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On the other hand the same BSE solution [$]C ( t , x ; T , K )[$] when ( t , x ) are fixed and T , K are variable is a solution of the eq(2). Recall that underlying of the BS call is the process[$]S_r ( t )[$] which is solution of the eq(3). If the developers of the LV concept paid a little attention to what they actually have got they probably could discover that the initial condition for eq(2) can be find putting T tends to t and therefore to get

C ( t , x ; t , K ) = max ( x - K , 0 ) (4)

Given some probability experience we can write probabilistic representation of the Dupire eq with boundary condition (4). One can denote this random function as k ( u ; T , K ) where variable u changes from T to t. Correspondent SDE Ito can be written in different form. One is represented in https://www.slideshare.net/list2do/supplement-to-local-voatility formula (10). It is wrong to mixed stock price S ( t ) or its risk neutral counterpart [$]S_r[$] with math abstract random process k ( t ; T , K ) which does not exist in the market.

3. Both BS and LV approaches deal with the same call option function determined by BSE. Therefore, one could not arrive at new value or property of the call option by using LV that does not implied by BSE. The new is only behavior C ( t , x ; T , K ) as a function of T , K when t , x are fixed. When one is trying to write initially stock price with nonlinear diffusion which value is defined from Durire eq he mixed (t , x ) variables with ( T , K ) variables this is wrong as these two pares variables do not free they are bounded by the function C. Such mixing probably exists in the paper when author on the top of the p8 consider stock with arbitrary [$]\mu ( t ) [$] and at the end p17 with [$]\mu(t)[$] denotes risk-neutral drift and local volatility

Statistics: Posted by list1 — 46 minutes ago

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Can't wait. The excitement is killing me, it's almost as good as having peak gamma.

Will this result in a new paper by you?

Statistics: Posted by frolloos — Today, 4:39 pm

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1. Noting that the local volatility will in general be a function of the current stock price [$]S_0[$], we write this process as

[$]\frac{dS}{S} = \mu ( t ) dt + \sigma ( S , t ; S_0 ) dZ [$] (1)

If one follows right way he should write stock eq without any reference on whether options are exist or not. In other words we assume that stock price follows (1) on original probability space [$]( \Omega, F , P ) [$]. Assumption that [$]\sigma ( S , t ; S_0 ) [$] depends on [$]S_0[$] indeed can be done nevertheless then S ( t ) will not a Markov process and its density does not satisfy Fokker-Planck equation. Thus this assumption is incorrect in order to apply it for derivation of the PDE for call option in ( T , K ) coordinate space. On the other hand given (1) we assume also that eq (1) is considered on [ 0 , + [$]\infty [$] ) and coefficients [$]\mu , \sigma [$] are known deterministic or random functions. Otherwise it does not make sense to consider known random process S ( t ) with unknown drift and diffusion coefficients.

2. Next we arrive at the eq (4) p8

[$] \frac{\partial C } { \partial T} = \frac{\sigma^2 K^2}{2} \frac{ \partial^2 C}{\partial K^2} + ( r ( T ) - D ( T ) ) ( C ( T ) - K \frac{ \partial C} { \partial K} ) [$] (2)

where r(t) is the risk-free rate, D(t) is the dividend yield and C is short for [$] C (S_0; K; T )[$]. Though eq (2) here is correct but something here is incorrect.

Following the original Dupire idea on the end of the p7 we start with BS call option [$]C ( S_0 , K , T )[$]. This notation does not full but it implies that we actually start with call [$]C ( 0 , S_0 ; T , K )[$] which underlying stock follows standard GBM

[$]\frac{dS}{S} = \mu ( t ) dt + \sigma ( t ) dZ [$]

BS derivation leads us to the fact that

[$]C ( 0 , S_0 ; T , K ) = E max [ S_r ( T ; 0 , S_0 ) - K , 0 ] [$]

where

[$]\frac{dS_r}{S_r} = r ( t ) dt + \sigma ( t ) dZ [$] (3)

The common trick to hide the fact that BS concept implies that real underlying is [$]S_r[$] but not S ( t ) is so-called risk-neutral concept which is called for reconcile contradiction between general statement that option price as well as any derivatives takes its value from underlying stock and the fact that BS option price takes its value from heuristic risk neutral underlying [$]S_r[$] which does not exist on the market.

The original Dupire derivation deals with case r = 0. Thus we consider eq (3) putting r = 0 and therefore he also assumed that discount is equal to zero too.

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Statistics: Posted by list1 — Today, 3:35 pm

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a) On top of the page 8in the first formula [$]\sigma ( S , t ; S_0 )[$] . One can assume that volatility depends on initial value of S but in this case function S does not markovian process and therefore Fokker -Plank eq does not makes sense.

b) I could not see written boundary condition for [$]C ( T , K ) = C ( t , S_0 ; T , K )[$]

Statistics: Posted by list1 — Yesterday, 6:04 pm

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One more confusion moment.

No, it's confucian.

Sorry, that wasn't very nice of me. Please continue.

Statistics: Posted by frolloos — Yesterday, 5:26 pm

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Statistics: Posted by list1 — Yesterday, 8:39 am

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Please provide an example of a Martingale not being a Markov, as applicable in derivatives financial modelling.

To elaborate on outrun's stoch. vol. suggestion(s):

Simplest example is perhaps an ARCH/GARCH model:

[$]x_t = \sigma_t \epsilon_t[$], where [$]\sigma_t^2 = \sum_{n \ge 1} \alpha^n \epsilon^2_{t-n}[$], the [$]\epsilon_t[$]'s are i.i.d. std normal variates, and the putative martingale is [$]S_t = S_0 + x_1 + x_2 + \cdots + x_t, \quad (t \ge 0)[$].

Now [$]S_t[$] is non-Markovian as a one-dimensional process. In this example, once you promote the system to, say, [$](S_t,\sigma_t)[$], then it becomes a two-dimensional Markovian process with one component a martingale.

Sometimes it is harder/impossible to cure the non-Markovian problem by proper observation. In continuous-time [$]\sigma_t[$] is not (directly) observable.

It is also not observable in discrete-time s.v. models when a separate source of volatility noise is introduced. So, then the system is Markov, but better terminology is 'hidden Markov', when one of the system components is not observed. Further confusing the issue, sometimes the hidden component can be indirectly observed via prices of some securities which trade.

Statistics: Posted by Alan — March 22nd, 2017, 7:14 pm

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The original Dupire eq for lv was derived when [$]\mu = r = 0[$]T am interested how the similar derivation takes place when stock and risk free rates do not equal to 0 and [$]\mu \neq r [$]. I looked at the JG's lecture1 page 17. He used undiscounted call option, ie r = 0 and when he wrote Fokker-Planck equation for density of the underlying of the call he used [$] \mu[$] while underlying of the call is the process [$]S_r ( t )[$] which is solution of the sde

[$]dS_r = r S dt + \sigma S dw ( t )[$]

on the other hand he assumed that r = 0 when he supposed that call is undiscounted. Subjectively it does not look sufficiently good?

Statistics: Posted by list1 — March 22nd, 2017, 4:59 pm

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Your notation confused me; It looked like "u" was some state that only depends on time.

when i wrote [$]1 + \int_0^t f ( u ) dw ( u )[$] I thought about martingale without Markovian property. When you asked about Markovian property it rather related to math and not to finance. In math we can use any exotic representation for f, ie for example

f ( u ) = f ( [$] \int_0^u \xi ( v , \omega ) dv ) [$] where [$]\xi[$] can be arbitrary random process

Statistics: Posted by list1 — March 22nd, 2017, 3:41 pm

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list1 wrote:outrun wrote:why is this non-Markovian?

That was an example of a martingale which can be volatility. If f ( u ) is a path dependent or of it depends on a point of the past say t = 0 then all finite distributions will depends on a point in the past and the process will not a Markovian one.

Ah, so "u" is a historical realization path?

It might be does not match for financial needs but in general the martingale property does not related to the markovian property.

Statistics: Posted by list1 — March 22nd, 2017, 3:02 pm

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outrun wrote:list1 wrote:1+ [$]\int_0^t f ( u ) dw ( u )[$] is a martingale and can be an example of stoch vol.

why is this non-Markovian?

That was an example of a martingale which can be volatility. If f ( u ) is a path dependent or of it depends on a point of the past say t = 0 then all finite distributions will depends on a point in the past and the process will not a Markovian one.

Ah, so "u" is a historical realization path?

Statistics: Posted by outrun — March 22nd, 2017, 2:53 pm

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list1 wrote:outrun wrote:Stoch vol.

1+ [$]\int_0^t f ( u ) dw ( u )[$] is a martingale and can be an example of stoch vol.

why is this non-Markovian?

That was an example of a martingale which can be volatility. If f ( u ) is a path dependent or of it depends on a point of the past say t = 0 then all finite distributions will depends on a point in the past and the process will not a Markovian one.

Statistics: Posted by list1 — March 22nd, 2017, 2:36 pm

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