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s654351


Total Posts: 1
Joined: Aug 2019
 
Posted: 2019-08-23 04:26
Hi guys,

Assuming you have 2 assets that follow geometric brownian motions: S1 and S2
The spread is defined as H=S1-S2
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: 3457
Joined: Jun 2004
 
Posted: 2019-08-23 05:38
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: 52
Joined: Feb 2018
 
Posted: 2019-08-23 11:03
You may approximate sum of log-normals 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: 4
Joined: Dec 2015
 
Posted: 2019-08-23 15:56
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
 
Posted: 2019-09-10 17:31
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
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