ONE FACTOR GAUSSIAN SHORT RATE MODEL IMPLEMENTATION. 1. model

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1 ONE FACTOR GAUSSIAN SHORT RATE MODEL IMPLEMENTATION P. CASPERS First Version March 1, This Version March 1, 2013 Abstract. We collect some results in Piterbarg, Interest Rate Modelling, needed for the implementation of a GSR model. We develop explicit formulas for pieceise constant volatility and reversion parameters under the forard measure. The short rate dynamics is given by 1. model (1.1) dr(t) = κ(t)(θ(t) r(t))dt + σ r (t)dw (t) under the risk neutral measure. κ, σ r are pieceise constant. Setting x(t) := r(t) f(t, t) ith f(t, T ) denoting the instanteous forard rate observed at t for T > t, the dynamics can be reritten (1.2) dx(t) = (y(t) κ(t)x(t))dt + σ r (t)dw (t) ith deterministic (1.3) y(t) = t 0 e 2 t u κ(s)ds σ r (u) 2 du In the T forard measure the dynamics becomes (1.4) dx(t) = (y(t) σ r (t) 2 G(t, T ) κ(t)x(t))dt + σ r (t)dw T (t) ith (1.5) G(t, t ) = (1.6) (1.7) (1.8) t t e u t κ(s)ds du This fits into the general treatment under 2.1 ith a(t) = κ(t) b(t) = y(t) σ r (t) 2 G(t, T ) c(t) = σ r (t) Zero bond prices can be expressed as follos Date: March 1, Electronic copy available at:

2 2 P. CASPERS (1.9) P (t, t ) = P (0, t ) P (0, t) e x(t)g(t,t ) 1 2 y(t)g(t,t ) 2 Consider the SDE 2. Basic SDE Integration (2.1) dx(t) = (a(t)x(t) + b(t))dt + c(t)dw (t) ith deterministic scalar functions a, b, c. The folloing is an explicit solution of 2.1. (2.2) x(t) = e t 0 a(u)du (x(0) + t 0 e s 0 a(u)du b(s)ds + t 0 ) e s 0 a(u)du c(s)dw (s) That means that for < t x(t) conditional on x() is normally distributed ith mean and variance given by (2.3) (2.4) ith short hand notation E(x(t) x()) = A(, t)x() + Var(x(t) x()) = t t A(s, t)b(s)ds A(s, t) 2 c(s) 2 ds (2.5) A(s, t) = e t s a(u)du 3. Formulas for the pieceise constant case Under the assumption of pieceise constant κ, σ r e are left ith the computation of some integrals for hich e derive closed form formulas here. We fix a grid 0 = t 0 < t 1 <... < t n = T such that κ and σ r are constant on each interval [t i, t i+1 ) and equal to κ i, σ r,i. We introduce the folloing notation: l(t) denotes the largest index such that t l(t) t. Likeise h(t) denotes the smallest index such that t h(t) t. Moreover e set t i,s := max(t i, s) and t s,i := min(t i, s) Formula for G(t, t ). We start ith 1.5, the formula for G(t, t ). The integrand is (3.1) e h(u) 1 u t κ(s)ds = By this (3.2) G(t, t ) = hich is h(t ) 1 tt,i+1 t i,t tu,i+1 e κi t i,t i 1 j=l(t) h(u) 1 ds = e κj(tj+1 tj,t) e κi(tu,i+1 ti,t) e κi(u ti,t) du Electronic copy available at:

3 ONE FACTOR GAUSSIAN SHORT RATE MODEL IMPLEMENTATION 3 (3.3) h(t ) 1 ( 1 e κi(t t,i+1 ti,t) κ i ) i 1 j=l(t) e κj(tj+1 tj,t) We abbreviate this by (3.4) G(t, t ) = h(t ) 1 η i i 1 j=l(t) γ j is dependent on t if and only if j = l(t). In this case i > l(t) necessarily. η i is dependent on t if and only if i = l(t). In these cases (3.5) (3.6) γ j dt = e κj(tj+1 t) κ j γ j η i dt = tκ i e κi(t t,i+1 t) κ 2 i If j > l(t) resp. i > l(t) these integrals can be computed trivially by multiplying γ j resp. η i by the interval length over hich the integral is computed Formula for y(t). We continue ith 1.3. The integrand here is (3.7) e 2 t u κ(s)ds σ r (u) 2 = We get (3.8) y(t) = hich is i=0 i=l(u) σ 2 r,ie 2κi(tt,i+1 ti,u) tt,i+1 e 2κi(tt,i+1 u) σr,je 2 2κj(tt,j+1 tj) t i j=i+1 du (3.9) i=0 σ2 r,i 2κ i [ 1 e 2κi(tt,i+1 ti)] e 2κj(tt,j+1 tj) j=i+1 and hich e abbreviate by (3.10) y(t) = i=0 α i β j j=i+1 α i resp. β j is dependent on t if and only if i = h(t) 1 resp. j = h(t) 1. In the latter case i < h(t) 1 necessarily. In these cases

4 4 P. CASPERS (3.11) (3.12) α i dt = σ2 r,i (tκ i + e 2κi(t ti) ) 2κ 2 i β j dt = e 2κj(t tj) 2κ j If i < h(t) 1 resp. j < h(t) 1 these integrals can trivially computed by multiplying α i resp. β j by the interval length over hich the integral is computed Formula for A(s, t). No e continue ith the formulas for conditional expectation and variance 2.3. First of all e notice that (3.13) A(s, t) = i=l(s) e κi(tt,i+1 ti,s) = i=l(s) ζ i is dependent on s if and only if i = l(s). In this case (3.14) ζ i ds = 1 κ i e κi(tt,i+1 s) If i > l(s) the integral can be computed trivially by multiplying ζ i by the interval length over hich the integral is computed Formula for E(x(t) x()) The easy part. The first term A(, t)x() in the conditional expectation can easily be computed ith the result obtained so far. The second term is (3.15) t A(s, t)(y(s) σ r (s) 2 G(s, T ))ds First not so easy part of the integral. Let s start ith the integral over A(s, t)y(s), hich is (3.16) k=l() tt,k+1 t k, k l=0 α l ζ i i=k k j=l+1 β j ζ i ds We integrate each single summand, i.e. e fix k and l. ζ i depends on s iff i = k. beta depends on s iff j = k. α l depends on s iff l = k. Consider the case l < k first. Then α l does not depend on s. Amongst the factors in round brackets exactly ζ k and β k are depending on s. In essence e are left ith computation of (3.17) ζ k β k ds hich is (3.18) e κ k(t t,k+1 s) e (s t k ) ds

5 ONE FACTOR GAUSSIAN SHORT RATE MODEL IMPLEMENTATION 5 This again is explicitly (3.19) 1 κ k e κ ks+κ k (2t k t t,k+1 ) No consider the case l = k. In this case exactly the factors alpha k and ζ k depend on s. Note that β k does not occur in the product in this case. We have therefore to evaluate (3.20) ζ k α k ds = This simplifies to e κ k(t t,k+1 s) σ2 r,k [ 1 e (s t k ) ] ds σ 2 r,k (3.21) e κ k(t t,k+1 s) e κ ks+κ k (2t k t t,k+1 ) ds hich is in explicit terms (3.22) σ 2 r,k 2κ 2 k (e κ ks+κ k (2t k t t,k+1 ) + e κ k(t t,k+1 s) ) Second not so easy part of the integral. Similarly the integral over A(s, t)σ r (s) 2 G(s, T ) can be ritten (3.23) k=l() σ 2 k tt,k+1 t k, h(t ) 1 l=k η l l 1 ζ i γ j i=k j=k ds As above, fix k and l. η l is dependent on s iff l = k, ζ i is dependent on s iff i = k, γ j is dependent on s iff j = k. Again e start it the case l > k. As above e are left ith ζ k γ k, hich is (3.24) and explicitly (3.25) e κ k(t t,k+1 s) e κ k(t k+1 s) ds e s κ k (t t,k+1 +t k+1 ) If on the other hand l = k, e face η k ζ k hich can be computed as (3.26) and further e κ k(t t,k+1 s) ( 1 e κ k (t T,k+1 s) κ k ) ds (3.27) 2e κ k(t t,k+1 s) e s κ k (t T,k+1 +t t,k+1 ) 2κ 2 k

6 6 P. CASPERS 3.5. Formula for Var(x(t) x()). Finally e analyze the integral representing the conditional variance, hich is (3.28) As before e rite t A(s, t) 2 σ r (t) 2 ds (3.29) k=l() σ 2 r,k tt,k+1 t k, i=k ζ 2 i ds For fixed k the term not covered yet is ζk 2 ds, hich is (3.30) e (t t,k+1 s) ds = e(s t t,k+1 ) address, P. Caspers: pcaspers1973@googl .com