From 8384889cf92459ae26eeeac70bde46f7cffa79d7 Mon Sep 17 00:00:00 2001
From: casellimarco7 <marco.caselli@mindfoundry.ai>
Date: Fri, 28 Apr 2023 17:03:28 +0200
Subject: [PATCH] Fix coefficient example Montecarlo

---
 montecarlo.Rmd | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/montecarlo.Rmd b/montecarlo.Rmd
index 0e2dc23..e13e6fe 100644
--- a/montecarlo.Rmd
+++ b/montecarlo.Rmd
@@ -80,7 +80,7 @@ The central limit theorem can be used to construct confidence intervals for our
       (\#eq:cltconfidence)
 \end{align}
 where $Z$ is a standard normal random variable.
-We can find values of $N$ such that the probability to have a good estimate $\widetilde{\mu}_N$ up to fixed additive error $\epsilon$ is almost 1. Usual values for this probability are $95\%$ or $99\%$. So for example if we want $P(-z_c\le Z \le z_c) = 0.99$ where $z_c = \epsilon\sqrt{N}/\sigma$ then $z_c =\epsilon\sqrt{N}/\sigma= 2.58$ because of the properties of the normal distribution. Estimating $\mu$ with additive error $\epsilon$ would require $N = 2.58\, \sigma^2/\epsilon^2$ samples.
+We can find values of $N$ such that the probability to have a good estimate $\widetilde{\mu}_N$ up to fixed additive error $\epsilon$ is almost 1. Usual values for this probability are $95\%$ or $99\%$. So for example if we want $P(-z_c\le Z \le z_c) = 0.99$ where $z_c = \epsilon\sqrt{N}/\sigma$ then $z_c =\epsilon\sqrt{N}/\sigma= 2.58$ because of the properties of the normal distribution. Estimating $\mu$ with additive error $\epsilon$ would require $N = 6.66\, \sigma^2/\epsilon^2$ samples.
 
 You can feel the power of Monte Carlo!! All of the above does not depend on the dimension of the sample space of the random variables $X_i$, but just on the number of repetitions $N$ and on the variance $\sigma^2$.