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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).

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To keep things simple let's assume you have a perfect random number generator (i.e. I will discuss only the statistics not the numerics of the problem). I will also focus on the practical matter and gloss over some mathematical details.

From a practical perspective "convergence" means that you will never get an exact answer from Monte-Carlo but increasingly good approximations. Try out your 100'000 paths example. The two values for the price of your option will be slightly different everytime you use a fresh, i.e. independent, sample.

Two mathematical theorems are relevant to describe convergence: First, the law of large numbers, which says that the average of independent samples converges to the expected value (i.e. price) and the central limit theorem, which tells you that the distribution of the error converges to a properly scaled normal distribution. This justifies what Mark Joshi is alluding to in his post.

You mention a typical and very relevant question: What size samples do I need to achieve a certain prescribed accuracy? If you assume normal distribution of errors you can calculate a confidence interval and solve this expression for the sample size. You will often hear people say that Monte-Carlo "converges very slowly" or "converges with $\sqrt{n}$". This is because to achieve a tenfold increase in accuracy you need a hundredfold increase in number of paths. For a serious study of this important topic I recommend the book by Paul Glasserman

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  • $\begingroup$ a brief explanation of what converges to what would help understand convergence better $\endgroup$
    – Saravanabalagi Ramachandran
    May 16, 2018 at 10:23
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the output of an MC simulation depends on the random numbers used and if the distribution used is not too weird, after 10,000 runs you will get an answer that is distributed $$ \mu + \frac{\sigma}{\sqrt{n}} Z, $$ with $Z$ a standard normal. Here $n=10,000.$

With $\mu$ the quantity you want and $\sigma$ the standard deviation. So you won't get precisely the same answer as you will get different values of $Z$ but if they are close together it is a good sign.

The usual approach is to estimate $\sigma$ as well as $\mu$ and then take your error magnitude to be $\sigma/\sqrt{n}.$

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  • $\begingroup$ Thanks Mark. Where does convergence come in? $\endgroup$
    – Karthik Balasubramaniam
    Apr 1, 2015 at 21:51
  • $\begingroup$ well convergence is how much the answer can change if you run more paths it's the random part that can change. The size of the error is how much convergence may be left. $\endgroup$
    – Mark Joshi
    Apr 1, 2015 at 22:08
  • $\begingroup$ @MarkJoshi, I came cross a similar issue. Does your method has a citation stats.stackexchange.com/questions/289044/…? $\endgroup$
    – TTT
    Jul 6, 2017 at 2:32
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"Monte Carlo convergence" means that you've sampled enough individuals to represent (and understand) a general population. If the probability models behind your Monte Carlo simulation are accurate, then your results will match reality as you increase your sampling size.

Monte Carlo convergence becomes difficult when you try to study a low-probability sub-population. As you sample the general population and only count those outcome that match your low-probability tally, you find convergence to be slow. Very few individuals arrive to your desire outcome. So, you sample and sample, but converge never comes.

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With Monte Carlo, which is nothing more than a numerical method to approximate a definite integral, convergence to K significant digits using N samples means that you will obtain the same K significant digits regardless of the random number sequence used for the N samples.

When you say that you obtained the same option price from two Monte Carlo runs using 100,000 samples, I am presuming that you are truncating or rounding your Monte Carlo result to cents, or possibly dollars. Using 100,000 samples for a Monte Carlo, you can have numerical method errors ranging from at least +/- 6% of the answer you get. So, if you truncate your MC answers to 10% or more precision (to deciles), then it is very possible that you could get the same answers every time you run your MC using 100,000 samples. If you truncated your MC answers to less than a 10% precision (to say percentiles), you got lucky. It will be just a fluke that you got the same answers twice in a row.

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  • $\begingroup$ Where does you 6% come from? $\endgroup$
    – lehalle
    Sep 4, 2016 at 8:53

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