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