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Re: Solving for equations with CDF functions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67798] Re: Solving for equations with CDF functions
  • From: "Valeri Astanoff" <astanoff at yahoo.fr>
  • Date: Sat, 8 Jul 2006 04:56:00 -0400 (EDT)
  • References: <e8lgau$r6e$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi Matt,

Given that K is supposed to be not very far from S,
here is a first order approximate solution :


In[1]:=<< Statistics`

In[2]:=normdist = NormalDistribution[0, 1] ;
d1 = (Log[(S/K)] + (r + delta + 1/2*sigma^2)*T)/(sigma*Sqrt[T]) ;
d2 = d1 - sigma*Sqrt[T] ;
price = S*Exp[-delta*T]*CDF[normdist, d1] - K*Exp[-r*T]*CDF[normdist,
d2] ;
Solve[C == price, K]

>From In[2]:=
[...]

In[3]:=K1=K /. First@Solve[C == (Series[price, {K,S,1}]//Normal),K]
//FullSimplify

Out[3]=(Sqrt[2]*E^(r*T+((-2*(delta+r)+sigma^2)^2*T)/
(8*sigma^2))*S-Sqrt[2]*E^(delta*T+((2*(delta+r)+sigma^2)^2*T)/
(8*sigma^2))*S-2*C*E^(((2*(delta+r)+sigma^2)^2*T)/
(4*sigma^2))*Sqrt[Pi]*sigma*Sqrt[T]+E^(((4*(delta+r)^2+
4*r*sigma^2+sigma^4)*T)/(4*sigma^2))*Sqrt[Pi]*S*sigma*Sqrt[T]*
(1+Erf[((2*(delta+r)+sigma^2)*Sqrt[T])/(2*Sqrt[2]*sigma)]))/
(Sqrt[2]*(E^(r*T+((-2*(delta+r)+sigma^2)^2*T)/(8*sigma^2))-
E^(delta*T+((2*(delta+r)+sigma^2)^2*T)/(8*sigma^2)))+
E^(((4*(delta+r)^2+4*delta*sigma^2+sigma^4)*T)/(4*sigma^2))*
Sqrt[Pi]*sigma*Sqrt[T]*(1+Erf[((2*(delta+r)-sigma^2)*Sqrt[T])/
(2*Sqrt[2]*sigma)]))

Check example :

In[4]:=val={C -> 6, S -> 60, r -> 0.04, delta -> 0, sigma -> 0.3,
      T -> 0.5};

In[5]:=FindRoot[C == price /. val,{K,50}]

Out[5]={K -> 59.2639}

In[6]:=K1 /. val

Out[6]=59.2467


Not too bad, with that example...


V.Astanoff


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