Enoncé du TD

double n(double)n(x) = e−
x2
2
x0x1
v0v1vi= 2xi−1
s=x2
0+x2 1
s
v0q−2ln(s)
s
Stt
K r σ
(ST−K)+= max(ST−K; 0)
C=E[P ayoff ×e−rT ]
C(S, K, r, t, σ) = SN(d1)−Ke−rt N(d2)
(K− ST)+= max(K−ST; 0)
P(S, K, r, t, σ) = −SN(−d1) + Ke−rtN(−d2)
v1q−2ln(s)
s

N N (0,1)
d1=1
σ√tln S
K+r+1 2σ2t
d2=d1−σ√t
double N (double)
c0
c0=c(S, K, r, σ, T −t)
1.0e−5
0.3