From cc7b9ba5a49bce52933ecb8c714fbea8c806342f Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Gi=C3=B9=20Marcer?= <giuliamarcer@yahoo.it>
Date: Tue, 10 Mar 2020 22:59:25 +0100
Subject: [PATCH] ex-5: complete Miser section

---
 notes/sections/5.md | 65 ++++++++++++++++++++++++++++-----------------
 1 file changed, 41 insertions(+), 24 deletions(-)

diff --git a/notes/sections/5.md b/notes/sections/5.md
index 3396008..2d6cf47 100644
--- a/notes/sections/5.md
+++ b/notes/sections/5.md
@@ -147,47 +147,61 @@ population.
 
 The MISER technique aims to reduce the integration error through the use of
 recursive stratified sampling.  
-Consider two disjoint regions $a$ and $b$ with volumes $V_a$ and $V_b$ and Monte
-Carlo estimates $I_a = V_a \cdot \langle f \rangle_a$ and $I_b = V_b \cdot
-\langle f \rangle_b$ of the integrals, where $\langle f \rangle_a$ and $\langle
-f \rangle_b$ are the means of $f$ of the points sorted in those regions, and
-variances $\sigma_a^2$ and $\sigma_b^2$ of those points. If the weights $N_a$
-and $N_b$ of $I_a$ and $I_b$ are unitary, then the variance $\sigma_I^2$ of the
-combined estimate $I$:
-
-\textcolor{red}{QUI}
+As stated before, according to the law of large numbers, for a large number of
+extracted points, the estimation of the integral $I$ can be computed as:
 
 $$
-  I = \frac{1}{2} (I_a + I_b)
+  I= V \cdot \langle f \rangle
+$$
+
+
+Since $V$ is known (in this case, $V = 1$), it is sufficient to estimate
+$\langle f \rangle$.
+
+Consider two disjoint regions $a$ and $b$, such that $a \cup b = \Omega$, in
+which $n_a$ and $n_b$ points were uniformely sampled. Given the Monte Carlo
+estimates of the means $\langle f \rangle_a$ and $\langle f \rangle_b$ of those
+points and their variances $\sigma_a^2$ and $\sigma_b^2$, if the weights $N_a$
+and $N_b$ of $\langle f \rangle_a$ and $\langle f \rangle_b$ are chosen unitary,
+then the variance $\sigma^2$ of the combined estimate $\langle f \rangle$:
+
+$$
+  \langle f \rangle = \frac{1}{2} \left( \langle f \rangle_a
+                    + \langle f \rangle_b \right)
 $$
 
 is given by:
 
 $$
-  \sigma_I^2 = \frac{\sigma_a^2}{N_a} + \frac{\sigma_b^2}{N_b}
+  \sigma^2 = \frac{\sigma_a^2}{4n_a} + \frac{\sigma_b^2}{4n_b}
 $$
 
 It can be shown that this variance is minimized by distributing the points such
 that:
 
 $$
-  \frac{N_a}{N_a + N_b} = \frac{\sigma_a}{\sigma_a + \sigma_b}
+  \frac{n_a}{n_a + n_b} = \frac{\sigma_a}{\sigma_a + \sigma_b}
 $$
 
 Hence, the smallest error estimate is obtained by allocating sample points in
-proportion to the standard deviation of the function in each sub-region.
+proportion to the standard deviation of the function in each sub-region.  
+The whole integral estimate and its variance are therefore given by:
 
-such that $a \cup b = \Omega$
+$$
+  I = V \cdot \langle f \rangle \et \sigma_I^2 = V^2 \cdot \sigma^2
+$$
 
 When implemented, MISER is in fact a recursive method. With a given step, all
 the possible bisections are tested and the one which minimizes the combined
-variance of the two sub-regions is selected. The same procedure is then repeated
-recursively for each of the two half-spaces from the best bisection. At each
-recursion step, the integral and the error are estimated using a plain Monte
-Carlo algorithm.  
-After a given number of calls, the final individual values and their error
-estimates are then combined upwards to give an overall result and an estimate of
-its error.
+variance of the two sub-regions is selected. The variance in the sub-regions is
+estimated with a fraction of the total number of available points. The remaining
+sample points are allocated to the sub-regions using the formula for $n_a$ and
+$n_b$, once the variances are computed.  
+The same procedure is then repeated recursively for each of the two half-spaces
+from the best bisection. At each recursion step, the integral and the error are
+estimated using a plain Monte Carlo algorithm. After a given number of calls,
+the final individual values and their error estimates are then combined upwards
+to give an overall result and an estimate of its error.
 
 Results for this particular sample are shown in @tbl:MISER.
 
@@ -201,10 +215,13 @@ $\sigma$             0.0000021829      0.0000001024       0.0000000049
 diff                 0.0000032453      0.0000000858       000000000064
 -------------------------------------------------------------------------
 
-Table: MISER results with different numbers of function calls. {#tbl:MISER}
+Table: MISER results with different numbers of function calls. Be careful:
+       while in @tbl:MC the number of function calls stands for the number of
+       total sampled poins, in this case it stands for the times each section
+       is divided into subsections. {#tbl:MISER}
 
-The error, altough it lies always in the same order of magnitude of diff, seems
-to seesaw around the correct value as $N$ varies.
+This time the error, altough it lies always in the same order of magnitude of
+diff, seems to seesaw around the correct value.
 
 
 ## VEGAS \textcolor{red}{WIP}