ex-3: revised and typo-fixed
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ex-3/main.c
12
ex-3/main.c
@ -85,12 +85,16 @@ int main(int argc, char **argv) {
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fputs("# event sampling\n\n", stderr);
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fprintf(stderr, "generating %ld events...", opts.num_events);
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struct event *e;
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// for (size_t i = 0; i < s.size; i++){
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// e = &s.events[i];
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// do {
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// e->th = acos(1 - 2*gsl_rng_uniform(r));
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// e->ph = 2 * M_PI * gsl_rng_uniform(r);
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// } while(0.2 * gsl_rng_uniform(r) > distr(def_par, e));
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for (size_t i = 0; i < s.size; i++){
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e = &s.events[i];
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do {
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e->th = acos(1 - 2*gsl_rng_uniform(r));
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e->ph = 2 * M_PI * gsl_rng_uniform(r);
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} while(0.2 * gsl_rng_uniform(r) > distr(def_par, e));
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e->th = acos(1 - 2*gsl_rng_uniform(r));
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e->ph = 2 * M_PI * gsl_rng_uniform(r);
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// update the histogram
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gsl_histogram2d_increment(hist, e->th, e->ph);
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@ -9,7 +9,8 @@ bins = tuple(map(int, sys.stdin.readline().split()))
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xedges, yedges, counts = loadtxt(sys.stdin, unpack=True, usecols=[0,2,4])
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counts = counts.reshape(bins)
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plt.rcParams['font.size'] = 15
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plt.rcParams['font.size'] = 30
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tight_layout()
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suptitle('Angular decay distribution')
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#subplot2grid((1, 3), (0, 0), colspan=2, aspect='equal')
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@ -27,4 +28,6 @@ norm = colorbar(fraction=0.023, pad=0.04).norm
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# x, y, z, rstride=1, cstride=1,
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# facecolors=cm.viridis(norm(counts)))
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#axis('off')
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show()
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@ -89,3 +89,20 @@
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year={2008},
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publisher={Citeseer}
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}
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@misc{painless94,
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title={An introduction to the conjugate gradient method without the
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agonizing pain},
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author={Shewchuk, Jonathan Richard and others},
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year={1994},
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pages={42},
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publisher={Carnegie-Mellon University. Department of Computer Science}
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}
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@article{Lou05,
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title={A brief description of the levenberg-marquardt algorithm
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implemened by levmar},
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author={M. I. A. Lourakis},
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year={2005},
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journal={Matrix}
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}
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@ -116,7 +116,7 @@ approx: 0.57225\ 72410\ 34058
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diff: 0.00495\ 84238\ 67473
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--------- -----------------------
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Table: Best esitimation of $\gamma$ using
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Table: Best estimation of $\gamma$ using
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the alternative formula. {#tbl:second}
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Here, the problem lies in the binomial term: computing the factorial of a
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@ -4,37 +4,34 @@
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A number of $N = 50'000$ points on the unit sphere, each representing a
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particle detection event, must be generated according to then angular
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probability distribution function $F_0$:
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probability distribution function $F$:
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\begin{align*}
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F_0 (\theta, \phi) = \frac{3}{4 \pi} \Bigg[
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\frac{1}{2} (1 - \alpha_0) + \frac{1}{2} (3\alpha_0 - 1) \cos^2(\theta)& \\
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- \beta_0 \sin^2(\theta) \cos(2 \phi)
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- \gamma_0 \sqrt{2} \sin(2 \theta) \cos(\phi)& \Bigg]
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F (\theta, \phi) = &\frac{3}{4 \pi} \Bigg[
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\frac{1}{2} (1 - \alpha) + \frac{1}{2} (3\alpha - 1) \cos^2(\theta) \\
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&- \beta \sin^2(\theta) \cos(2 \phi)
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- \gamma \sqrt{2} \sin(2 \theta) \cos(\phi) \Bigg]
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\end{align*}
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where $\theta$ and $\phi$ are, respectively, the polar and azimuthal angles, and
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$$
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\alpha_0 = 0.65 \et \beta_0 = 0.06 \et \gamma_0 = -0.18
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$$
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To generate the points a *try and catch* method was employed:
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1. generate a point $(\theta , \phi)$ uniformly on a unit sphere
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2. generate a third value $Y$ uniformly in $[0, 1]$
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3. if @eq:requirement is satisfied save the point
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1. generate a point $(\theta , \phi)$ uniformly on a unit sphere,
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2. generate a third value $Y$ uniformly in $[0, 1]$,
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3. if @eq:requirement is satisfied save the point,
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4. repeat from 1.
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$$
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Y < F_0(\theta, \phi)
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Y < F(\theta, \phi)
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$$ {#eq:requirement}
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To increase the efficiency of the procedure, the $Y$ values were actually
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generated in $[0, 0.2]$, since $\max F_0 = 0.179$ for the given parameters
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(an other option would have been generating the numbers in the range $[0 ; 1]$,
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but all the points with $Y_{\theta, \phi} > 0.179$ would have been rejected with
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the same probability, namely 1).
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generated in $[0, 0.2]$, since $\max F = 0.179$ for the given parameters
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(by generating the numbers in the range $[0 ; 1]$, all the points with
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$Y_{\theta, \phi} > 0.179$ would have been rejected with the same
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probability, namely 1).
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While $\phi$ can simply be generated uniformly between 0 and $2 \pi$, for
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$\theta$ one has to be more careful: by generating evenly distributed numbers
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@ -48,21 +45,20 @@ easily generated by the GSL function `gsl_rng_uniform()`.
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<div id="fig:compare">
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{width=45%}
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0 and $\pi$.](images/histo-i-u.pdf){width=50%}
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{width=45%}
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surface.](images/histo-p-u.pdf){width=50%}
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{width=45%}
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0 and $\pi$.](images/histo-i-F.pdf){width=50%}
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{width=45%}
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distributed.](images/histo-p-F.pdf){width=50%}
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Examples of samples. On the left, points with $\theta$ evenly distributed
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between 0 and $\pi$; on the right, points with $\theta$ properly distributed.
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</div>
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The transformation can be found by imposing the angular PDF to be a constant:
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\begin{align*}
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\frac{d^2 P}{d \Omega^2} = \text{const} = \frac{1}{4 \pi}
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\hspace{20pt} &\Longrightarrow \hspace{20pt}
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@ -86,7 +82,6 @@ The transformation can be found by imposing the angular PDF to be a constant:
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If $\theta$ is chosen to grew together with $x$, then the absolute value can be
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omitted:
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\begin{align*}
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\frac{d \theta}{dx} = \frac{2}{\sin(\theta)}
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\hspace{20pt} &\Longrightarrow \hspace{20pt}
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@ -108,10 +103,11 @@ a single point, the effect of this omission is negligible.
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\vspace{30pt}
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## Parameter estimation
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## Parameters estimation
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The sample must now be used to estimate the parameters $\alpha_0$,
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$\beta_0$ and $\gamma_0$ of the angular distribution $F_0$.
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The sample must now be used to estimate the parameters $\alpha$, $\beta$ and
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$\gamma$ of the angular distribution $F$. The correct set will be referred to
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as {$\alpha_0$, $\beta_0$, $\gamma_0$}.
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### Maximum Likelihood method {#sec:MLM}
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@ -119,9 +115,7 @@ $\beta_0$ and $\gamma_0$ of the angular distribution $F_0$.
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The Maximum Likelihood method (MLM) is based on the observation that the best
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estimate {$\bar{\alpha}$, $\bar{\beta}$, $\bar{\gamma}$} of the parameters
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{$\alpha$, $\beta$, $\gamma$} should maximize the Likelihood function $L$
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defined in @eq:Like0, where the index $i$ runs over the sample and $F$ is the
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function $F_0$ with free parameters $\alpha$, $\beta$ and $\gamma$:
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defined in @eq:Like0, where the index $i$ runs over the sample:
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$$
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L = \prod_i F(\theta_i, \phi_i) = \prod_i F_i
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$$ {#eq:Like0}
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@ -134,9 +128,8 @@ $\beta$, $\gamma$} = {$\alpha_0$, $\beta_0$, $\gamma_0$}. Thus, by viewing $F_i$
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as a function of the parameters, by maximizing $P = L$ with respect
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to $\alpha$, $\beta$ and $\gamma$, a good estimate should be found.
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Instead of actually maximising $L$, the function $-\ln(L)$
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was minimised, as minimisation methods as usually described in literature
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and $\ln$ since it simplifies the math:
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was minimised, as minimisation methods are usually described in literature
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and the logarithm simplifies the math:
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$$
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L = \prod_i F_i \thus - \ln(L) = - \sum_{i} \ln(F_i)
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$$ {#eq:Like}
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@ -148,36 +141,36 @@ conjugate gradient Fletcher-Reeves algorithm, used in the solution, which requir
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the implementation of $-\ln(L)$ and its gradient.
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To minimise a function $f$, given an initial guess point $x_0$, the gradient
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$\nabla f$ is used to find the initial direction $v_0$ (the steepest descent) along
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which a line minimisation is performed: the minimum occurs where the
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directional derivative along $v_0$ is zero:
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$\nabla f$ is used to find the initial direction $v_0$ (the steepest descent)
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along which a line minimisation is performed: the minimum $x_1$ occurs where
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the directional derivative along $v_0$ is zero:
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$$
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\frac{\partial f}{\partial v_0} = \nabla f \cdot v_0 = 0
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$$
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or, equivalently, at the point where the gradient is orthogonal to $v_0$.
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After this first step the following procedure is iterated:
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After this first step, the following procedure is iterated:
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1. Find the steepest descent $v_n$.
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2. Compute the *conjugate* direction: $w_n = v_n + \beta_n w_{n-1}$
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3. Perform a line minimisation along $w_n$
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4. Update the position $x_{n+1} = x_n + \alpha_n w_n$
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1. find the steepest descent $v_n$ at $x_n$,
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2. compute the *conjugate* direction: $w_n = v_n + \beta_n w_{n-1}$,
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3. perform a line minimisation along $w_n$ and update the minimum $x_{n+1}$,
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where $\alpha_n$ is line minimum and $\beta_n$ is given by the Fletcher-Reeves
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formula:
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Different formulas for $\beta_n$ have been developed, all equivalent for a
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quadratic function, but not for nonlinear optimization: in such instances, the
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best formula is a matter of heuristics [@painless94]. In this case, $\beta_n$ is
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given by the Fletcher-Reeves formula:
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$$
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\beta = \frac{\|\nabla f(x_n)\|^2}{\|\nabla f(x_{n-1})\|^2}
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\beta_n = \frac{\|\nabla f(x_n)\|^2}{\|\nabla f(x_{n-1})\|^2}
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$$
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The accuracy of each line minimisation is controlled the parameter `tol`,
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The accuracy of each line minimisation is controlled with the parameter `tol`,
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meaning:
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$$
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w \cdot \nabla f < \text{tol} \, \|w\| \, \|\nabla f\|
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$$
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The minimisation is quite unstable and this forced the starting point to be
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taken very close to the solution, namely:
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$$
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\alpha_{sp} = 0.79 \et
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\beta_{sp} = 0.02 \et
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@ -187,29 +180,28 @@ $$
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The overall minimisation ends when the gradient module is smaller than $10^{-4}$.
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The program took 25 of the above iterations to reach the result shown in @eq:Like_res.
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The Cramér-Rao bound states that the covariance matrix of parameters estimated
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by MLM is greater than the the inverse of the Hessian matrix of $-\log(L)$
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at the minimum. Thus, the Hessian matrix was computed
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analytically and inverted by a Cholesky decomposition, which states that
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every positive definite symmetric square matrix $H$ can be written as the
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product of a lower triangular matrix $L$ and its transpose $L^T$:
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As regards the error estimation, the Cramér-Rao bound states that the covariance
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matrix of parameters estimated by MLM is greater than the inverse of the
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Hessian matrix of $-\log(L)$ at the minimum. Thus, the Hessian matrix was
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computed analytically and inverted by a Cholesky decomposition, which states
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that every positive definite symmetric square matrix $H$ can be written as the
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product of a lower triangular matrix $\Delta$ and its transpose $\Delta^T$:
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$$
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H = L \cdot L^T
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H = \Delta \cdot \Delta^T
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$$
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The Hessian matrix is, indeed, both symmetric and positive definite, when
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computed in a minimum. Once decomposed the inverse is given by
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computed in a minimum. Once decomposed, the inverse is given by
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$$
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H^{-1} = (\Delta \cdot \Delta^T)^{-1} = (\Delta^{-1})^T \cdot \Delta^{-1}
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$$
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$$
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H^{-1} = (L \cdot L^T)^{-1} = (L^{-1})^T \cdot L^{-1}
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$$
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The inverse of $L$ is easily computed since it's a triangular matrix.
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where the inverse of $\Delta$ is easily computed, since it's a triangular
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matrix.
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The GSL library provides two functions `gsl_linalg_cholesky_decomp()` and
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`gsl_linalg_cholesky_invert()` to decompose and invert in-place a matrix.
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The result is shown below:
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$$
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\bar{\alpha_L} = 0.6541 \et
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\bar{\beta_L} = 0.06125 \et
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@ -231,22 +223,24 @@ Hence:
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\gamma_L &= -0.1763 \pm 0.0016
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\end{align*}
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See @sec:res_comp for results compatibility.
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### Least squares estimation
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Another method that can be used to estimate {$\alpha$, $\beta$, $\gamma$} are
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the least squares (LSQ), which is a multidimensional minimisation
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specialised to the case of sum of squared functions called residuals.
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The residuals can be anything in general but are usually interpreted as the
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In general, the residuals can be anything but are usually interpreted as the
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difference between an expected (fit) and an observed value. The least squares
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then correspond to the expected values that best fit the observations.
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To apply the LSQ method, the data must be properly binned, meaning that each
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bin should contain a significant number of events (say greater than four or
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five): in this case, 30 bins for $\theta$ and 60 for $\phi$ turned out
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to be satisfactory. The expected values were given as the product of $N$, $F$
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computed in the geometric center of the bin and the solid angle enclosed by the
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bin itself, namely:
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to be satisfactory. The expected values $E_{\theta, \phi}$ were given as the
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product of $N$, $F$ computed in the geometric center of the bin and the solid
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angle enclosed by the bin itself, namely:
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$$
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E_{\theta, \phi} = N F(\theta, \phi) \, \Delta \theta \, \Delta \phi \,
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@ -254,11 +248,8 @@ $$
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$$
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Once the data are binned, the number of events in each bin plays the role of the
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observed value, while the expected one is given by the probability of finding
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an event in that bin multiplied by the total number of points, $N$. Then, the
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$\chi^2$, defined as follow, must be minimized with respect to the three
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parameters to be estimated:
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observed value. Then, the $\chi^2$, defined as follow, must be minimized with
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respect to the three parameters to be estimated:
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$$
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\chi^2 = \sum_i f_i^2 = \|\vec f\|^2 \with f_i = \frac{O_i - E_i}{\sqrt{E_i}}
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$$
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@ -270,12 +261,11 @@ trust region method was used to minimize the $\chi^2$.
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In such methods, the $\chi^2$, its gradient and its Hessian matrix are computed
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in a series of points in the parameters space till the gradient and the matrix
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reach given proper values, meaning that the minimum has been well approached,
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where "well" is related to that values.
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where "well" is related to those values.
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If {$x_1 \dots x_p$} are the parameters which must be estimated, first $\chi^2$
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is computed in the starting point $\vec{x_k}$, then its value in a point
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$\vec{x}_k + \vec{\delta}$ is modelled by a function $m_k (\vec{\delta})$ which
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is the second order Taylor expansion around the point $\vec{x}_k$, namely:
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is computed in a point $\vec{x_k}$, then its value in a point $\vec{x}_k +
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\vec{\delta}$ is modelled by a function $m_k (\vec{\delta})$ which is the
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second order Taylor expansion around the point $\vec{x}_k$, namely:
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$$
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\chi^2 (\vec{x}_k + \vec{\delta}) \sim m_k (\vec{\delta}) =
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\chi^2 (\vec{x}_k) + \nabla_k^T \vec{\delta} + \frac{1}{2}
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@ -284,41 +274,48 @@ $$
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where $\nabla_k$ and $H_k$ are the gradient and the Hessian matrix of $\chi^2$
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in the point $\vec{x}_k$ and the superscript $T$ stands for the transpose. The
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initial problem is reduced the so called trust-region subproblem (TRS), which
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is the minimisation of $m_k(\vec{\delta})$ in a region where
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$m_k(\vec{\delta})$ should be a good approximation of $\chi^2 (\vec{x}_k
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+ \vec{\delta})$, given by the condition:
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initial problem is reduced into the so called trust-region subproblem (TRS),
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which is the minimisation of $m_k(\vec{\delta})$ in a region where
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$m_k(\vec{\delta})$ should be a good approximation of $\chi^2 (\vec{x}_k +
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\vec{\delta})$, given by the condition:
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$$
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\|D_k \vec\delta\| < \Delta_k
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$$
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If $D_k = I$ the region is a hypersphere of radius $\Delta_k$
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centered at $\vec{x}_k$ but can be deformed according to $D_k$. This is
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necessary in the case one or more parameters have scale very different scales.
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Given an initial point $x_0$, radius $\Delta_0$, scale matrix $D_0$
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and step $\vec\delta_0$ the LSQ algorithm consist in the following iteration:
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If $D_k = I$, the region is a hypersphere of radius $\Delta_k$ centered at
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$\vec{x}_k$. In case one or more parameters have very different scales, the
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region should be deformed according to a proper $D_k$.
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1. Construct the function $m_k(\vec\delta)$.
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2. Solve the TRS for step $\vec\delta_k$.
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3. Check whether the solution actually decreases $\chi^2$
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1. If positive increase the trust region radius $\Delta_{k+1} =
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\alpha\Delta_k$ and shift the position $\vec x_{k+1} = \vec x_k +
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\vec \delta_k$.
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2. If negative decrease the radius $\Delta_{k+1} = \Delta_k/\beta$.
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4. Repeat
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Given an initial point $x_0$, the radius $\Delta_0$, the scale matrix $D_0$
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and step $\vec\delta_0$, the LSQ algorithm consist in the following iteration:
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1. construct the function $m_k(\vec\delta)$,
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2. solve the TRS and find $x_k + \vec\delta_k$ which corresponds to the
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plausible minimum,
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3. check whether the solution actually decreases $\chi^2$:
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1. if positive and converged (see below), stop;
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1. if positive but not converged, it means that $m_k$ still well
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approximates $\chi^2$ in the trust region which is therefore enlarged
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by increasing the radius $\Delta_{k+1} = \alpha\Delta_k$ for a given
|
||||
$\alpha$ and the position of the central point is shifted: $\vec
|
||||
x_{k+1} = \vec x_k + \vec \delta_k$;
|
||||
2. if negative, it means that $m_k$ does not approximate properly
|
||||
$\chi^2$ in the trust region which is therefore decreased by reducing
|
||||
the radius $\Delta_{k+1} = \Delta_k/\beta$ for a given $\beta$.
|
||||
4. Repeat.
|
||||
|
||||
This method is advantageous compared to a general minimisation because
|
||||
the TRS usually amounts to solving a linear system. There are many algorithms
|
||||
to solve the problem, in this case the Levenberg-Marquardt was used. It is based
|
||||
on a theorem that proves the existence of a number $\mu_k$ such that
|
||||
\begin{align*}
|
||||
\Delta_k \mu_k = \|D_k \vec\delta_k\| &&
|
||||
(H_k + \mu_k D_k^TD_k) \vec\delta_k = -\nabla_k
|
||||
\end{align*}
|
||||
to solve the problem, in this case the Levenberg-Marquardt was used. It is
|
||||
based on a theorem which proves the existence of a number $\mu_k$ such that:
|
||||
$$
|
||||
\mu_k (\Delta_k - \|D_k \vec\delta_k\|) = 0 \et
|
||||
(H_k + \mu_k D_k^TD_k) \vec\delta_k = -\nabla_k
|
||||
$$
|
||||
|
||||
Using the approximation[^2] $H\approx J^TJ$, obtained by computing the Hessian
|
||||
of the first-order Taylor expansion of $\chi^2$, $\vec\delta_k$ can
|
||||
be found by solving the system
|
||||
|
||||
be found by solving the system:
|
||||
$$
|
||||
\begin{cases}
|
||||
J_k \vec{\delta_k} + \vec{f_k} = 0 \\
|
||||
@ -326,10 +323,13 @@ $$
|
||||
\end{cases}
|
||||
$$
|
||||
|
||||
For more informations, see [@Lou05].
|
||||
|
||||
[^2]: Here $J_{ij} = \partial f_i/\partial x_j$ is the Jacobian matrix of the
|
||||
vector-valued function $\vec f(\vec x)$.
|
||||
|
||||
The algorithm terminates if on of the following condition are satisfied:
|
||||
The algorithm terminates if one of the following convergence conditions is
|
||||
satisfied:
|
||||
|
||||
1. $|\delta_i| \leq \text{xtol} (|x_i| + \text{xtol})$ for every component
|
||||
of $\vec \delta$.
|
||||
@ -337,11 +337,10 @@ The algorithm terminates if on of the following condition are satisfied:
|
||||
3. $\|\vec f(\vec x+ \vec\delta) - \vec f(\vec x)|| \leq \text{ftol}
|
||||
\cdot \max(\|\vec f(\vec x)\|, 1)$
|
||||
|
||||
These tolerance have all been set to \SI{1e-8}{}. The program converged in 7
|
||||
iterations giving the results below. The covariance of the parameters can again
|
||||
been estimated through the Hessian matrix at the minimum. The following results
|
||||
were obtained:
|
||||
|
||||
Where xtol, gtol and ftol are tolerance values all been set to \SI{1e-8}{}. The
|
||||
program converged in 7 iterations giving the results below. The covariance of
|
||||
the parameters can again been estimated through the Hessian matrix at the
|
||||
minimum. The following results were obtained:
|
||||
$$
|
||||
\bar{\alpha_{\chi}} = 0.6537 \et
|
||||
\bar{\beta_{\chi}} = 0.05972 \et
|
||||
@ -363,12 +362,14 @@ Hence:
|
||||
&\gamma_{\chi} = -0.1736 \pm 0.0016
|
||||
\end{align*}
|
||||
|
||||
See @sec:res_comp for results compatibility.
|
||||
|
||||
### Results compatibility
|
||||
|
||||
### Results compatibility {#sec:res_comp}
|
||||
|
||||
In order to compare the values $x_L$ and $x_{\chi}$ obtained from both methods
|
||||
with the correct ones (namely {$\alpha_0$, $\beta_0$, $\gamma_0$}), the
|
||||
following compatibility t-test was applied:
|
||||
with the correct ones ({$\alpha_0$, $\beta_0$, $\gamma_0$}), the following
|
||||
compatibility t-test was applied:
|
||||
|
||||
$$
|
||||
p = 1 - \text{erf}\left(\frac{t}{\sqrt{2}}\right)\ \with
|
||||
@ -380,6 +381,8 @@ where $i$ stands either for the MLM or LSQ parameters and $\sigma_{x_i}$ is the
|
||||
uncertainty of the value $x_i$. At 95% confidence level, the values are
|
||||
compatible if $p > 0.05$.
|
||||
|
||||
\vspace{30pt}
|
||||
|
||||
Likelihood results:
|
||||
|
||||
----------------------------
|
||||
@ -417,8 +420,8 @@ arrangement of $F$ would be required in order to justify this outcome.
|
||||
## Isotropic hypothesis testing
|
||||
|
||||
What if the probability distribution function was isotropic? Could it be
|
||||
compatible with the found results? If $F$ was isotropic, then $\alpha_I$,
|
||||
$\beta_I$ and $\gamma_I$ would be $1/3$, 0, and 0 respectively, since this
|
||||
gives $F_I = 1/{4 \pi}$. The t-test gives a $p$-value approximately zero for all
|
||||
the three parameters, meaning that there is no compatibility at all with this
|
||||
hypothesis.
|
||||
compatible with the found results?
|
||||
If $F$ was isotropic, then $\alpha_I$, $\beta_I$ and $\gamma_I$ would be $1/3$
|
||||
, 0, and 0 respectively, since this gives $F_I = 1/{4 \pi}$. The t-test gives a
|
||||
$p$-value approximately zero for all the three parameters, meaning that there is
|
||||
no compatibility at all with this hypothesis.
|
||||
|
||||
@ -1,3 +1,4 @@
|
||||
- cambiare simbolo convoluzione
|
||||
- aggiungere citazioni e referenze
|
||||
- rifare grafici senza bordino
|
||||
- leggere l'articolo di Lucy
|
||||
|
||||
Loading…
Reference in New Issue
Block a user