Hi all:Can someone show me how to find Vega PDE for American option?I know for European option, I can get the pde by taking derivative wrt sigma on the whole BS PDE:At this point, I could solve it to get vega by FDM.How can I adjust this for american options?

The 'standard' way to get vega for the amer-style would be to take [f(t,S;sig+dsig)) - f(t,S; sig)]/dsigwhere f is the option value. In other words solve the amer. problem *twice*. In principle, I supposeyou could:(i) solve the american problem once for f(t,S;sig) and the critical exercise boundary b(t)(ii) solve your PDE for vega v in the continuation region, with the terminal condition v(T,S) = 0 and the b.c. v(t,b(t)) = 0Off-hand this second approach seems like more work, but I don't see why it shouldn't work if youreally want to do it that way.

>> (ii) solve your PDE for vega v in the continuation region, with the terminal condition v(T,S) = 0 and the b.c. v(t,b(t)) = 0Alan, I'm wondering if it's correct to impose the 2nd b.c. because the boundary itself, b(t), implicitly depends on sig.

Certainly vega(t,S) = 0 for S in the stopping region. If vega is continuous across the exercise boundary, thenthat would justify it. My intuition (well, wild guess, really) was that there was enough smoothness from smooth pasting tohave vega continuous. But, you're right to challenge this as (i) I certainly don't have a proof,and (ii) I agree that the boundary moves with sigma. So, straightforward issue: is vega continuous across the exercise boundary for american options under GBM? Somebody must know for sure. Or, failing that, can somebody with a good american solver please post a plot of vega(t,S) vs S for a range of S that crosses the exercise boundary? Any discontinuity should be clear from the plot. A final thought: it should be very easy to settle this continuity question for perpetual american puts. Perhaps accelas can report on that case. This would be suggestive for the finite horizon case.

Well, since no response, I just did this perpetual put case myself. I found that vega is indeed zero as S approaches B from above for the perpetual american put.Here B is the critical exercise value.Specifically, with K = strike price, I found that for S >= B and r > 0, This clearly vanishes at S = B(v) because of the log term.So, vega is continuous across the boundary and my boundary condition is justified for the perpertual case, anyway.

Hi,Referring to the equation by accelas (vega pde with source/sink term): ,for standard european options, what would be the upper and lower boundaries? ( Vega(Smin,t) and Vega(Smax,t) )I used BS vega at the boundaries but it gave me almost 0 vegas for the whole grid. What would you use as boundary conditions?

QuoteOriginally posted by: DinDecHi,Referring to the equation by accelas (vega pde with source/sink term): ,for standard european options, what would be the upper and lower boundaries? ( Vega(Smin,t) and Vega(Smax,t) )I used BS vega at the boundaries but it gave me almost 0 vegas for the whole grid. What would you use as boundary conditions?How do you know that vega satisfies a PDE? What about taking the BCs for the perpetual American in the current PDE? (?)

Prof. Duffy,This vega pde comes from the Lagnado and Osher paper, where they used this to compute the variational derivative.For the european call boundary conditions, I just took the vega I computed using direct differences. And the result I mentioned where I had nul everywhere was due to a bug in my code.Din

So, since this thread is revived, can anybody answer the open question:Is vega continuous across the critical boundary for the finite maturity American case?Or, in the absence of a proof (or the cite of one), what does a numerical pde solution suggest?

QuoteOriginally posted by: AlanSo, since this thread is revived, can anybody answer the open question:Is vega continuous across the critical boundary for the finite maturity American case?Or, in the absence of a proof (or the cite of one), what does a numerical pde solution suggest?How would one formuate such a pde, i.e. what does it look like? In particular, the boundary?

Well, take a look at my orig. post here.As I said there, there are two ways to proceed.In method one, you simply take the regular (free boundary) pde problem for the american-style put, and solve it twice. (with sigma & sigma + eps)In method two, the pde is simply the one you get by taking d/d sigma of the first one, solvedon (B(t),infty) where B(t) is the critical boundary of the first one. If, indeed, vega is continuous acrossB(t), then this is a very simple problem. B(t) is known (up to numerical errors) and the unknown vega = u(t,x) vanishes at both B(t) and infinity (and initially) So it's a *very* simple Dirichlet problem,once you've taken the trouble to get B(t) and the american put soln to good accuracy.Then, you could check that this Dirichlet solution is numerically very close to the method one soln.

QuoteOriginally posted by: AlanWell, take a look at my orig. post here.As I said there, there are two ways to proceed.In method one, you simply take the regular (free boundary) pde problem for the american-style put, and solve it twice. (with sigma & sigma + eps)In method two, the pde is simply the one you get by taking d/d sigma of the first one, solvedon (B(t),infty) where B(t) is the critical boundary of the first one. If, indeed, vega is continuous acrossB(t), then this is a very simple problem. B(t) is known (up to numerical errors) and the unknown vega = u(t,x) vanishes at both B(t) and infinity (and initially) So it's a *very* simple Dirichlet problem,once you've taken the trouble to get B(t) and the american put soln to good accuracy.Then, you could check that this Dirichlet solution is numerically very close to the method one soln.For method two I think an extra BC (?) is needed at the B(t) because B(t) is unknown. My mathematical handwaving wrt the smooth pasting conditions leads tovega = 0 (same conclusion as Alan)d (vega) / dS = 0 (but it looks this does _not_ hold in the perpetual case looking at your perpetual formula)on B(t).This would be a 3rd alternative.But I think you are suggesting to compute B(t) before computing vega. This would entail running the solver twice. Hopefully, errors will not cause problems between the solvers.

Yes, B(t) is not supposed to be an unknown for method 2. You compute it separately, first, and then solve method 2 pde with B(t) known.

QuoteOriginally posted by: AlanYes, B(t) is not supposed to be an unknown for method 2. You compute it separately, first, and then solve method 2 pde with B(t) known.Computationally, it will be twice as slow as the normal case.It is also possible to solve 'normal' early exercise PDE without having to compute B(t) (e.g. penalty). Would this work also for vega? So differentiate the PDE and then add the penalty term?(?)

I don't know -- method 4?