ex-1: revised and typo-fixed

ex-1/main.c

@@ -94,6 +94,7 @@ int main(int argc, char** argv) {
   pdf.function = &landau_pdf;
   pdf.params = NULL;
 
+  // number of bootstrap samples
   const size_t boots = 100;
 
   double mode_e = numeric_mode(min, max, &pdf, 1);

makefile

@@ -6,11 +6,13 @@ CCOMPILE = \
 	mkdir -p $(@D); \
 	$(CC) $(CFLAGS) $^ -o $@
 
-ex-1: ex-1/bin/main ex-1/bin/pdf
+ex-1: ex-1/bin/main ex-1/bin/pdf ex-1/bin/histo
 ex-1/bin/main: ex-1/main.c ex-1/landau.c ex-1/tests.c ex-1/bootstrap.c
 	$(CCOMPILE)
 ex-1/bin/pdf: ex-1/pdf.c
 	$(CCOMPILE)
+ex-1/bin/histo: ex-1/histo.c
+	$(CCOMPILE)
 
 ex-2: ex-2/bin/fancy ex-2/bin/fancier ex-2/bin/limit ex-2/bin/naive ex-2/bin/recip
 ex-2/bin/%: ex-2/%.c

notes/sections/1.md

@@ -4,7 +4,6 @@
 
 The Landau distribution is a probability density function which can be defined
 as follows:
-
 $$
   f(x) = \int \limits_{0}^{+ \infty} dt \, e^{-t \log(t) -xt} \sin (\pi t)
 $$
@@ -19,8 +18,8 @@ For the purpose of visualizing the resulting sample, the data was put into
 an histogram and plotted with matplotlib. The result is shown in @fig:landau.
 
 ![Example of N = 10'000 points generated with the `gsl_ran_landau()`
-function and plotted in a 100-bins histogram ranging from -10 to
-80.](images/landau-hist.png){#fig:landau}
+function and plotted in a 100-bins histogram ranging from -20 to
+80.](images/landau-histo.pdf){#fig:landau}
 
 
 ## Randomness testing of the generated sample
@@ -33,7 +32,6 @@ distance between the cumulative distribution function of the Landau distribution
 and the one of the sample. The null hypothesis is that the sample was
 drawn from the reference distribution.  
 The KS statistic for a given cumulative distribution function $F(x)$ is:
-
 $$
   D_N = \text{sup}_x |F_N(x) - F(x)|
 $$
@@ -53,14 +51,12 @@ found in GSL.
 
 Under the null hypothesis, the distribution of $D_N$ is expected to
 asymptotically approach a Kolmogorov distribution:
-
 $$
   \sqrt{N}D_N \xrightarrow{N \rightarrow + \infty} K
 $$
 
-where $K$ is the Kolmogorov variable, with cumulative
-distribution function given by:
-
+where $K$ is the Kolmogorov variable, with cumulative distribution function
+given by [@marsaglia03]:
 $$
   P(K \leqslant K_0) = 1 - p = \frac{\sqrt{2 \pi}}{K_0}
   \sum_{j = 1}^{+ \infty} e^{-(2j - 1)^2 \pi^2 / 8 K_0^2}
@@ -75,8 +71,6 @@ $u$-transform with the `gsl_sum_levin_utrunc_accel()` function. The algorithm
 terminates when the difference between two successive extrapolations reaches a
 minimum.
 
-\clearpage
-
 For $N = 50000$, the following results were obtained:
 
   - $D = 0.004$
@@ -88,16 +82,16 @@ Hence, the data was reasonably sampled from a Landau distribution.
 Contrary to what one would expect, the $\chi^2$ test on a histogram is not very
 useful in this case. For the test to be significant, the data has to be binned
 such that at least several points fall in each bin. However, it can be seen
-(@fig:landau) that many bins are empty both in the right and left side of the
+in @fig:landau that many bins are empty both in the right and left side of the
 distribution, so it would be necessary to fit only the region where the points
 cluster or use very large bins in the others, making the $\chi^2$ test
 unpractical.
 
 
-## Parameters comparison
+### Parameters comparison
 
 When a sample of points is generated in a given range, different tests can be
-applied in order to check whether they follow an even distribution or not. The
+applied in order to check whether they follow a given distribution or not. The
 idea which lies beneath most of them is to measure how far the parameters of
 the distribution are from the ones measured in the sample.  
 The same principle can be used to verify if the generated sample effectively
@@ -105,9 +99,7 @@ follows the Landau distribution. Since it turns out to be a very pathological
 PDF, very few parameters can be easily checked: mode, median and full width at
 half maximum (FWHM).
 
-### Mode
-
-![Landau distribution with emphatized mode and
+![Landau distribution with emphatized mode $m_e$ and
   FWHM = ($x_+ - x_-$).](images/landau.pdf)
 
 \begin{figure}
@@ -130,6 +122,10 @@ half maximum (FWHM).
 \end{tikzpicture}}
 \end{figure}
 
+\vspace{30pt}
+
+#### Mode
+
 The mode of a set of data values is defined as the value that appears most
 often, namely: it is the maximum of the PDF. Since there is no closed form for
 the mode of the Landau PDF, it was computed numerically by the *Brent*
@@ -145,7 +141,6 @@ an upper bound $x_\text{max}$, with an estimate of the location of the minimum
 $x_e$.  The value of the function at $x_e$ must be less than the value of the
 function at the ends of the interval, in order to guarantee that a minimum is
 contained somewhere within the interval.
-
 $$
   f(x_\text{min}) > f(x_e) < f(x_\text{max})
 $$
@@ -165,26 +160,26 @@ f(b)$ between $f(x_\text{min})$, $f(x_\text{min})$ and $f(x_e)$. The interval is
 reduced until it encloses the true minimum to a desired tolerance.
 The error of the result is estimated by the length of the final interval.
 
-On the other hand, to compute the mode of the sample the half-sample mode (HSM)
+On the other hand, to compute the mode of the sample, the half-sample mode (HSM)
 or *Robertson-Cryer* estimator was used. This estimator was chosen because makes
-no assumptions on the underlying distribution and is not computationally expensive.
-The HSM is obtained by iteratively identifying the half modal interval, which
-is the smallest interval containing half of the observation. Once the sample is
-reduced to less that three points the mode is computed as the average. The
-special case $n=3$ is dealt with by averaging the two closer points.
+no assumptions on the underlying distribution and is not computationally
+expensive. The HSM is obtained by iteratively identifying the half modal
+interval, which is the smallest interval containing half of the observation.
+Once the sample is reduced to less that three points the mode is computed as the
+average. The special case $n=3$ is dealt with by averaging the two closer points
+[@robertson74].
 
 To obtain a better estimate of the mode and its error the above procedure was
-bootstrapped. The original sample is treated as a population and used to build
-other samples, of the same size, by *sampling with replacements*. For each one
-of the new samples the above statistic is computed. By simply taking the
-mean of these statistics the following estimate was obtained
+bootstrapped. The original sample was treated as a population and used to build
+100 other samples of the same size, by *sampling with replacements*. For each one
+of the new samples, the above statistic was computed. By simply taking the
+mean of these statistics the following estimate was obtained:
 $$
   \text{observed mode: } m_o = \SI{-0.29 \pm 0.19}{}
 $$
 
-In order to compare the values $m_e$ and $x_0$, the following compatibility
+In order to compare the values $m_e$ and $m_0$, the following compatibility
 t-test was applied:
-
 $$
   p = 1 - \text{erf}\left(\frac{t}{\sqrt{2}}\right)\ \with
   t = \frac{|m_e - m_o|}{\sqrt{\sigma_e^2 + \sigma_o^2}}
@@ -198,29 +193,28 @@ In this case:
   - p = 0.311
 
 Thus, the observed mode is compatible with the mode of the Landau distribution,
-however the result is quite imprecise.
+although the result is quite imprecise.
 
 
-### Median
+#### Median
 
 The median is a central tendency statistics that, unlike the mean, is not
 very sensitive to extreme values, albeit less indicative. For this reason
-is well suited as test statistic in a pathological case such as the
-Landau distribution.  
-The median of a real probability distribution is defined as the value
-such that its cumulative probability is $1/2$. In other words the median
-partitions the probability in two (connected) halves.
-The median of a sample, once sorted, is given by its middle element if the
-sample size is odd, or the average of the two middle elements otherwise.
+is well suited as test statistic in a pathological case such as the Landau
+distribution.  
+The median of a probability distribution is defined as the value such that its
+cumulative probability is $1/2$. In other words, the median partitions the
+probability in two (connected) halves. The median of a sample, once sorted, is
+given by its middle element if the sample size is odd, or the average of the two
+middle elements otherwise.
 
 The expected median was derived from the quantile function (QDF) of the Landau
 distribution[^1].
-Once this is know, the median is simply given by $\text{QDF}(1/2)$.
-Since both the CDF and QDF have no known closed form they must
-be computed numerically. The cumulative probability has been computed by
-quadrature-based numerical integration of the PDF (`gsl_integration_qagiu()`
-function in GSL). The function calculate an approximation of the integral
-
+Once this is know, the median is simply given by $\text{QDF}(1/2)$. Since both
+the CDF and QDF have no known closed form, they must be computed numerically.
+The cumulative probability was computed by quadrature-based numerical
+integration of the PDF (`gsl_integration_qagiu()` function in GSL). The function
+calculate an approximation of the integral:
 $$
   I(x) = \int_x^{+\infty} f(t)dt
 $$
@@ -229,58 +223,57 @@ $$
 because the quantile had been already implemented and was initially
 used for reverse sampling.
 
-
-The $CDF$ is then given by $p(x) = 1 - I(x)$. This was done to avoid the
+The CDF is then given by $p(x) = 1 - I(x)$. This was done to avoid the
 left tail of the distribution, where the integration can sometimes fail.
-The integral $I$ is actually mapped beforehand onto $(0, 1]$ by
-the change of variable $t = a + (1-u)/u$, because the integration
+The integral $I$ was actually mapped beforehand onto $(0, 1]$ by
+the change of variable $t = x + (1-u)/u$, because the integration
 routine works on definite integrals. The result should satisfy the following
 accuracy requirement:
 $$
   |\text{result} - I| \le \max(\varepsilon_\text{abs}, \varepsilon_\text{rel}I)
 $$
 
-The tolerances have been set to \SI{1e-10}{} and \SI{1e-6}{}, respectively.  
-As for the QDF, this was implemented by numerically inverting the CDF. This is
-done by solving the equation
+where the absolute and relative tolerances $\varepsilon_\text{abs}$ and
+$\varepsilon_\text{rel}$ were set to \SI{1e-10}{} and \SI{1e-6}{},
+respectively.  
+As for the QDF, this was implemented by numerically inverting the CDF. This was
+done by solving the equation;
 $$
   p(x) = p_0
 $$
 
-for x, given a probability value $p_0$, where $p(x)$ is again the CDF.
-The (unique) root of this equation is found by a root-finding routine
-(`gsl_root_fsolver_brent` in GSL) based on the Brent-Dekker method. This
-algorithm consists in a bisection search, similar to the one employed in the
-mode optimisation, but improved by interpolating the function with a parabola
-at each step. The following condition is checked for convergence:
+for x, given a probability value $p_0$, where $p(x)$ is the CDF. The (unique)
+root of this equation was found by a root-finding routine
+(`gsl_root_fsolver_brent` in GSL) based on the Brent-Dekker method it too.
+The following condition was checked for convergence:
 $$
   |a - b| < \varepsilon_\text{abs} + \varepsilon_\text{rel} \min(|a|, |b|)
 $$
 
 where $a,b$ are the current interval bounds. The condition immediately gives an
 upper bound on the error of the root as $\varepsilon = |a-b|$.  The tolerances
-here have been set to 0 and \SI{1e-3}{}.
-
+here were set to 0 and \SI{1e-3}{}.  
 The result of the numerical computation is:
 $$
   \text{expected median: } m_e = \SI{1.3557804 \pm 0.0000091}{}
 $$
 
-while the sample median, obtained again by bootstrapping, was found to be
+while the sample median, obtained again by bootstrapping, was found to be:
 $$
   \text{observed median: } m_e = \SI{1.3605 \pm 0.0062}{}
 $$
 
-Applying again the t-test from before to this statistic:
+As stated above, the median is less sensitive to extreme values with respect to
+the mode: this lead the result to be much more precise. Applying again the
+aforementioned t-test to this statistic:
 
   - $t=0.761$
   - $p=0.446$
 
-This result is much more precise than the mode and the two values show
-a good agreement.
+Hence, the two values show a good agreement.
 
 
-### FWHM
+#### FWHM
 
 For a unimodal distribution (having a single peak) this statistic is defined as
 the distance between the two points at which the PDF attains half the maximum
@@ -292,58 +285,55 @@ $$
   f(x_\pm) = \frac{f_\text{max}}{2}
 $$
 
-The function $f'(x)$ was minimized using the same minimization method
-used for finding $m_e$. Once $f_\text{max}$ is known, the equation
+The function derivative $f'(x)$ was minimized using the same minimization method
+used for finding $m_e$. Once $f_\text{max}$ was known, the equation:
 $$
   f'(x) = \frac{f_\text{max}}{2}
 $$
 
-is solved by performing the Brent-Dekker method (described before) in the
-ranges  $[x_\text{min}, m_e]$ and $[m_e, x_\text{max}]$ yielding the two
-solutions $x_\pm$. With a relative tolerance of \SI{1e-7}{} the following
+was solved by performing the Brent-Dekker method (described before) in the
+ranges $[x_\text{min}, m_e]$ and $[m_e, x_\text{max}]$ yielding the two
+solutions $x_\pm$. With a relative tolerance of \SI{1e-7}{}, the following
 result was obtained:
-
 $$
 \text{expected FWHM: } w_e = \SI{4.0186457 \pm 0.0000001}{}
 $$
 \vspace{-1em}
 
-![Example of a Moyal distribution density obtained by
-  the KDE method described above. The rug plot shows the original
-  sample used in the reconstruction.](images/landau-kde.pdf)
+![Example of a Moyal distribution density obtained by the KDE method. The rug
+  plot shows the original sample used in the reconstruction. The 0.6 factor
+  compensate for the otherwise peak height reduction.](images/landau-kde.pdf)
 
 On the other hand, obtaining a good estimate of the FWHM from a sample is much
-more difficult. In principle it could be measured by binning the data and
+more difficult. In principle, it could be measured by binning the data and
 applying the definition to the discretised values, however this yields very
 poor results and depends on an completely arbitrary parameter: the bin width.  
-A more refined method to construct an nonparametric empirical PDF function from
+A more refined method to construct a nonparametric empirical PDF function from
 the sample is a kernel density estimation (KDE). This method consist in
-convolving the (ordered) data with a smooth symmetrical kernel, in this cause a
-standard gaussian function. Given a sample $\{x_i\}_{i=1}^N$, the empirical PDF
-is thus constructed as
+convolving the (ordered) data with a smooth symmetrical kernel: in this case a
+standard Gaussian function. Given a sample of values $\{x_i\}_{i=1}^N$, the
+empirical PDF is defined as:
 $$
   f_\varepsilon(x) = \frac{1}{N\varepsilon} \sum_{i = 1}^N
     \mathcal{N}\left(\frac{x-x_i}{\varepsilon}\right)
 $$
 
-where $\varepsilon$ is called the *bandwidth* and is a parameter that controls
-the strength of the smoothing. This parameter can be determined in several
-ways: bootstrapping, cross-validation, etc. For simplicity it was chosen
-to use Silverman's rule of thumb, which gives
+where $\mathcal{N}$ is the kernel and the parameter $\varepsilon$, called
+*bandwidth*, controls the strength of the smoothing. This parameter can be
+determined in several ways: bootstrapping, cross-validation, etc. For
+simplicity, it was chosen to use Silverman's rule of thumb [@silverman86],
+which gives:
 $$
   \varepsilon = 0.63 S_N
     \left(\frac{d + 2}{4}N\right)^{-1/(d + 4)}
 $$
 
-where 
-
-  - $S_N$ is the sample standard deviation.
-
-  - $d$ is ne number of dimensions, in this case $d=1$
-
-The $0.63$ factor was chosen to compensate for the distortion that
+where the $0.63$ factor was chosen to compensate for the distortion that
 systematically reduces the peaks height, which affects the estimation of the
-mode.
+mode, and:
+
+  - $S_N$ is the sample standard deviation,
+  - $d$ is the number of dimensions, in this case $d=1$.
 
 With the empirical density estimation at hand, the FWHM can be computed by the
 same numerical method described for the true PDF. Again this was bootstrapped