I have read about convergence in terms of MC simulation for derivative pricing, but I am not clear on what it exactly means. Let us suppose I price an option 100,000 paths twice and both result in the same option price. Does that mean 100,000 paths has resulted in convergence?Also, in determining the number of paths to use for pricing, is getting the same option price with 2 different runs a factor? (Assumption is I am not reseeding the Random Number. So the sequence of Random Numbers between the two runs is different).

balasukWell yes and no. Typically mean that "convergence" is a model price has reached some stable proximity to a target price. For example, choosing parameter values for Heston results in a Heston option price that is close enough to a Black-Scholes price to make you happy. Then in step two you use these parameters to run a Heston MC simulation which in turn then prices some exotic option. Also if you want to choose an optimal number of MC paths then I would try to keep the random seed the same for each test.HTH

Thank you Orbit. How does keeping the random number seed the same for each test help in determining the optimal number of paths? In production, the model will run with a different seed every time. To ensure that I get a consitent price , wouldn't I want to confirm that even with different random numbers I get the same price for the number of paths chosen?

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There was a discussion on pricing a call with r = 0, sig = 1, sodS = dWHow many time steps NT do you need to get any kind of decent 'convergence' with Euler (hint: NT >> 10^15).

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QuoteOriginally posted by: outrunHuh? Euler should perform good in that case.Perhaps it wasdS = SdW?I am sure he meant to say SdW. Otherwise, exactly one step is required. To expand a little on the earlier question, if you run a basic MC simulation, the standard approach is to simulate the square of the payoff along with the actual payoff and then estimate the simulation error from the full sample standard deviation. Unfortunately, this no longer works reliably as soon as you start modifying your paths in order to speed up convergence. If runtime speed is an issue at all I don't think you will ever want to run a naive MC. There are hundreds of papers and dozens of books on how to approach this. Paul Glasserman's book would be a good place to start.

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QuoteOriginally posted by: outrunHuh? Euler should perform good in that case.Perhaps it wasdS = SdW?Yes, sorry. You are correct.There was a thread on this topic (I can't find it any more, maybe @Alan knows) but Euler needed ~ 10^20 steps (At least ).Maybe Richardson extrapolation Euler is better but in general fd schemes for SDE are unpredictable and rather disappointing. The lack of monotonicity is an issue. Decreasing dt and/or increasing NSIM means nothing.edit: thanks, here// Engineering uses Karhunen-Loeve to solve SDEs. Not in finance?

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QuoteDecreasing dt and/or increasing NSIM means nothing.We do, in fact, have some theory for this; see "Efficient Monte Carlo Simulation of Security Prices" by Darrell Duffie and Peter Glynn.PDF: http://www.darrellduffie.com/uploads/pu ... pdfQuoteWe can also view the result in terms of the asymptotic relation between the number k of Monte Carlo simulations and the size h of a time interval.The theorem implies that, asymptotically, it is optimal to have k increasing at the order of h^(-2p).For instance, with the Euler scheme (p = 1), the number of simulations should quadruple with each doubling of the number of time intervals.With a second-order scheme, the number of simulations should be on the order of the number of time intervals to the fourth power and so on.Similarly, with an optimal Euler scheme, asymptotically speaking, for each doubling of the number of time intervals the root-mean-squared estimation error is halved.For an asymptotically optimal second-order scheme such as the Milshtein (1978) scheme (for N = 1) or Talay (1984, 1986) scheme (for N >= 1), for each doubling of the number of time intervals, the root-mean-squared estimation error is reduced by a factor of 4.The option pricing example shown in the next section has estimation errors for ﬁnite samples that are consistent with this asymptotic error behavior. Numerical examples given by Kloeden and Platen (1992) are also consistent with this predicted behavior.There's a but though: In general, the optimal constant of proportionality is not known. That being said, in some cases it can be found experimentally, AFAIR there's also been some research with an analytic tretment.

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I supervised an MSc student from Birmingham last year. It's sharper than normal MC, i.e. multigrid approach.MLMC Nick Webber in his Wiley VBA book on MC has interesting conclusions on MC convergence..

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As Paul once said "Monte Carlo is a blunt instrument".

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QuoteWhen you discretize a non-linear SDE the step distributions won't be normal/closed form in general.I don't understand this statement. So, if the SDE is constant or linear in the volatility term then you don't have that problem?Of course, it is possible to transforming a NL SDE using the Nelson-Ramaswamy method to get an SDE with constant volatility term and then use a constant mesh size.

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