ex-1: rewrite the chapters for bootstrapping and kde
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@ -75,10 +75,12 @@ $u$-transform with the `gsl_sum_levin_utrunc_accel()` function. The algorithm
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terminates when the difference between two successive extrapolations reaches a
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minimum.
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For $N = 1000$, the following results were obtained:
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\clearpage
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- $D = 0.020$
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- $p = 0.79$
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For $N = 50000$, the following results were obtained:
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- $D = 0.004$
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- $p = 0.38$
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Hence, the data was reasonably sampled from a Landau distribution.
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@ -100,8 +102,8 @@ idea which lies beneath most of them is to measure how far the parameters of
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the distribution are from the ones measured in the sample.
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The same principle can be used to verify if the generated sample effectively
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follows the Landau distribution. Since it turns out to be a very pathological
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PDF, very few parameters can be easily checked: mode, median and
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full-width-half-maximum (FWHM).
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PDF, very few parameters can be easily checked: mode, median and full width at
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half maximum (FWHM).
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### Mode
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@ -129,12 +131,12 @@ full-width-half-maximum (FWHM).
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\end{figure}
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The mode of a set of data values is defined as the value that appears most
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often, namely: it is the maximum of the PDF. Since there is no closed form
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for the mode of the Landau PDF, it was computed numerically by the *golden
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section* method (`gsl_min_fminimizer_goldensection` in GSL), applied to $-f$
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with an arbitrary error of $10^{-2}$, giving:
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often, namely: it is the maximum of the PDF. Since there is no closed form for
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the mode of the Landau PDF, it was computed numerically by the *Brent*
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algorithm (`gsl_min_fminimizer_brent` in GSL), applied to $-f$ with a relative
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tolerance of $10^{-7}$, giving:
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$$
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\text{expected mode: } m_e = \SI{-0.2227830 \pm 0.0000001}{}
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\text{expected mode: } m_e = \SI{-0.22278298 \pm 0.00000006}{}
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$$
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This is a minimization algorithm that begins with a bounded region known to
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@ -148,23 +150,36 @@ $$
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f(x_\text{min}) > f(x_e) < f(x_\text{max})
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$$
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On each iteration the interval is divided in a golden section (using the ratio
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($(3 - \sqrt{5})/2 \approx 0.3819660$) and the value of the function at this new
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point $x'$ is calculated. If the new point is a better estimate of the minimum,
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namely if $f(x') < f(x_e)$, then the current estimate of the minimum is
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updated.
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On each iteration the function is interpolated by a parabola passing though the
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points $x_\text{min}$, $x_e$, $x_\text{max}$ and the minimum is computed as the
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vertex of the parabola. If this point is found to be inside the interval it's
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taken as a guess for the true minimum; otherwise the method falls
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back to a golden section (using the ratio $(3 - \sqrt{5})/2 \approx 0.3819660$
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proven to be optimal) of the interval. The value of the function at this new
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point $x'$ is calculated. In any case if the new point is a better estimate of
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the minimum, namely if $f(x') < f(x_e)$, then the current estimate of the
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minimum is updated.
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The new point allows the size of the bounded interval to be reduced, by choosing
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the most compact set of points which satisfies the constraint $f(a) > f(x') <
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f(b)$ between $f(x_\text{min})$, $f(x_\text{min})$ and $f(x_e)$. The interval is
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reduced until it encloses the true minimum to a desired tolerance.
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The error of the result is estimated by the length of the final interval.
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In the sample, on the other hand, once the data were binned, the mode can be
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estimated as the central value of the bin with maximum events and the error
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is the half width of the bins. In this case, with 40 bins between -20 and 20,
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the following result was obtained:
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On the other hand, to compute the mode of the sample the half-sample mode (HSM)
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or *Robertson-Cryer* estimator was used. This estimator was chosen because makes
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no assumptions on the underlying distribution and is not computationally expensive.
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The HSM is obtained by iteratively identifying the half modal interval, which
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is the smallest interval containing half of the observation. Once the sample is
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reduced to less that three points the mode is computed as the average. The
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special case $n=3$ is dealt with by averaging the two closer points.
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To obtain a better estimate of the mode and its error the above procedure was
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bootstrapped. The original sample is treated as a population and used to build
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other samples, of the same size, by *sampling with replacements*. For each one
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of the new samples the above statistic is computed. By simply taking the
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mean of these statistics the following estimate was obtained
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$$
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\text{observed mode: } m_o = \SI{0 \pm 1}{}
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\text{observed mode: } m_o = \SI{-0.29 \pm 0.19}{}
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$$
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In order to compare the values $m_e$ and $x_0$, the following compatibility
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@ -179,11 +194,11 @@ where $\sigma_e$ and $\sigma_o$ are the absolute errors of $m_e$ and $m_o$
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respectively. At 95% confidence level, the values are compatible if $p > 0.05$.
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In this case:
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$$
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p = 0.82
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$$
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- t = 1.012
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- p = 0.311
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Thus, the observed value is compatible with the expected one.
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Thus, the observed mode is compatible with the mode of the Landau distribution,
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however the result is quite imprecise.
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### Median
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@ -200,9 +215,9 @@ sample size is odd, or the average of the two middle elements otherwise.
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The expected median was derived from the quantile function (QDF) of the Landau
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distribution[^1].
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Once this is know, the median is simply given by $QDF(1/2)$.
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Once this is know, the median is simply given by $\text{QDF}(1/2)$.
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Since both the CDF and QDF have no known closed form they must
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be computed numerically. The comulative probability has been computed by
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be computed numerically. The cumulative probability has been computed by
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quadrature-based numerical integration of the PDF (`gsl_integration_qagiu()`
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function in GSL). The function calculate an approximation of the integral
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@ -247,58 +262,101 @@ upper bound on the error of the root as $\varepsilon = |a-b|$. The tolerances
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here have been set to 0 and \SI{1e-3}{}.
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The result of the numerical computation is:
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$$
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\text{expected median: } m_e = \SI{1.3557804 \pm 0.0000091}{}
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$$
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while the sample median was found to be
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while the sample median, obtained again by bootstrapping, was found to be
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$$
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\text{observed median: } m_e = \SI{1.3605 \pm 0.0062}{}
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$$
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$$
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\text{observed median: } m_e = \SI{1.3479314}{}
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$$
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Applying again the t-test from before to this statistic:
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- $t=0.761$
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- $p=0.446$
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This result is much more precise than the mode and the two values show
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a good agreement.
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### FWHM
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The same approach was taken as regards the FWHM. This statistic is defined as
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the distance between the two points at which the function assumes half times
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the maximum value. Even in this case, there is not an analytic expression for
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it, thus it was computed numerically ad follow.
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First, some definitions must be given:
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For a unimodal distribution (having a single peak) this statistic is defined as
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the distance between the two points at which the PDF attains half the maximum
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value. For the Landau distribution, again, there is no analytic expression
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known, thus the FWHM was computed numerically as follows. First of all, some
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definitions must be given:
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$$
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f_{\text{max}} := f(m_e) \et \text{FWHM} = x_+ - x_- \with
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f(x_{\pm}) = \frac{f_{\text{max}}}{2}
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f_{\text{max}} = f(m_e) \et \text{FWHM} = x_+ - x_- \with
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f(x_\pm) = \frac{f_\text{max}}{2}
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$$
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then the function $f'(x)$ was minimized using the same minimization method
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used for finding $m_e$, dividing the range into $[x_\text{min}, m_e]$ and
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$[m_e, x_\text{max}]$ (where $x_\text{min}$ and $x_\text{max}$ are the limits
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in which the points have been sampled) in order to be able to find both the
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minima of the function:
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The function $f'(x)$ was minimized using the same minimization method
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used for finding $m_e$. Once $f_\text{max}$ is known, the equation
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$$
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f'(x) = \left|f(x) - \frac{f_{\text{max}}}{2}\right|
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f'(x) = \frac{f_\text{max}}{2}
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$$
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resulting in:
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is solved by performing the Brent-Dekker method (described before) in the
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ranges $[x_\text{min}, m_e]$ and $[m_e, x_\text{max}]$ yielding the two
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solutions $x_\pm$. With a relative tolerance of \SI{1e-7}{} the following
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result was obtained:
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$$
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\text{expected FWHM: } w_e = \SI{4.0186457 \pm 0.0000001}{}
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$$
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\vspace{-1em}
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On the other hand, the observed FWHM was computed as the difference between
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the center of the bins with the values closer to $\frac{f_{\text{max}}}{2}$
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and the error was taken as twice the width of the bins, obtaining:
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On the other hand, obtaining a good estimate of the FWHM from a sample is much
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more difficult. In principle it could be measured by binning the data and
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applying the definition to the discretised values, however this yields very
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poor results and depends on an completely arbitrary parameter: the bin width.
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A more refined method to construct an nonparametric empirical PDF function from
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the sample is a kernel density estimation (KDE). This method consist in
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convolving the (ordered) data with a smooth symmetrical kernel, in this cause a
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standard gaussian function. Given a sample $\{x_i\}_{i=1}^N$, the empirical PDF
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is thus constructed as
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$$
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\text{observed FWHM: } w_o = \SI{4 \pm 2}{}
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f_\varepsilon(x) = \frac{1}{N\varepsilon} \sum_{i = 1}^N
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\mathcal{N}\left(\frac{x-x_i}{\varepsilon}\right)
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$$
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This two values turn out to be compatible too, with:
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where $\varepsilon$ is called the *bandwidth* and is a parameter that controls
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the strength of the smoothing. This parameter can be determined in several
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ways: bootstrapping, cross-validation, etc. For simplicity it was chosen
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to use Silverman's rule of thumb, which gives
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$$
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p = 0.99
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\varepsilon = 0.63 S_N
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\left(\frac{d + 2}{4}N\right)^{-1/(d + 4)}
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$$
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where
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- $S_N$ is the sample standard deviation.
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- $d$ is ne number of dimensions, in this case $d=1$
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The $0.63$ factor was chosen to compensate for the distortion that
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systematically reduces the peaks height, which affects the estimation of the
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mode.
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With the empirical density estimation at hand, the FWHM can be computed by the
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same numerical method described for the true PDF. Again this was bootstrapped
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to estimate the standard error giving:
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$$
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\text{observed FWHM: } w_o = \SI{4.06 \pm 0.08}{}
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$$
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Applying the t-test to these two values gives
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- $t=0.495$
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- $p=0.620$
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which shows a very good agreement and proves the estimator is robust.
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For reference, the initial estimation based on an histogram gave a rather
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inadequate \si{4 \pm 2}.
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