Consider the Brownian motion given by
S(u)=Seμu+σW(u) 0≤u≤t……(a)
Payoff function is given by
R(t)=[S(3t)S(32t)S(t)31……(b)
where t is the expiration time.
For this first find the value of S at u=0,t,(3t),(32t)
At u=0
equation(a) becomes
S(0)=Se0+σW(0)
S(0)=SeσW(0)……(c)
at
u=t
S(t)=Seμt+σW(t)……(d)
atu=3t
S(3t)=Seμ(3t)+σW(3t)……(e)
atu=32t
S(32t)=Seμ(32t)+σW(32t)……(f)
Put the values from equation(c) ,equation(d) ,equation(e) and equation(f)in equation (b)
R(t)=[S(3t)S(32t)S(t)]31
R(t)=[(Seμ(3t)+σW(3t))(Seμ(32t)+σW(32t))(Seμt+σW(t))]31
=[(S3eμ(3t)+σW(3t))(eμ(32t)+σW(32t))(eμt+σW(t))]31
=S[(eμ(3t)+σW(3t))(eμ(32t)+σW(32t))(eμt+σW(t))]31
=S[eμ(3t+32t+t)eσW(3t+32t+t)]31
=S[eμ(2t)eσW(2t)]31
R(t)=S[eμ(2t)+σW(2t)]31……(g)
Now we find no-arbitrage price of a derivative .
differentiate equation(g) with respect to t.
R′(t)=3S[eμ(2t)+σW(2t)]−32(eμ(2t)+σW(2t)(2μ+2σW)
=3S[eμ(2t)+σW(2t)]31(2μ+2σW)
R′(t)=32S[eμ(2t)+σW(2t)]31(μ+σW)
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