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Cuchulainn
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American options -- Reference prices

January 27th, 2016, 6:37 pm

What about?CASE D (BTW A to C were from the spursfan)S = K = 100T = 1r = 0.01q = 0.02sig = 0.01CASE ES = K = 100T = 1r = 0.01q = 0.01sig = 0.01All in favour?
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Alan
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American options -- Reference prices

January 27th, 2016, 11:45 pm

QuoteOriginally posted by: CuchulainnWhat about?CASE D (BTW A to C were from the spursfan)S = K = 100T = 1r = 0.01q = 0.02sig = 0.01CASE ES = K = 100T = 1r = 0.01q = 0.01sig = 0.01All in favour?Those are good. My answer: Case D: 1.0671904578521593019042848, where I expect all 26 digits are good. Timing is less than 1/1000 of a second for `almost' unlimited digits.For example:Timing[N[pbs[100,100,1/100,2/100,1/100,1],100]]{0.,1.067190457852159301904284849092240828401660777025395245899802301901728023665508859046891074937644206} New break-through in American-style pricing?No -- if you think about it, Case D should be the same as the Euro-style value up to a correction of [$]O(10^{-1000})[$]
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mj
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American options -- Reference prices

January 28th, 2016, 2:44 am

I'd like to see some benchmarks for S is 120,K is 100, r=0.08, the volatilityis 0.2, the dividend rate q is 0.08, and the expiry T is 3This was studied in Quantitative Finance, Vol. 12, No. 1, January 2012, 17?20On the analytical/numerical pricing of American putoptions against binomial tree pricesMARK JOSHI and MIKE STAUNTONand was the subject of a certain amount of dispute. The price obtained there was5.929805
 
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January 28th, 2016, 3:07 pm

Case F (mj and spursfan)S = 120, K = 100r = 0.08, q = 0.08sig = 0.2T = 3mj/spursfan 5.929805///For FDM using the penalty methodNS, NT, P1000,1000,5.92579310000,10000, 5.92764100000,100000, 5.927768For FDM/ADE with 1st order correctionNS, NT, P1000,1000,5.92651710000,10000, 5.927727100000,100000, 5.927775For Tian BMNT,P1000, 5.9289295000,5.92949010000,5.929836913000, 5.9297679 (monotonicity breaks down, don't know..)
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Alan
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American options -- Reference prices

January 28th, 2016, 5:57 pm

QuoteOriginally posted by: outrunInteresting, an online BS calculator gives 5.62 and you mentioning a dispute means that my intuition that it's just a European BS put on a future is wrong? Ah,.. because of K!=S ? There is this nice American call/put parity formula:C(S,K,r,q,sigma,T)=P(K,S,q,r,sigma,T)mj's case, using straight binomial but that dual call:Tsteps Put------- -------------1 x 10^6 5.92980483 x 10^6 5.9298050If I have time, I'll try to update later with 3 x 10^6 steps. (done)It's strange. On my system, the value is essentially the same for the put vs the dual call,but the latter runs about 6 times faster. (27 mins for the 10^6 case). Weird.
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Cuchulainn
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January 29th, 2016, 9:34 am

QuoteIn my code I do an early exercise check at every node -which is costly-,What is the accuracy of this approximation? With FDM it is first order.
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January 29th, 2016, 9:56 am

QuoteOriginally posted by: outrunInteresting, an online BS calculator gives 5.62 and you mentioning a dispute means that my intuition that it's just a European BS put on a future is wrong? Ah,.. because of K!=S ? There is this nice American call/put parity formula:C(S,K,r,q,sigma,T)=P(K,S,q,r,sigma,T)Does this parity hold for perpetuals, for which a closed formula is known (in Collector's book page 108)..//thinking out loud: is a perpetual case equivalent to saying T -> infinity?
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Cuchulainn
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January 29th, 2016, 10:49 am

QuoteOriginally posted by: AlanQuoteOriginally posted by: CuchulainnWhat about?CASE D (BTW A to C were from the spursfan)S = K = 100T = 1r = 0.01q = 0.02sig = 0.01CASE ES = K = 100T = 1r = 0.01q = 0.01sig = 0.01All in favour?Those are good. My answer: Case D: 1.0671904578521593019042848, where I expect all 26 digits are good. Timing is less than 1/1000 of a second for `almost' unlimited digits.For example:Timing[N[pbs[100,100,1/100,2/100,1/100,1],100]]{0.,1.067190457852159301904284849092240828401660777025395245899802301901728023665508859046891074937644206} New break-through in American-style pricing?No -- if you think about it, Case D should be the same as the Euro-style value up to a correction of [$]O(10^{-1000})[$]Case DFDM with NS = NT = 10^5 P = 1.067185BinomialNT = 10,000 P = 1.06719450NT = 13,000 P = 1.06719342Case EFDM with NS = NT = 10^5 P = 0.395621BinomialNT = 10,000 P = 0.3956365NT = 13,000 P = 0.3956339
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Alan
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American options -- Reference prices

January 29th, 2016, 7:46 pm

QuoteOriginally posted by: outrunQuoteOriginally posted by: CuchulainnQuoteIn my code I do an early exercise check at every node -which is costly-,What is the accuracy of this approximation? With FDM it is first order.I was talking about the trees here, (not the paper)after updating a node C(i,t) I force the value to be at least parity.C(i,t) = max[ C(i,t), K-S(i,t) ]I check at every node too. But, you're right -- it might be more efficient to be smarter about this. It would easy to justcheck at the "in-the-money" nodes and probably yield a speed-up (for trees). Beyond that, I wonder how much would be gained, if any, by just checking at nodes that are at-or-below (puts) or at-or-above (calls)the last known exercise boundary node. This would be 'valid' in the continuum -- since the boundary is monotone in time.But (i) keeping track of the extra info might not be worth it, plus (ii) you probably need to let the tree boundary bounce up byone node (producing a sawtooth) to keep the procedure un-biased. So now the "efficient-checking" rule gets quite messy: probably not worth it ...
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Cuchulainn
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American options -- Reference prices

January 29th, 2016, 8:05 pm

Quote* The initial guess taken from an other paper is based on root finding and the current root finder I've used (sort of Halley's method) overshoot in some cases. I need to use a more robust method...This method is a bit esoteric. I have never used it. Just sayin' Another nice solution is to write f(x) = 0 as a least square min f^2(x) and use Boost Brent.
Last edited by Cuchulainn on January 28th, 2016, 11:00 pm, edited 1 time in total.
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