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katastrofa
Posts: 9853
Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

### Re: Proof of the risk-neutral assumption

Lie Trotter .

Just so happens that I am implementing  LT as we speak. It's first-order. or 2nd order, go for Strang-Marchuk operator Splitting. Nice thing is splitting {diffusion, convection}, {potential, kinetic energy} and not {x,y}.

Wow! I didnie know you physicists knew BCH et al.

https://www.asc.tuwien.ac.at/~ewa/semin ... ag_Exl.pdf
I'm making candied orange skin.

I know it all mostly from quantum spin systems theory (earlier applications in physics are related to Brownian motions). What's different about physics and other hard sciences is that the Nature actually does the job of performing efficiently under strong random fluctuations. You guys just crash and crash, whenever the heat goes up.

What's BCH?

platinum
Posts: 174
Joined: May 6th, 2009, 11:08 am

### Re: Proof of the risk-neutral assumption

katastrofa
Posts: 9853
Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

### Re: Proof of the risk-neutral assumption

Thanks, platinum.

Cuchulainn, splitting methods are commonly used for solving Schroedinger equations of featured physical systems (e.g. Bose-Einstein condensate). They can be numerically cumbersome when the solution oscillates strongly (they're not precise enough), so more sophisticated methods are being developed.

Cuchulainn
Posts: 63816
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
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### Re: Proof of the risk-neutral assumption

Thanks, platinum.

Cuchulainn, splitting methods are commonly used for solving Schroedinger equations of featured physical systems (e.g. Bose-Einstein condensate). They can be numerically cumbersome when the solution oscillates strongly (they're not precise enough), so more sophisticated methods are being developed.
Pathological numerical problems? What causes the oscillations? Overshoot?

Splitting error is a fixed cost, the other error is truncation error for the separate operators.

$u(t) = e^{(A+B)t}$ = $e^{At}e^{Bt}$ if $[A,B] = AB - BA = 0$ for matrices $A$ and $B$. $t$ is a real number.

is it also if and only if (iff)?
My C++ Boost code gives
262537412640768743.999999999999250072597198185688879353856337336990862707537410378210647910118607313

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

katastrofa
Posts: 9853
Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

### Re: Proof of the risk-neutral assumption

"Pathological numerical problems? What causes the oscillations? Overshoot?"

Electric field or multiparticle interactions. They sometimes can be removed by hand-waving, sorry coarse graining procedures.

It's iff, methinks. It's obvious that if A and B commute, higher orders of BCHD are zero. To show the opposite implication, calculate the second derivative of exp(A+B)t and exp(B+A)t, which you can do in Lie group. But that's only for small |t| and bounded operators. I think unboundedness can be covered by expanding the operator into Taylor series or a similar operation... It's been 10 years ago, anyone here remembers more?

Cuchulainn
Posts: 63816
Joined: July 16th, 2004, 7:38 am
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### Re: Proof of the risk-neutral assumption

To show the opposite implication, calculate the second derivative of exp(A+B)t and exp(B+A)t, which you can do in Lie group.
This is gonna a bit time-consuming.

We want matrices A and B to commute

1. If AB = BA then exp(A+B) = exp(A)exp(B).
2. If the Golden-Thompson inequality is an equality then AB = BA (easier than a sledgehammer approach).

$tr(e^{A+B}) \le tr(e^Ae^B)$.

$tr(A) =\displaystyle \sum_{i = 1}^ n a_{ii}$
QED
My C++ Boost code gives
262537412640768743.999999999999250072597198185688879353856337336990862707537410378210647910118607313

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

Mars
Posts: 115
Joined: November 13th, 2002, 5:10 pm

### Re: Proof of the risk-neutral assumption

"easier than a sledgehammer approach" Really?
Second order term in time of exp(A+B)t exp(At) and exp(Bt) are quite easy to calculate using known Taylor expansion of exponentials, then second order polynomial multiplication (non commutative one) show that exponential equality implies that matrix A and B commute.

katastrofa
Posts: 9853
Joined: August 16th, 2007, 5:36 am
Location: Alpha Centauri

### Re: Proof of the risk-neutral assumption

To show the opposite implication, calculate the second derivative of exp(A+B)t and exp(B+A)t, which you can do in Lie group.
This is gonna a bit time-consuming.

We want matrices A and B to commute

1. If AB = BA then exp(A+B) = exp(A)exp(B).
2. If the Golden-Thompson inequality is an equality then AB = BA (easier than a sledgehammer approach).

$tr(e^{A+B}) \le tr(e^Ae^B)$.

$tr(A) =\displaystyle \sum_{i = 1}^ n a_{ii}$
QED
The $tr(e^{A+B}) \le tr(e^Ae^B)$ holds regardless of the commutativity. I have no idea what you QED-ed up there.

Cuchulainn
Posts: 63816
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: Proof of the risk-neutral assumption

To show the opposite implication, calculate the second derivative of exp(A+B)t and exp(B+A)t, which you can do in Lie group.
This is gonna a bit time-consuming.

We want matrices A and B to commute

1. If AB = BA then exp(A+B) = exp(A)exp(B).
2. If the Golden-Thompson inequality is an equality then AB = BA (easier than a sledgehammer approach).

$tr(e^{A+B}) \le tr(e^Ae^B)$.

$tr(A) =\displaystyle \sum_{i = 1}^ n a_{ii}$
QED
The $tr(e^{A+B}) \le tr(e^Ae^B)$ holds regardless of the commutativity. I have no idea what you QED-ed up there.
Read the 'equality' clause in 2.
// A and B are Hemitian, which is very restrictive.
My C++ Boost code gives
262537412640768743.999999999999250072597198185688879353856337336990862707537410378210647910118607313

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

Cuchulainn
Posts: 63816
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: Proof of the risk-neutral assumption

"easier than a sledgehammer approach" Really?
Second order term in time of exp(A+B)t exp(At) and exp(Bt) are quite easy to calculate using known Taylor expansion of exponentials, then second order polynomial multiplication (non commutative one) show that exponential equality implies that matrix A and B commute.
I agree.
But If we have two specific splitting operators A and B how can we see at a glance if AB = BA?
In general, Strang-Marchuk doesn't need commutativity, but it's nice for the 1st-order splitting schemes.
My C++ Boost code gives
262537412640768743.999999999999250072597198185688879353856337336990862707537410378210647910118607313

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