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