Update: Turn BS PDE into a parabolic PDE system and _then_ discretis (in contrast to discretising BS PDE and then trying to recover the Greeks by jumping through hoops and messing).

QED

- Cuchulainn
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Update: Turn BS PDE into a parabolic PDE system and _then_ discretis (in contrast to discretising BS PDE and then trying to recover the Greeks by jumping through hoops and messing).

QED

QED

Last edited by Cuchulainn on April 19th, 2019, 1:17 pm, edited 2 times in total.

Step over the gap, not into it. Watch the space between platform and train.

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- Cuchulainn
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Stressed data

Step over the gap, not into it. Watch the space between platform and train.

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Actually, in finance we write x = log S, S = exp(x) and use them in calculations as if there is no relationship between them. They look like constants.

BUT they are functions!

S = f(x) == exp(x)

x = g(S) == log(S)

Transformations are functions, not numbers.

BUT they are functions!

S = f(x) == exp(x)

x = g(S) == log(S)

Transformations are functions, not numbers.

Step over the gap, not into it. Watch the space between platform and train.

http://www.datasimfinancial.com

http://www.datasim.nl

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http://www.datasim.nl

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Paul (and maybe others as well!) knows the answer to this one:

1. Consider a continuous Asian PDE[$](S,A)[$]

2. Take {Float, Fixed}{Put, Call} Strike payoff

3. Arithmetic average [$]A[$] defined on [$](0,1)[$] after scaling,

Hypothesis 1 is that any numerical BC on [$]A=0[$] and [$]A=1[$] have no bearing on the solution in the interior of the domain (e.g. Anchor PDE as we discussed), i.e. no source information coming from the [$]A[$] boundaries.

And the same conclusions should hold for Cheyette model (Hypothesis 2).

Justify this. Or disprove.

1. Consider a continuous Asian PDE[$](S,A)[$]

2. Take {Float, Fixed}{Put, Call} Strike payoff

3. Arithmetic average [$]A[$] defined on [$](0,1)[$] after scaling,

Hypothesis 1 is that any numerical BC on [$]A=0[$] and [$]A=1[$] have no bearing on the solution in the interior of the domain (e.g. Anchor PDE as we discussed), i.e. no source information coming from the [$]A[$] boundaries.

And the same conclusions should hold for Cheyette model (Hypothesis 2).

Justify this. Or disprove.

Last edited by Cuchulainn on January 27th, 2020, 2:56 pm, edited 1 time in total.

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Here's another one regarding elliptic PDEs that can be transformed to *canonical form* (aka get rid of them pesky mixed derivatives that people so unhappy but still an area of active experimentation).

https://en.wikipedia.org/wiki/Elliptic_partial_differential_equation

Some remarks/questions:

1. It works and is easy once you get the hang of it.

2. What would be the rationale for using it?

3. What would be/are the reasons for not using it?

4. Has anyone published the method for 2-factor pdes with mixed derivatives like Heston, BS, CB?

// There are about 20 million ways to approximate [$]\frac{\partial^2 V}{\partial x \partial y} [$]

https://en.wikipedia.org/wiki/Elliptic_partial_differential_equation

Some remarks/questions:

1. It works and is easy once you get the hang of it.

2. What would be the rationale for using it?

3. What would be/are the reasons for not using it?

4. Has anyone published the method for 2-factor pdes with mixed derivatives like Heston, BS, CB?

// There are about 20 million ways to approximate [$]\frac{\partial^2 V}{\partial x \partial y} [$]

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Some guy on LI claims for integer [$]n[$]

[$]\frac {dn!}{dn} = n[$]

Vote

1. True

2. False

[$]\frac {dn!}{dn} = n[$]

Vote

1. True

2. False

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That works.Via gamma fn, looks messy!

Since [$]\frac{d}{dz} \log \Gamma[z] = \psi(z)[$], the Digamma function (A&S, 6.3.1), then

[$] \frac{dn!}{dn} = \psi(n+1) \, n![$], where of course [$]n![$] is interpreted everywhere as [$]\Gamma(n+1)[$].

I think it's just recognizing a "named function".

So, take n=6

In Mathematica, what I posted evaluates to [$]1764 - 720 \, \gamma \approx 1348.4[$], using Euler's constant [$]\gamma[$].

Since [$]1348.4 \not= 6[$] that answer's Daniel's question. If it was truly circular, I don't think one could get a numerical result.

So, take n=6

In Mathematica, what I posted evaluates to [$]1764 - 720 \, \gamma \approx 1348.4[$], using Euler's constant [$]\gamma[$].

Since [$]1348.4 \not= 6[$] that answer's Daniel's question. If it was truly circular, I don't think one could get a numerical result.

The named function just happens to be the answer!

Exam question 1: Find the solution of ...

Answer: Define [$]\mbox{Lewis}_1[$] as the answer to question 1. Trivially and wlog the answer to question 1 is [$]\mbox{Lewis}_1[$].

Exam question 2: Etc.

!!!

Exam question 1: Find the solution of ...

Answer: Define [$]\mbox{Lewis}_1[$] as the answer to question 1. Trivially and wlog the answer to question 1 is [$]\mbox{Lewis}_1[$].

Exam question 2: Etc.

!!!

But we do now know that the answer to Cuch's question is no!

Well, I'm glad the thread title is called "Silly questions". That way, I feel free to belabor the issue.The named function just happens to be the answer!

Exam question 1: Find the solution of ...

Answer: Define [$]\mbox{Lewis}_1[$] as the answer to question 1. Trivially and wlog the answer to question 1 is [$]\mbox{Lewis}_1[$].

Exam question 2: Etc.

!!!

The more careful analogy to the question would be:

Q1. Some guy claims that "function expression blah blah (x)" = 3 x. True or False?

Answer: "function expression blah blah (x)" is actually a named special function, called the Lewis function. Like all special functions, it's well-studied. It is known that Lewis(x) does not equal 3 x. In fact, at x=3, 3 x = 9, and Lewis(3) = 1036, which I get from Mathematica, which has it built-in.

Last edited by Alan on February 13th, 2020, 5:48 am, edited 1 time in total.

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!

I demand a recount! Anyway, it's not even 6am -- don't you people sleep?

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