Keith A. Lewis
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 Dec 3 comment Arbitragefree Pricing: Q vs. P There is no physical measure P. The measure Q is not the probability of anything. The FTAP is a geometric result, not a probabilistic. See kalx.net/fms/fms.html for the details. Dec 3 awarded Commentator Dec 3 comment Equivalent (true) Martingale Measures and no-arbitrage conditions See kalx.net/fms/fms.html for my latest thinking. It is mathematical folly to waste time proving the hard direction of the FTAP. Dec 3 comment Equivalent (true) Martingale Measures and no-arbitrage conditions Trader: Great. How much do I make up front? Quant: Nothing. Trader: Okay. How much do I make on the back end? Quant: A positive non-zero amount of money. Trader: Um, how positive are we talking here? Quant: I can't tell you that. Trader: What are the odds I make this positive non-zero amount? Quant: I can't tell you that. Trader: Well, when do I make it? Quant: I can't tell you that. Trader: You call that an arbitrage? Quant: Yes. It is the mathematical definition! Trader: Get the !@#$%^&* off my trading floor. Dec 3 comment Equivalent (true) Martingale Measures and no-arbitrage conditions Quant: Hi, Mr. Trader. I have an arbitrage for you! Nov 29 answered FTAP a-la Harrison, Kreps and Pliska Nov 29 answered Equivalent (true) Martingale Measures and no-arbitrage conditions Nov 29 awarded Supporter Nov 29 comment risk-neutral valuation implies no arbitrage? Use the fact exp(-sigma^2 t/2 + sigma W_t) is a martingle. Nov 29 awarded Critic Nov 29 comment risk-neutral valuation implies no arbitrage? It's wrong. As is your SDE for S. Should be dS/S = r dt + sigma dW, where W is Brownian motion. See the link I gave you for all the details. Nov 29 answered risk-neutral valuation implies no arbitrage? Jul 25 comment Paradoxes in quantitative finance Any function with f'(sd) = 0 = f'(su) will have dv/ds = 0. May 11 comment How to get greeks using Monte-Carlo for arbitrary option? I beleive AD refers to techniques for automatically generating functions for the derivatives. Dual numbers don't do that for you. May 9 comment Is the binomial model wrong? Relax, I'm a friendly troll. :-) Here is another one that I don't have a good answer for. Let$F = fe^{-\sigma^2t/2 + \sigma B_t}$. The (forward) value of a log contract is$v = E[\log F]$and so$dv/df = 1/f$. No surprise. Now consider the payof$v = E[\log F/f]$. Now$dv/df = 0$! It is easy to see what is going on in this example, but how do you know with more complicated parameterizations when the correct delta is not the derivative of value with respect to underlying? Hope you find this puzzle more interesting. May 8 awarded Student May 8 comment Is the binomial model wrong? But in a binomial model where$s$can go to$S^-$or$S^+$the option value is$v = ((S^+ - Rs)V(S^-) + (Rs - S^-)V(S^+)/R(S^+ - S^-)$and$dv/ds = (V(S^+) - V(S^-))/(S^+ - S^-)$. I have to confess I am trolling a little here. See kalx.net/ftapd.pdf for the explanation. Also note that$d(Rv)/dR = s(V^+ - V^-)/(S^+ - S^-)\$ is the dollar delta in both models. May 8 awarded Editor May 8 revised Is the binomial model wrong? edited body May 8 asked Is the binomial model wrong?