Chap.8, pb 4

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Chap.8, pb 4

Postby akbar » Tue Oct 05, 2010 10:01 am

This problem deals with binary tree construction given time-dependent volatility.
The issue here is to determine the vector of times so that we get the same fraction of total volatility at each time step. Given a general volatility parameter, the obvious approach seems to use a solver to find the time t such that the integral of variance from the previous time step to t matches a given volatility value.
Using for example a simple dichotomy solution, this slows down considerably the contruction of the BinaryTree object, at least for the vector of discounts. This even if it is done only once (for each set number of steps).
But the use of solvers is only detailled in the following chapter, not a this point. Is there any other valid approach to that problem?
By advance, thanks for your answer.
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Re: Chap.8, pb 4

Postby mj » Wed Oct 06, 2010 9:09 am

That is the standard solution.
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Re: Chap.8, pb 4

Postby chrissbliss » Sun Feb 06, 2011 11:53 am

So this is a teaser for the next chapter; we're supposed to come up with a non-optimal solution now and then see how much smoother it can be taken care of by using the techniques in the following chapter? This seems to be the general theme of the book; I quite like it! :)
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Re: Chap.8, pb 4

Postby emza0114 » Wed Mar 06, 2013 11:49 pm

Hi,

so I have managed to partially solve this problem. I have essentially determined the
appropriate time steps as described in the original problem.

The issue Im facing now is how to determine the Stock price. The \sigma^2 term in the exponent is easy we can just replace this by the integral of the square. However what does one do with discretized brownian motion term? is this replaced by the integral of \int_0^t \sigma/T?

This seems to work if I work with the Jarrod-Rudd parametrization but not with a martingale measure...

thanks

Mark
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Re: Chap.8, pb 4

Postby mj » Thu Mar 07, 2013 10:41 pm

you have to abandon the underlying original Brownian motion. Think of replacing the vol for the step with a constant vol that makes the integral come out right.
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