236 lines
8.0 KiB
Markdown
236 lines
8.0 KiB
Markdown
# Exercise 4
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## Kinematic dip PDF derivation
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Consider a great number of non-interacting particles, each of which with a
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random momentum $\vec{P}$ with module between 0 and $P_{\text{max}}$ randomly
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angled with respect to a coordinate system {$\hat{x}$, $\hat{y}$, $\hat{z}$}.
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Once the polar angle $\theta$ is defined, the momentum vertical and horizontal
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components of a particle, which will be referred as $\vec{P_v}$ and $\vec{P_h}$
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respectively, are the ones shown in @fig:components.
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If $\theta$ is evenly distributed on the sphere and the same holds for the
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module $P$, which distribution will the average value of $|P_v|$ follow as a
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function of $P_h$?
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\begin{figure}
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\hypertarget{fig:components}{%
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\centering
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\begin{tikzpicture}[font=\scriptsize]
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% Axes
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\draw [thick, ->] (5,2) -- (5,8);
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\draw [thick, ->] (5,2) -- (2,1);
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\draw [thick, ->] (5,2) -- (8,1);
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\node at (1.5,0.9) {$x$};
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\node at (8.5,0.9) {$y$};
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\node at (5,8.4) {$z$};
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% Momentum
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\definecolor{cyclamen}{RGB}{146, 24, 43}
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\draw [ultra thick, ->, cyclamen] (5,2) -- (3.8,6);
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\draw [thick, dashed, cyclamen] (3.8,0.8) -- (3.8,6);
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\draw [thick, dashed, cyclamen] (5,7.2) -- (3.8,6);
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\draw [ultra thick, ->, pink] (5,2) -- (5,7.2);
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\draw [ultra thick, ->, pink] (5,2) -- (3.8,0.8);
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\node at (4.8,1.1) {$\vec{P_h}$};
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\node at (5.5,6.6) {$\vec{P_v}$};
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\node at (3.3,5.5) {$\vec{P}$};
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% Angle
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\draw [thick, cyclamen] (4.4,4) to [out=80,in=210] (5,5);
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\node at (4.7,4.2) {$\theta$};
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\end{tikzpicture}
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\caption{Momentum components.}\label{fig:components}
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}
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\end{figure}
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Since the aim is to compute $\langle |P_v| \rangle (P_h)$, the conditional
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distribution probability of $P_v$ given a fixed value of $P_h = x$ must first
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be determined. It can be computed as the ratio between the probability of
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getting a fixed value of $P_v$ given $x$ over the total probability of
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getting that $x$:
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$$
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f (P_v | P_h = x) = \frac{f_{P_h , P_v} (x, P_v)}
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{\int_{\{ P_v \}} d P_v f_{P_h , P_v} (x, P_v)}
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= \frac{f_{P_h , P_v} (x, P_v)}{I}
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$$
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where $f_{P_h , P_v}$ is the joint pdf of the two variables $P_v$ and $P_h$ and
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the integral $I$ runs over all the possible values of $P_v$ given a certain
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$P_h$.
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$f_{P_h , P_v}$ can simply be computed from the joint pdf of $\theta$ and $P$
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with a change of variables. For the pdf of $\theta$ $f_{\theta} (\theta)$, the
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same considerations done in @sec:3 lead to:
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$$
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f_{\theta} (\theta) = \frac{1}{2} \sin{\theta} \chi_{[0, \pi]} (\theta)
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$$
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whereas, being $P$ evenly distributed:
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$$
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f_P (P) = \chi_{[0, P_{\text{max}}]} (P)
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$$
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where $\chi_{[a, b]} (y)$ is the normalized characteristic function which value
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is $1/N$ between $a$ and $b$ (where $N$ is the normalization term) and 0
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elsewhere. Being a couple of independent variables, their joint pdf is simply
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given by the product of their pdfs:
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$$
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f_{\theta , P} (\theta, P) = f_{\theta} (\theta) f_P (P)
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= \frac{1}{2} \sin{\theta} \chi_{[0, \pi]} (\theta)
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\chi_{[0, P_{\text{max}}]} (P)
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$$
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Given the new variables:
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$$
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\begin{cases}
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P_v = P \cos(\theta) \\
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P_h = P \sin(\theta)
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\end{cases}
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$$
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with $\theta \in [0, \pi]$, the previous ones can be written as:
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$$
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\begin{cases}
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P = \sqrt{P_v^2 + P_h^2} \\
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\theta = \text{atan}_2 ( P_h/P_v ) :=
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\begin{cases}
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\text{atan} ( P_h/P_v ) &\incase P_v > 0 \\
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\text{atan} ( P_h/P_v ) + \pi &\incase P_v < 0
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\end{cases}
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\end{cases}
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$$
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which can be shown having Jacobian:
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$$
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J = \frac{1}{\sqrt{P_v^2 + P_h^2}}
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$$
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Hence:
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$$
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f_{P_h , P_v} (P_h, P_v) =
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\frac{1}{2} \sin[ \text{atan}_2 ( P_h/P_v )]
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\chi_{[0, \pi]} (\text{atan}_2 ( P_h/P_v )) \cdot \\
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\frac{\chi_{[0, p_{\text{max}}]} \left( \sqrt{P_v^2 + P_h^2} \right)}
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{\sqrt{P_v^2 + P_h^2}}
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$$
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from which, the integral $I$ can now be computed. The edges of the integral
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are fixed by the fact that the total momentum can not exceed $P_{\text{max}}$:
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$$
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I = \int
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\limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
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dP_v \, f_{P_h , P_v} (x, P_v)
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$$
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after a bit of maths, using the identity:
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$$
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\sin[ \text{atan}_2 ( P_h/P_v )] = \frac{P_h}{\sqrt{P_h^2 + P_v^2}}
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$$
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and the fact that both the characteristic functions play no role within the
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integral limits, the following result arises:
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$$
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I = 2 \, \text{atan} \left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)
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$$
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from which:
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$$
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f (P_v | P_h = x) = \frac{x}{P_v^2 + x^2} \cdot
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\frac{1}{2 \, \text{atan}
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\left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)}
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$$
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Finally, putting all the pieces together, the average value of $|P_v|$ can be
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computed:
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$$
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\langle |P_v| \rangle = \int
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\limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
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f (P_v | P_h = x) = [ \dots ]
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= x \, \frac{\ln \left( \frac{P_{\text{max}}}{x} \right)}
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{\text{atan} \left( \sqrt{ \frac{P^2_{\text{max}}}{x^2} - 1} \right)}
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$$ {#eq:dip}
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Namely:
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{#fig:plot}
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## Monte Carlo simulation
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The same distribution should be found by generating and binning points in a
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proper way. A number of $N = 50000$ points were generated as a couple of values
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($P$, $\theta$), with $P$ evenly distributed between 0 and $P_{\text{max}}$ and
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$\theta$ given by the same procedure described in @sec:3, namely:
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$$
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\theta = \text{acos}(1 - 2x)
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$$
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with $x$ uniformly distributed between 0 and 1.
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The data binning turned out to be a bit challenging. Once $P$ was sampled and
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$P_h$ was computed, the bin containing the latter's value must be found. If $n$
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is the number of bins in which the range $[0, P_{\text{max}}]$ is divided into,
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then the width $w$ of each bin is given by:
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$$
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w = \frac{P_{\text{max}}}{n}
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$$
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and the $i^{th}$ bin in which $P_h$ goes in is:
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$$
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i = \text{floor} \left( \frac{P_h}{w} \right)
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$$
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where 'floor' is the function which gives the bigger integer smaller than its
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argument and the bins are counted starting from zero.
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Then, a vector in which the $j^{\text{th}}$ entry contains both the sum $S_j$
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of all the $|P_v|$s relative to each $P_h$ fallen into the $j^{\text{th}}$ bin
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itself and the number num$_j$ of the bin entries was reiteratively updated. At
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the end, the average value of $|P_v|_j$ was computed as $S_j / \text{num}_j$.
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For the sake of clarity, for each sampled couple the procedure is the
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following. At first $S_j = 0 \wedge \text{num}_j = 0 \, \forall \, j$, then:
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- the couple $(P, \theta)$ is generated,
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- $P_h$ and $P_v$ are computed,
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- the $j^{\text{th}}$ bin containing $P_h$ is found,
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- num$_j$ is increased by 1,
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- $S_j$ is increased by $|P_v|$.
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For $P_{\text{max}} = 10$ and $n = 50$, the following result was obtained:
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In order to check whether the expected distribution (@eq:dip) properly matches
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the produced histogram, a chi-squared minimization was applied. Being a simple
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one-parameter fit, the $\chi^2$ was computed without a suitable GSL function
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and the error of the so obtained estimation of $P_{\text{max}}$ was given as
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the inverse of the $\chi^2$ second derivative in its minimum, according to the
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Cramér-Rao bound.
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The following results were obtained:
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$$
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P^{\text{oss}}_{\text{max}} = 10.005 \pm 0.018 \with \chi_r^2 = 0.071
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$$
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where $\chi_r^2$ is the $\chi^2$ per degree of freedom, proving a good
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convergence.
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In order to compare $P^{\text{oss}}_{\text{max}}$ with the expected value
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$P_{\text{max}} = 10$, the following compatibility t-test was applied:
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$$
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p = 1 - \text{erf}\left(\frac{t}{\sqrt{2}}\right)\ \with
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t = \frac{|P^{\text{oss}}_{\text{max}} - P_{\text{max}}|}
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{\Delta P^{\text{oss}}_{\text{max}}}
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$$
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where $\Delta P^{\text{oss}}_{\text{max}}$ is the $P^{\text{oss}}_{\text{max}}$
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uncertainty. At 95% confidence level, the values are compatible if $p > 0.05$.
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In this case:
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- t = 0.295
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- p = 0.768
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which allows to assert that the sampled points actually follow the predicted
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distribution. In @fig:fit, the fit function superimposed on the histogram is
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shown.
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{#fig:fit}
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