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I would be interested in the interpolation methods predominantely used in the street. imO one should always keep it as simple as possible, i.e. use the most simplistic method that does not produce any undesired effects. Luckily live becomes much easier if you assume you have a library that clearly distinguishes between discount curves (e.g. FedFund) and forward curves (e.g. USDLIB3M):

- For forward curves I would always bootstrap and then work directly on forwardLiborRates (instead of pseudo discount factors) and interpolate linearly on these rates. For me there is no reason why to use a more sophisticated method (like spline interpolation) unless my traders tell me that it is not in line with what the market does

- For pure discounting curves I'm indifferent to using linear Interpolation on DFs, log DF or zeroRates. I do not care about how weird the forwardLiborRates would look on these curves as I simply never derive any forwardLiborRates from any such curve.

I'm still a bit uncertain about the FedFund case. This curves is in fact not a pure discounting curve but both a discount as well as a forward curve. However, I will never use it to calculate forwards from while not using it to discount these forwards at the same time (so the weird looking forwards will cancel out). So I think there shouldn't be any problem with using a simple interpolation method here either.

I would be very intersted in your thoughts on that.

Thanks,

Bernd

Statistics: Posted by BerndSchmitz — Today, 4:34 pm

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Amin, have you looked at polynomial chaos expansion methods?

I do not know very much in detail about those methods. I do recall reading some papers a few years ago. I do think this method is more natural.

Statistics: Posted by Amin — March 22nd, 2017, 7:46 pm

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I will try to present a completely worked out solution for simple diffusion equation in a few days. The idea is very simple. Write the dependent function in terms of integrals that depend on its arguments and then solve the integrals by method of iterated integrals after substituting the PDE for time derivatives. Following is a brief sketch that might change a bit in the final version. I copied the equations from a previous post and some notation might be slightly off.

In p(x,t) when we start from p(x_0,t_0)

We can write

[$]\zeta (x,t)=\zeta \left(x_0,t_0\right)+\int_{t_0}^t \frac{\partial }{\partial s}\zeta (x(s),s) \, ds+\int _{t_0}^t\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}ds[$]

From integration by parts

[$]\int _{t_0}^t\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}ds=x(t)\frac{\partial }{\partial x}\zeta (x(t),t)-x\left(t_0\right)\frac{\partial }{\partial x}\zeta \left(x\left(t_0\right),t_0\right)[$]

[$]-\int _{t_0}^tx(s)\frac{\partial }{\partial x}\left[\frac{\partial }{\partial s}\zeta (x(s),s)\right]ds[$]

I really believe that all terms and relevant integrals can be solved by the method of iterated integrals to get the final solution as a two dimensional series. In order to be able to convince Cuch and Paul, I will give a completely worked out numerical example so they can verify analytics with real numbers.

When we find an integral of a function of two variables of the sort we usually find in PDEs, the right expansion for the terms inside the integral should be

[$]\zeta (x(s),s)=\zeta \left(x(t_0),s\right)+\int _{t_0}^s\frac{\partial }{\partial x}\zeta (x(u),s)\frac{\partial x}{\partial u}du[$] Equation(1)

which can be further expanded as

[$]\zeta (x(s),s)=\zeta \left(x(t_0),t_0\right)+\int_{t_0}^s \frac{\partial }{\partial u}\zeta (x(t_0),u) \, du+\int _{t_0}^s\frac{\partial }{\partial x}\zeta (x(u),s)\frac{\partial x}{\partial u}du[$] Equation (2)

We could in fact use this last equation (2) for the solution of first order odes but we preferred to use equation (1) since it was far more convenient there. Obviously equation (2) and equation (1) are equivalent. In case of PDEs, we would have to prefer to use equation (2) in most general cases. We would have to continue to further expand each integrand term in terms of higher derivatives by representing the integrand in terms of time zero value(initial time) and its derivatives just like we did in equation (1) or equation (2). we would also have to substitute actual PDE wherever we encounter time derivatives. Sorry for slightly bad notation but partial derivatives should be replaced by dx/du in equation(2).

Trying to quickly write the right version of last equation in previous post that I have quoted. From integration by parts

[$]\int _{t_0}^s\frac{\partial }{\partial x}\zeta (x(u),s)\frac{\partial x}{\partial u}du=x(s)\frac{\partial }{\partial x}\zeta (x(s),s)-x\left(t_0\right)\frac{\partial }{\partial x}\zeta \left(x\left(t_0\right),s\right)[$]

[$]-\int _{t_0}^s x(u)\frac{\partial }{\partial x}\left[\frac{\partial }{\partial u}\zeta (x(u),s)\right]du[$]

In the final expansion solution all terms will be expanded until the the terms inside the integrals are in the form of [$]\zeta (x(t_0),t_0)[$] or its x derivatives at time t0 if x has dependence on t. If x does not have a dependence on t, the solution could be easier.

Statistics: Posted by Amin — March 22nd, 2017, 10:54 am

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Thanks

Statistics: Posted by vespaGL150 — March 22nd, 2017, 8:23 am

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In p(x,t) when we start from p(x_0,t_0)

We can write

[$]\zeta (x,t)=\zeta \left(x_0,t_0\right)+\int_{t_0}^t \frac{\partial }{\partial s}\zeta (x(s),s) \, ds+\int _{t_0}^t\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}ds[$]

From integration by parts

[$]\int _{t_0}^t\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}ds=x(t)\frac{\partial }{\partial x}\zeta (x(t),t)-x\left(t_0\right)\frac{\partial }{\partial x}\zeta \left(x\left(t_0\right),t_0\right)[$]

[$]-\int _{t_0}^tx(s)\frac{\partial }{\partial x}\left[\frac{\partial }{\partial s}\zeta (x(s),s)\right]ds[$]

I really believe that all terms and relevant integrals can be solved by the method of iterated integrals to get the final solution as a two dimensional series. In order to be able to convince Cuch and Paul, I will give a completely worked out numerical example so they can verify analytics with real numbers.

Statistics: Posted by Amin — March 21st, 2017, 10:18 pm

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- even if one just focuses on major currencies, the number of conventional bucketed DV01 risk sensitivities (covering: OIS; 1M; 3M; 6M; and xccy basis risks for a given currency) is large

- in the absence of AAD* enabled technology, computing these sensitivities in the context of XVA seems time consuming and expensive

Question:

- is there any merit in applying PCA to reduce dimensionality to factor style deltas and associated hedge baskets?

(* there may also be a range of cross-gamma risks that are important to include, from what little I know on the subject I'm told AAD has no computational advantage for second order risks)

Statistics: Posted by vespaGL150 — March 21st, 2017, 4:31 am

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This is not maths. This is like what kids learn in junior cert school.

Statistics: Posted by Cuchulainn — March 20th, 2017, 8:15 pm

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https://ocw.mit.edu/courses/mathematics ... /calendar/

But i would do this first

https://ocw.mit.edu/courses/mathematics ... -lectures/

And do the exercises/exam. The only way to walk the talk!

Statistics: Posted by Cuchulainn — March 19th, 2017, 9:33 pm

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