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Cuchulainn
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### Re: Silly questions

Well, if we are going to be so literal, then since the original post did not want an answer but a vote I am going to vote yes!! This is maths 21st century style!
That's exactly what happened on LI... everyone voted on some beautiful formula by RAMANUJAN (as in the movie).
Most people on LI were not interested in accuracy, but in beauty. It has become art, like physics 60 years ago. Even I have a t-shirt with E = mc^2 !
Hollywood has destroyed mathematics.

How can you differentiate wrt an integer?

n! is a special case of gammaish function.(z) when z is an integer, yes?
Last edited by Cuchulainn on February 13th, 2020, 1:51 pm, edited 1 time in total.
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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Cuchulainn
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### Re: Silly questions

Here's the formula! Who can explain it?

"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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http://www.datasim.nl

FaridMoussaoui
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### Re: Silly questions

That's exactly what happened on LI... everyone voted on some beautiful formula by RAMUJAN (as in the movie).
Ramanujan

Cuchulainn
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### Re: Silly questions

I demand a recount! Anyway, it's not even 6am -- don't you people sleep?
Isaiah 57:20
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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http://www.datasim.nl

Cuchulainn
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### Re: Silly questions

That's exactly what happened on LI... everyone voted on some beautiful formula by RAMUJAN (as in the movie).
Ramanujan
Corrected! It's the first time that I wrote this name. Number theory..
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

http://www.datasimfinancial.com
http://www.datasim.nl

Cuchulainn
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### Re: Silly questions

Does the solution of

$(1 - \tau)^2 \frac{\partial u}{\partial \tau} = \Delta u$ (where $\Delta$ is the Laplacian) with appropriate initial and boundary conditions converge as $\tau \rightarrow 1$? ($\tau$ is false/pseudo time)

And what does it converge to?

// aka
$\bigtriangledown^2 \equiv \Delta$
nabla ...
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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http://www.datasim.nl

Alan
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### Re: Silly questions

Pretty sure: in 1D, the transition density $p(x,t)$, starting from an arbitrary density at t=0, converges to 0 for all real x, as $t \uparrow 1$. Loosely, all the mass went to $\pm \infty$.

Proof: the Green function $G(x,T|x_0) \propto \frac{e^{-(x-x_0)^2/(2 v_T)}}{\sqrt{v_T}}$, where $v_T = T/(1-T)$.
Last edited by Alan on March 1st, 2020, 9:28 pm, edited 1 time in total.

katastrofa
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### Re: Silly questions

Does the solution of

$(1 - \tau)^2 \frac{\partial u}{\partial \tau} = \Delta u$ (where $\Delta$ is the Laplacian) with appropriate initial and boundary conditions converge as $\tau \rightarrow 1$? ($\tau$ is false/pseudo time)

And what does it converge to?

// aka
$\bigtriangledown^2 \equiv \Delta$
nabla ...
For tau =0 you have a time-independent heart/wave equation. I'm not sure why it wouldn't converge?

Alan
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### Re: Silly questions

Well, if my solution is correct, the Green function cannot be continued beyond $\tau = 1$, so there is an issue there.

katastrofa
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### Re: Silly questions

If you change variables to z = 1/(1-tau), you obtained a standard heat equation. If tau goes from 0 to 1, z goes from 0 to infty. The physical solution for u at infty is a uniform heat distribution, modulo boundary conditions.
Just some quick thoughts.

Alan
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### Re: Silly questions

Nice! Another proof that, on the whole line, the density -> 0 as tau -> 1.

Plus, shows the answer is the same (0) in D dimensions!

Cuchulainn
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### Re: Silly questions

Elliptic PDE are more difficult to solve than parabolic PDE.

In my example I add a false time derivative to give

$\frac{\partial u}{\partial t} = \Delta u$

When $t \rightarrow \infty$ the solution converges to the time-dependent solution. But $\infty$ is big so I transform $\tau = t/(1+t)$ and discretise up to (almost) $\tau = 1$ (works well for other bespoke cases) The empty quarter $\tau \gt 1$ leads to solution explosion, i.e. solution does not exist. This is easy to check by looking at the exact solution of $\frac {du}{dt} + u = 0$ or more generally Picard-Lindelof iteration.

kats almost recovered (typo?) my heat equation for $t = \tau/(1 - \tau)$. The mapping $t = 1/(1 - \tau)$ is not monotonic.

There's not much out there on this False Transient MOL but I can imagine that it would be more efficient than SSOR and that bunch.

.. that's the background ..
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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Cuchulainn
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### Re: Silly questions

Alan,
I have revisited the Anchor pde; I have several remarks and questions.
Anchor-Style PDE (See Anchor article)
2020-3-16

https://onlinelibrary.wiley.com/doi/epd ... wilm.10366

(A)  For large asymptotic t values the solution of the parabolic  PDE(x,y,t) converges to the solution of the corresponding elliptic PDE(x,y) (Friedman 1992).
(B)  The MOL solution leads to  system of ODEs $\frac{\partial u}{\partial t} + Au = F$. If $A$ is an M-matrix then the solution for large $t$ is the solution of $Au = F$(Varga 1962).
(C)  The Saul’yev ADE method is accurate and very fast. For example, in the case T = 1, run-time is 0.15 seconds on a normal laptop. The MOL works well also but in general 60 times slower than ADE. I haven’t checked, but I suspect that other FD methods would also be slower, e.g. ADE B&C is [2.3] times slower than Saul’yev  ADE.
(D)  I have no idea how to measure the rate of convergence of the results in (A) and (B), although in case (B) there might be a glimpse of hope if we knew what the smallest eigenvalues of the matrix A are (slow transients).
(E)   For Anchor ADE, the boundary conditions in both x and y have little impact on accuracy.
(F)   At the time, why did the authors not use a similarity reduction R = S/A (admittedly, it remains a trick and breaks down in this case)?
(G) For T = 10, K = 140 (NX = NY = 300) MOL C = 28.7023, Time = 13734 seconds (Alan gets Call = 28.629 but no Time given). Table 2 of Anchor article. For ADE, we got
NX, NY, NT; Call, Timing (seconds)
300,300,2000;28.311,27
300,300,3000;28.442,?
300,300,5000;28.522,70
300,300,6000;28.58,?
300,300,7000;28.6003,97
300,300,8000;28.6155,114
300,300,9000;28.6275,125
300,300,10000;28.637,140 (rounding errors start to creep in?)
NX = NY = 300, NT = 9000 seems to be optimal (ADE is more accurate and 110 times faster than MOL in this case).

(H) $\tau = t/(1+t)$ turns a ODE into a stiff ODE ("no free lunch in maths").

Main Questions:

(Q1) Which value of T is infinity in the article (T = 100, T = 1000 etc.)?
(Q2) In section 10.2 I assume NDSolve is used for large t.  I assume it was used to generate prices in Table 4? Also, T = 1000 is essentially infinity?
(Q3) Instead of NDSolve, what about Gauss-Seidel/SSOR etc.  on the elliptic PDE (see remark (A))? It would probably 1) faster and 2) less rough-and-ready than choosing a random T? It might be interesting to try it out? (Someone mentioned that Mathematica not have elliptic solvers).
(Q4) How can we reproduce/find call and put prices in Table 2 for T = 100? Why are the values as they are, e.g. T = 100, K = 100, why is put = 0.1073?
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

http://www.datasimfinancial.com
http://www.datasim.nl

Cuchulainn
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### Re: Silly questions

This is probably an easy one to answer for the local stochastic experts here. Consider the n-dimensional (driftless) GBM

$dx_t = \alpha dw_t$ where $x_t \in \mathbb {R}^n$, $w_t$ is a standard Brownian motion and $x_0 \in \mathbb {R}^n$ is an arbitrary starting value. $\alpha$ is a positive constant.

If all else fails, I suppose I can use Euler?

My basic question: is there an easily computable/closed/numerical solution to give a solution for any $t \gt 0$. The value $n$ can be 'biggish' e..g. [1,300] for starters.
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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http://www.datasim.nl

Alan
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### Re: Silly questions

BM or GBM?

If you mean what you actually typed, the SDE solution is $x_t = x_0 + \alpha \, w_t$.

The pdf for locating $x_t$ is just the n-dim normal density with mean $x_0$ and diagonal variance-covariance matrix $\alpha^2 t$ times the identity matrix. In other words, it's just the product of the one-dimensional pdf's.

If you meant to type, component-wise, $dx^i_t = \alpha \, x^i_t \, dw^i_t$, then switch to $y^i_t = \log x^i_t$ and you'll get a similar solution with a drift. In terms of the orig. coords, the pdf is a product of n lognormals.