s654351


Total Posts: 2 
Joined: Aug 2019 


Hi guys,
Assuming you have 2 assets that follow geometric brownian motions: S1 and S2 The spread is defined as H=S1S2 rho is the correlation between the Weiner processes of S1 and S2 and sigma1 and sigma2 their individual volatilities. I am confused regarding the variance of H: Is it: variance_spread= sigma1^2 + sigma2^2  2*rho*sigma1*sigma2
or is it: variance_spread= S1*sigma1^2 + S2*sigma2^2  2*rho*sigma1*sigma2*S1*S2
Thanks! 



pj


Total Posts: 3497 
Joined: Jun 2004 


Neither, since the Geometric Brownian motions are not normally distributed (but lognormally)

The older I grow, the more I distrust the familiar doctrine that age brings wisdom
Henry L. Mencken 

bullero


Total Posts: 59 
Joined: Feb 2018 


You may approximate sum of lognormals using moment matching. You can find some papers which show how to do it in order to price basket options. I guess, one may tune the approach for a difference.
Also simple Google search revealed the following:
Sums of Lognormals by Daniel Dufresne:
"Almost nothing is known of the distribution of the sum of lognormals, although the lognormal distribution is a simple transformation of the very pleasant normal distribution."
However, some approximations are provided in the third section.
https://pdfs.semanticscholar.org/38a3/9907d5af01a6340c3b6f38dce39be3be5e8a.pdf




vertigo


Total Posts: 6 
Joined: Dec 2015 


assume the assets have the same reset time.
define y=sigma1 w1 + sigma2 w2, then y = alpha * b, where alpha^2 = sigma1^2 + sigma2^2 + 2 rho*sigma1*sigma2, and b is a wiener process. (you can prove this via the cholesky decomposition for wiener processes)
see that: E[S1(t)^2] = S1(0)^2 exp(sigma1^2 t), likewise for S2.
see that: E[exp(y)]=E[alpha*b]=exp[ (1/2)* alpha^2 * t]
hence: V[s1(t)]=E[s1(t)^2]E[s1(t)]^2=s1(0)^2 [ exp(sigma1^2 * t)  1]
likewise for s2.
see that: V[s1(t)s2(t)]=V[s1(t)]+V[s2(t)]2*Cov[s1(t)*s2(t)].
see that: Cov[s1(t)*s2(t)]=E[s1(t)*s2(t)]E[s1(t)]*E[s2(t)]
now,
E[s1(t)*s2(t)]= s1(0)*s2(0) *exp[ (1/2) * (sigma1^2+sigma2^2 *t)] *E[exp(y)] =s1(0)*s2(0) *exp[ (1/2) * (sigma1^2+sigma2^2 *t)] *exp( (1/2)* alpha^2 * t) =s1(0)*s2(0)*exp(rho*sigma1*sigma2*t)
hence,
V[s1(t)s2(t)] = s1(0)*exp[sigma1^2*t] +s2(0)*exp[sigma2^2*t] 2*s1(0)*s2(0)*( exp[rho*sigma1*sigma2*t] 1)
(i am sure i made a typo somewhere but you get the idea. no approximation needed) 
... maybe one day ... 

granchio


Total Posts: 1541 
Joined: Apr 2004 


also google the Margrabe formula for spread options aka outperformance options. very old stuff, but those papers should have what you need (either moment matching or gaussian approximations) 
"Deserve got nothing to do with it"  Clint 

