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Empirical Evaluation of Hybrid Defaultable Bond Pricing ... - risklab

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solution:<br />

B(t, T ) = e −âr(T −t) Z T −t<br />

= 1 ³<br />

1 − e −â r(T −t)´<br />

,<br />

â r<br />

C d (t, T ) = 1 ³<br />

−âs(T<br />

1 − e<br />

−t)´<br />

,<br />

â s<br />

Z T −t<br />

D d (t, T ) = e −âu(T −t) eâul b su C d (0,l)dl<br />

= e −âu(T −t) Z T −t<br />

0<br />

0<br />

0<br />

eârl dl = e −âr(T −t) 1 ³eâr(T ´<br />

−t) − 1<br />

â r<br />

1<br />

³<br />

−âs(l)´<br />

eâul b su 1 − e dl<br />

â s<br />

´<br />

= e −âu(T −t) b su<br />

1<br />

â s<br />

µ 1<br />

â u<br />

³<br />

eâu(T −t) − 1<br />

= b su<br />

1<br />

â s<br />

µ 1 − e<br />

−â u(T −t)<br />

â u<br />

−<br />

+ e−âu(T −t) <br />

−âs(T −t)<br />

− e<br />

,<br />

â u − â s<br />

Z T −t<br />

E d (t, T ) = e −âw(T −t) ¡ eâwl b r B(0,l) − b sw C d (0,l) ¢ dl<br />

A d (t, T ) =<br />

= −b sw<br />

1<br />

â s<br />

µ 1 − e<br />

−â w(T −t)<br />

â w<br />

µ<br />

1 1 − e<br />

−â w(T −t)<br />

+b r<br />

â r<br />

Z T<br />

0<br />

â w<br />

+ e−âw(T −t) <br />

−âs(T −t)<br />

− e<br />

â w − â s<br />

+ e−âw(T −t) <br />

−âr(T −t)<br />

− e<br />

,<br />

â w − â r<br />

1<br />

³<br />

´<br />

e (âu−âs)(T −t) − 1<br />

â u − â s<br />

1 ¡ σ<br />

2<br />

t 2 s C d (l, T) 2 + σ 2 uD d (l, T) 2 + σ 2 wE d (l, T) 2 + σ 2 rB(l, T) 2¢<br />

−θ r (l)B(l, T) − θ s C d (l, T) − θ w E d (l, T) − θ u D d (l, T)dl.<br />

C Pro<strong>of</strong> <strong>of</strong> Theorem 6<br />

By the theorem <strong>of</strong> Feynman Kac P d is the solution to the following partial<br />

differential equation:<br />

0 = 1 ¡ σ<br />

2<br />

2 u Puu d +2σ u ρ r,u σ r Pru d + σ 2 wPww d + σ 2 rPrr d +2σ r ρ r,w σ w Prw<br />

d ¢ +(w − âr r) Pr<br />

d<br />

+ (θ w − â w w) Pw d +(θ u − â u u) Pu d − (Λ 0 + Λ r r + Λ u u)P d + Pt d .<br />

If we assume that the structure <strong>of</strong> P d is <strong>of</strong> the type as in Equation (26), then<br />

for P d > 0 we get:<br />

0 = 1 ¡<br />

σ<br />

2<br />

2 u D d2 +2σ u ρ r,u σ r D d B d + σ 2 wE d2 + σ 2 rB d2 +2σ r ρ r,w σ w E d B d¢<br />

+r ¡ â r B d − Λ r − Bt d ¢ + w<br />

¡âw E d − B d − Et d ¢ + u<br />

¡âu D d − Λ u − Dt<br />

d ¢<br />

+A d t − θ u D d − θ w E d − Λ 0 .<br />

34

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