353 reputation
16
bio website kal@kalx.net
location New York, NY
age
visits member for 3 years, 8 months
seen Dec 3 at 13:05

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?