# Monte Carlo convergence sample size

I'm studying Monte Carlo analysis but I find very counter-intuitive the computation of the minimum sample size in order to reach a certain level of precision. As stated in Montecarlo methods in Financial Engineering, such sample size comes from the CLT and it can be computed according to

$n = \frac{z^2_{\delta/2} \sigma^2}{\epsilon^2}$

My experiment is the classic bernoulli variable, with $p$ probability of success. Suppose my level of precision is $0.01$, then my $z_{0.01/2} = 2.57$.

I consider three cases.

$p=0.1\%$

$p=1\%$

$p=10\%$

With $\sigma^2 = p(1-p)$, I get

$n_{10\%} = 2.57^2 \cdot 0.1 \cdot 0.9 / 0.01^2 = 5944$

$n_{1\%} = 2.57^2 \cdot 0.01 \cdot 0.99 / 0.01^2 = 654$

$n_{0.1\%} = 2.57^2 \cdot 0.001 \cdot 0.999 / 0.01^2 = 66$

Intuitevely I would think the rarer the event, the greater the number of the simulations needed in order to approach the theoretical mean. What am I getting wrong?

• You are holding the level of precision fixed, so it is easier to approximate whatever distribution has the smallest variance. The rarer the event, the smaller the absolute variance. – dm63 Sep 30 '17 at 14:19
• So, in order to scale the precision parameter to get always a 1% precision around the true value, should I put a) 1% of 10% = 0.001 b) 1% of 1% = 0.0001 c) 1% of 0.1% = 0.00001 in the denominator? – mg91 Oct 1 '17 at 14:30