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frolloos
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Re: Arbitrage conditions IV surface

January 4th, 2017, 9:52 pm

I am talking about the IV surf too, so I think we are talking about the same thing. I'll try to write all this down properly in a note.
 
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outrun
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Re: Arbitrage conditions IV surface

January 4th, 2017, 10:13 pm

Ok, in that case it should be simple to test! what happens if you plug in this:
[$]
\begin{equation*}
   \sigma(S_t,t) = \begin{cases}
              0               & t < 1\\
              1               & t \ge 1
          \end{cases}
\end{equation*}
[$]
IV should go up -for all strikes- for [$]t > 1[$]

I'll buy you a strong coffee tomorrow morning if you're tired from working late throughout the night! ;-)
 
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Alan
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Re: Arbitrage conditions IV surface

January 5th, 2017, 12:44 am

[$] \boxed{\int_t^T du \int_0^\infty \frac{dK}{K^2} \nu_{BS} \frac{\partial \Sigma}{\partial u} = 0} [$] (?)

which can only be the case if [$] \Sigma(K,T) [$] is not monotonically increasing or decreasing in [$] T [$] for all [$]K[$].

Happy to hear thoughts on the above from the board.
Since [$]T-t[$] can be arbitrarily small, it seems that (?) would also imply that the dK integral by itself also vanishes at a general time = u. But  I am suspicious of that implication under the case where [$]\sigma(t)[$] is a purely deterministic function of time only. In that case [$]\Sigma[$] also depends only on times and not K's.
 
frolloos
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Re: Arbitrage conditions IV surface

January 5th, 2017, 6:05 am

Hi Outrun, ha yes, coffee would be good. Had one of those twilight zone nights where integrals kept appearing in my mind and I was trying to figure out which of the integrals smelt rotten :)

Alan, I had exactly the same questions as you: why doesn't it work for a deterministic vol? This probably also why the examples by Outrun are (good) counter-examples. The only answers I can come up with are: either (?) is just wrong, or it works only for surfaces generated by true smile models.

Given a skew, which flattens at long time to maturities, then it does seem plausible that the smile have to twist in time to go from skewed to flat(tish). Also, as you pointed out, given T-t can be made (arbitrarily) small, if (?) were true it would hold for every time slice. This is odd. I'll re-do the whole thing and try to understand it better.
 
frolloos
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Re: Arbitrage conditions IV surface

January 5th, 2017, 11:46 am

Gave it some more though, the reason it doesn't appear to work for deterministic case is because in the deterministic case this condition indeed does not exist. Because in deterministic case C = C_{BS}, not only in terms of value at a certain (K,T), but also in functional form. So I don't see any issue here actually.
 
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Alan
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Re: Arbitrage conditions IV surface

January 5th, 2017, 3:06 pm

Still confusing to me. What steps in post #14 change?
 
frolloos
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Re: Arbitrage conditions IV surface

January 5th, 2017, 5:13 pm

Still confusing to me. What steps in post #14 change?
Sorry, yes you're right. I thought I understood it, but I am also still confused actually.

There must be something very silly about (?) :) but i can't see it yet.

EDIT:

Ok, I think I identified the evil. Instead of looking at the spot starting varstrike, consider the forward var:

[$] \int \frac{dK}{K^2} \left [ C(K,T_2) - C(K, T_1) \right ] [$]

Now,

[$] C(K,T_2) - C(K, T_1) = C_{BS} (K, T_2, \Sigma(T_2)) - C_{BS} (K, T_1, \Sigma(T_1)) [$]

[$] = C_{BS} (K, T_2, \Sigma(T_2)) - C_{BS} (K, T_1, \Sigma(T_2)) + \int du \, \nu_{BS} (\partial\Sigma / \partial u)  [$]

Clearly [$] C_{BS} (K, T_2, \Sigma(T_2)) - C_{BS} (K, T_1, \Sigma(T_2)) \neq C(K,T_2) - C(K, T_1) [$] and the integral term consisting of vega and term structure slope compensates for it. 

The problem was that I looked at [$] T_1 = t [$], in which case there is no time value left and the choice of implied vol does not matter, hence misleading me (us), sorry abt that folks!
 
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outrun
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Re: Arbitrage conditions IV surface

January 5th, 2017, 8:26 pm

Nice inspection froloos, smart find! You can now finally have calm dreams tonight
(so don't think about Brownian vol surfaces)
 
frolloos
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Re: Arbitrage conditions IV surface

September 21st, 2017, 6:50 pm

I've seen two nice proofs of no calendar spread arbitrage condition.

The first, in a well-known paper by Fengler: http://www.javaquant.net/papers/Fengler.pdf , see proposition 2.1 on page 11, and the second in an equally well-known paper by Gatheral: https://arxiv.org/pdf/1204.0646.pdf , see lemma 2.1 on page 3.

It seems to me though that the proof given in Fengler (which is actually due to Reiner) is more general than the Gatheral proof, since the latter explicitly assumes martingale property, whereas the former relies purely on no-arbitrage arguments and so the underlying can also be a local martingale? Anyone thoughts on this / (dis) agree with my conclusion?