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Let X and Y be two GBM’s, they have each a univariate log-normal distribution for some time t, that is $X_t\sim{LnN(µ_x, σ^2_x)}$, $Y_t\sim{LnN(µ_y, σ^2_y})$ and $Z_t=[X_t,Y_t]\sim{ MvLnN(μ, Σ)}$ where $µ_x, σ^2_x, µ_y, σ^2_y , Σ$ have the known expressions as functions of time and initial values..

We take the value of the basket as $B_t=X_t+Y_t$. The distribution of B is not log-normal but its density can be written as :

$f(b)=\int_{-\infty}^{+\infty}f(x,b-x)dx$

We write the multivariate density as the product of univariate densities and the copula density :

$f(x,y)=c(x,y)f(x)g(y)$

So we can write the density of the basket as :

$f(b)=\int_{-\infty}^{+\infty}c(x,b-x)f(x)g(b-x)dx$

Where c(x,y) is the density of the normal copula (since the copula remain the same under increasing transformations of rvs), which has a known form. Moreover we can generalise this to a sum of n variables by looking at the bivariate pairs of combinations.

Question

Is this used in pricing basket options? Since I haven’t seen such a procedure used I was wondering what is the reason for not using it?

Edit I modified the integral expressions above because they weren't totally correct. The question is still up, though I'm less convinced by this approach in the case where B is a sum of many rvs since the last integral above might become quite difficult to evaluate. Still, this would give an exact solution.

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  • $\begingroup$ Doesn't look correct to me.... $\endgroup$
    – Brian B
    Aug 5, 2013 at 14:40
  • $\begingroup$ Do you mean the whole thing or is there a specific part that you have troubles with? Could you please be more specific. $\endgroup$
    – KAT
    Aug 5, 2013 at 21:50
  • $\begingroup$ I don't understand the question. Copula and a correlation do the same thing - specify the joint density except a correlation is easier to mark intuitively. $\endgroup$
    – Arshdeep
    Aug 3, 2023 at 21:36

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