θu0 , because S is an open set containing the origin. We know that Yt ⇒ Y∞ and, by Lemma 1, that supt E exp(2u0 · Yt ) is finite. It follows from Theorem 4.5.2 of Chung [1] that E exp(2u · Yt ) → E exp(2u · Y∞ ), so we conclude n−m }. that S ⊆ {u : E exp(2u · Y∞ ) < ∞, ∀ Y0 ∈ Rm + ×R

We prove the opposite inclusion by contradiction. For this, suppose that u ∈ / S and that E exp(2u · Y∞ ) < ∞. Define θ∗ = sup{θ ∈ [0, 1] : θu ∈ S}; then θ∗ > 0 and θ∗ u is on ∂S, the topological boundary of S, because S is open and u ∈ / S. Fix a θ0 ∈ (0, θ∗ ), so that θ0 u ∈ S, and set g(t) = Φt (θ0 u)/θ0 . Lemma A.1 implies that Φt (θu) ≥ θg(t), for all t ≥ 0 and all θ ∈ [θ0 , θ∗ ). Consider the trajectory of Φt (θ∗ u). We claim that τ (θ∗ u) = ∞. If q = 0, then the blow-up time is always infinite (Lemma 3.2). Hence, we restrict out attention to the case q ≥ 1. To see this, choose a θ ∈ (θ0 , θ∗ ). Then, for each i ∈ {1, . . . , q}, x2i +

X

Aij xj + Qi (x) ≥ x2i + Aii xi + θ

j

X

Aij gj (t)

j6=i

≥ x2i + Aii xi + θM where x(t) = Φt (θu) and M is a lower bound of the summation. Next, we define a new function y

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starting at t0 by y˙ = y 2 + Aii y + θM,

y(t0 ) = xi (t0 ).

If y(t0 ) is sufficiently large, then y(t) blows up in finite time, and so does xi (t). Suppose τ (θ∗ u) < ∞. Then, there is some i ∈ {1, . . . , q} such that xi (t) blows up in finite time. Otherwise, we can readily show that Φt (u) is finite all the time by solving (1) for (xq+1 , . . . , xm ). Therefore, it is possible to choose θ close to θ∗ and t0 < τ such that some xi (t0 ) becomes large enough to make y(t) blow up in finite time. This is a contradiction to θu ∈ S. Therefore, we have limt→∞ Ψt (θ∗ u) = ∞ as shown in the proof of Lemma 1. On the other hand, we have Z

∞

Z v

Λ · 0

(Φvt (θ∗ u)

∗ v

∞

lim Λv · (Φvt (θu) − θg v (t))dt Z ∞ Z ∞ v v ∗ ≤ lim inf Λ · Φt (θu)dt − θ Λv · g v (t)dt ∗

− θ g (t))dt =

0

θ↑θ ∗

θ↑θ

0

0

where the equality comes from the continuity of the flow Φ and the inequality is from Fatou’s R∞ lemma. Since Λv · g v (t) and Φdt (θ∗ u) are integrable, limt→∞ Ψt (θ∗ u) = ∞ implies that 0 Λv · R∞ (Φvt (θ∗ u) − θ∗ g v (t))dt = ∞. Therefore, lim inf θ↑θ∗ 0 Λv · Φvt (θu)dt = ∞. But for θ ∈ (0, θ∗ ), θu ∈ S and utilizing Jensen’s inequality, ¡ ¢θ exp(2Ψ∞ (θu)) = E exp(2θu · Y∞ ) ≤ E exp(2u · Y∞ ) < ∞. Therefore, lim supθ↑θ∗ Ψ∞ (θu) < ∞ and this is a contradiction. To conclude the proof, we need to show that the limiting distribution is a stationary distribution. Suppose that Y0 is equal to Y∞ in distribution. Then, for any u ∈ S, by taking a conditional

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expectation,

E exp(2u · Yt ) = = =

=

Z t ³ Z t ´ E exp 2 Λ · Φs (u)ds + 2 Φds (u)> B0 Φds (u)ds + 2Φt (u) · Y0 0 0 Z t ³ Z t ´ exp 2 Λ · Φs (u)ds + 2 Φds (u)> B0 Φds (u)ds E exp(2Φt (u) · Y0 ) 0 0 Z t ³ Z t ´ exp 2 Λ · Φs (u)ds + 2 Φds (u)> B0 Φds (u)ds 0 0 Z ∞ ³ Z ∞ ´ Λ · Φs (Φt (u))ds + 2 Φds (Φt (u))> B0 Φds (Φt (u))ds × exp 2 0 0 Z ∞ ³ Z ∞ ´ d > d exp 2 Λ · Φt (u)dt + 2 Φs (u) B0 Φs (u)dt 0

(1)

0

= E exp(2u · Y∞ ). Because the distribution of a random vector is determined by the values of its moment generating function in a neighborhood of the origin, we conclude that Yt has the distribution of Y∞ whenever Y0 does.

References [1] Chung, K. L. (2001): A Course in Probability Theory, 3rd edition, Academic Press. [2] Filipovi´c, D. and E. Mayerhofer (2009) Affine Diffusion Processes: Theory and Applications, Radon Series on Computational and Applied Mathematics, 8, 1–40 [3] Glasserman, P. and K.-K. Kim (2008) Moment Explosions and Stationary Distributions in Affine Diffusion Models, Mathematical Finance, forthcoming [4] Kim, K.-K. (2009) Stability Analysis of Riccati Differential Equations related to Affine Diffusion Processes, Journal of Mathematical Analysis and Applications, forthcoming

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