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I want to figure out the following question

$ 1 = (10 - 9)^{100} = 10^{100}-100 \times 10^{99} \ 9 + \frac{100 \times 99}{2} 10^{98} \ 9^{2} - \frac{100 \times 99 \times 98 }{3}10^{97} 9^{3} \pm...$

is there any suggestion how to sample this expansion using Monte Carlo.

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    In fact, this is the question I want to figure out. The problem stated as above.2012-10-25

1 Answers 1

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Are you trying to sample the individual terms in the series? In other words, each term in the expansion looks like $(-1)^k\binom{100}{k}(10)^{n-k}(9)^k$ where you can temporarly drop the $(-1)^k$ to see that sampling a binomial distribution with $n=100$ and probability coefficient $p=10/19$ will give you want you want:

$P(B(n,p)=k)=\binom{100}{k}p^k(1-p)^{n-k}$

so

$(-1)^k\binom{100}{k}(10)^{n-k}(9)^k=(-1)^k\cdot 19^n\cdot P(B(n,p)=k)$

In other words, flip $n=100$ biased coins of probability $p=10/19$ and then count the number of heads that appear. Repeat over and over to get an approximation of the binomial distribution.

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    @Eric: I'm not sure I understand. By sampling the above distribution a ton of times, you'll get close approximations to the above probabilities. You then need to multiply them by $19^n$ and then by $(-1)^k$ to adjust for the minus sign.2012-10-26