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Cuchulainn
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Who understands Black Scholes Boundary Conditions, really

November 8th, 2008, 8:56 pm

What's the update on TBC, guys? I had a look but could not find anything.
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Cuchulainn
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November 10th, 2008, 7:56 pm

Alan,The following analysis is central to what I am doing. What I am doing different (for the moment) is 1) transformation only in y (== v), 2) w = tanh(y) instead of Cayley and 3) MeanRev in v (C_v)(1) - C_t + 1/2 y x^2 C_xx + 1/2 sig^2 y C_yy + K(theta = y)C_y = 0I examine the PDE when x = 0 and use w = tanh(y) to get:(2) - C_t + "ZERO" + AC_ww + BC_w = 0 on x = 0where A = 1/2 sig^2 H(w)F(w)^2B = {-sig^2 H(w) + K(theta - H(w))} F(w)and H(w) = 1/2 log ((1+w)/(1-w))F(w) = (1 - w^2)BC===w = 0, Fichera == K theta - sig^2/2 >= 0 ==> NO BCand PDE becomes -C_t + K theta C_w = 0 (same as PDE on all of w = 0 for 0 < x < Xmax)w =1, Fichera == 0 ==> NO BCand DE becomes C_t = 0 ==> C = initial payoffI seem to be happy with this.What do you think?on bdy w = 0 I get the usual 1st order PDE (as in Heston). FineOPEN QUESTION IS NOW!ON w = 1, the term H(w) is not finite and then Fichera == 0 and the PDE has a unbounded H(w) at w = 1. I CANNOT USE THE DEGENERATE PDE ON w = 1 (I suspect).So, do we say C = x? //Numericallly, we could take w = 1 - epsilon as upper boundary, but that's not the point, just yet.
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Alan
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November 10th, 2008, 8:41 pm

Yes, I keep saying C = S.Then, you change threads and I say C=S again. I'm not going to change my mind unless you either: (i) prove my book Ch. 10 argument wrong or(ii) give some compelling rationale for something else.
 
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November 10th, 2008, 8:52 pm

QuoteOriginally posted by: AlanYes, I keep saying C = S.Then, you change threads and I say C=S again. I'm not going to change my mind unless you either: (i) prove my book Ch. 10 argument wrong or(ii) give some compelling rationale for something else.I don't have an answer for this BC from my analysis!. Now, if we take C = S and the other boundary conditions, is this a well-posed problem? If so, we can then go to try out the numerical solution. I can test this prototype.Shall we do this now?
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November 10th, 2008, 9:10 pm

Well-posed? I would say -In the continuum problem, I don't think C = S is needed -- it's more of a consistency condition.-For your numerical method, (i) if you -need- a value, then that's the one to use.(ii) if you don't need a value, then your soln should show that C->S as w -> 1.
 
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November 11th, 2008, 7:08 pm

Alan,I now have a formulation of the problem and have started setting up the algorithms. So, my first prototype is for the transformed variant of: (1) - C_t + 1/2 y x^2 C_xx + 1/2 sig^2 y C_yy + K(theta = y)C_y = 0For V1, I put K = 0 (don't feel like solving a PDE&FDM on w = 0 just yet )The idea is to do explicit Euler and let it rip until I get convergence (or not). I have already solved the sub-problems on z = 0 and z = 1 last week (see Well-posed thread), so these PDEs behave properly. If the results are good, intend to up a gear to Soviet splitting and maybe a stiff ODE solver. Accuracy is the issue at the moment.At some stage we can input some relevant data to the solver when I get it working.QUESTION:I should transform in such a way that we can have 'hotspots' at (z,w) == (1/2, 1/2). It would be good for initial checking.//for OpenMP people, I can parallelise this problem (later) because the 4 boundary conditions at each time are independent.
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November 11th, 2008, 9:28 pm

What's the question?
 
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November 12th, 2008, 9:33 am

Transforming from all (z,w) space back to (x,y) == (S,v) space. I'll cross that bridge when I come to it. Now I just define special values S = 100, v = .2 etc. and get the price at (z,w) = (1/2, 1/2).
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May 3rd, 2009, 7:02 pm

QuoteTransparent boundary conditions (TBC) or artificial boundary conditons (ABC) are very used termsfor problems defined in unbounded domains. Called it what you want but this terminology is the right one.It gives you a non-local operator.In the case of BS, it is simply!!where and depend on r and I had a look at this thread again. But the issue of TBC and truncation is becoming less attactive. The CFD thread is discussing alternatives.
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March 24th, 2010, 9:42 am

Thought that this post was useful to teleport (from Student) to hereQuote@AlanThere are real financial products that have real boundary conditions; for example, an option that is knockedout (becomes worthless) if the underlying price touches 30 or 70 has C(S=30,t) = C(S=70,t) = 0.However, you want boundary conditons purely to satisfy the need for a finite difference method tohave boundary condtions on an artificially truncated domain. There are *many* ways to do this because,in the end, the boundary conditions don't matter as the truncation is removed. As a hint,just experiment with boundary conditions that behave like the exact solns or its derivatives as S becomes small or large.
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May 7th, 2014, 5:58 pm

Fichera II
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May 8th, 2014, 4:44 pm

QuoteOriginally posted by: CuchulainnFichera IIInteresting variation in the Student forum: [$]dX_t = \kappa (\theta(t) - X_t) + \sigma \sqrt{X_t} \, dW_t [$], where [$]\kappa > 0[$] and [$]\theta(t)[$] is periodic. Here is the threadFor example, [$]\theta(t)= a + \cos^2(t)[$], and [$]a[$] some real number. Then, it's possible that the need for a boundary condition at x=0 can change with time!
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July 21st, 2015, 1:26 pm

I have reduced an elliptic PDE(x,y) to canonical form PDE(v,w) (no mixed derivatives :)) . So far, so good.The question is how to 'translate;' the BC(x,y) to BC(v,w).Ex. the Heston PDE BC at v = 0 is fine. But how to formulate the BC in (v,w) world? My concern is you get a modified PDE on a non-rectanguar domain(?) We probably get a rhombus thing.// the transformation is linear; v = ax - y. w = bx
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July 21st, 2015, 11:23 pm

QuoteOriginally posted by: CuchulainnI have reduced an elliptic PDE(x,y) to canonical form PDE(v,w) (no mixed derivatives :)) . So far, so good.The question is how to 'translate;' the BC(x,y) to BC(v,w).Ex. the Heston PDE BC at v = 0 is fine. But how to formulate the BC in (v,w) world? My concern is you get a modified PDE on a non-rectanguar domain(?) We probably get a rhombus thing.// the transformation is linear; v = ax - y. w = bxMy recollection is that you can transform away the correlation term in the Heston operatorwithout altering the state space from R x R+On the other hand, for some other finance operators, it doesn't necessarily work so nicely or at all.
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August 10th, 2015, 8:50 am

QuoteOriginally posted by: AlanQuoteOriginally posted by: CuchulainnI have reduced an elliptic PDE(x,y) to canonical form PDE(v,w) (no mixed derivatives :)) . So far, so good.The question is how to 'translate;' the BC(x,y) to BC(v,w).Ex. the Heston PDE BC at v = 0 is fine. But how to formulate the BC in (v,w) world? My concern is you get a modified PDE on a non-rectanguar domain(?) We probably get a rhombus thing.// the transformation is linear; v = ax - y. w = bxMy recollection is that you can transform away the correlation term in the Heston operatorwithout altering the state space from R x R+On the other hand, for some other finance operators, it doesn't necessarily work so nicely or at all.I think there are two issues here:1. Reduction of PDE/quadratic form to canonical form by a homogeneous linear change of variables. This is OK, bug but the new domain is a rhombus.2. Approximating the boundary condition on the transformed boundary. This aspect could to be either or painful depending on the particular problem. We have to find neighbouring nodes on the boundary and corresponding BCs.Life has been made easy with 1 and difficult with 2.
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