ex-4: χ² minimization results reported
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@ -160,7 +160,10 @@ int main(int argc, char **argv)
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}
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double result = x;
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printf("p_max: %.7f\n", result);
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double res_chi = chi2(result, ¶ms);
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printf("Results:\n");
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printf("χ² = %.3f\n", res_chi);
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printf("p_max = %.3f\n", result);
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// Compute the second derivative of χ² in its minimum for the result error.
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//
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@ -189,7 +192,7 @@ int main(int argc, char **argv)
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error = error + A + B;
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};
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error = 1/error;
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printf("ΔP_max: %.7f\n\n", error);
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printf("ΔP_max = %.3f\n\n", error);
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// Free memory.
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@ -151,13 +151,16 @@ $$
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Finally, putting all the pieces together, the average value of $|P_v|$ can now
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be computed:
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$$
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\langle |P_v| \rangle = \int
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\begin{align*}
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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_{\text{max}}}{x^2} - 1} \right)}
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$$
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f (P_v | P_h = x)
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\\
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&= [ \dots ]
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\\
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&= x \, \frac{\ln \left( \frac{P_{\text{max}}}{x} \right)}
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{\text{atan} \left( \sqrt{ \frac{P^2_{\text{max}}}{x^2} - 1} \right)}
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\end{align*}
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Namely:
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@ -218,3 +221,18 @@ $$
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The following result was obtained:
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In order to check wheter the expected distribution properly metches the
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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 inverese of the $\chi^3$ 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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$$
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p_{\text{max}} = 10 \pm 0.016 \with \chi^2 = 0.072
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$$
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which allows to assert that the sampled points actually follow the predicted
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distribution.
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