Editing Location parameter (section)

==Proofs==

For the continuous univariate case, consider a probability density function <math>f(x | \theta), x \in [a, b] \subset \mathbb{R}</math>, where <math>\theta</math> is a vector of parameters. A location parameter <math>x_0</math> can be added by defining:
:<math>
g(x | \theta, x_0) = f(x - x_0 | \theta), \; x \in [a + x_0, b + x_0]
</math>
it can be proved that <math>g</math> is a p.d.f. by verifying if it respects the two conditions<ref name="Ross 2010 p. ">{{cite book | last=Ross | first=Sheldon | title=Introduction to probability models | publisher=Academic Press | publication-place=Amsterdam Boston | year=2010 | isbn=978-0-12-375686-2 | oclc=444116127 }}</ref> <math>g(x | \theta, x_0) \ge 0</math> and <math>\int_{-\infty}^{\infty} g(x | \theta, x_0) dx = 1</math>. <math>g</math> integrates to 1 because:
:<math>
\int_{-\infty}^{\infty} g(x | \theta, x_0) dx = \int_{a + x_0}^{b + x_0} g(x | \theta, x_0) dx = \int_{a + x_0}^{b + x_0} f(x - x_0 | \theta) dx
</math>
now making the variable change <math>u = x - x_0</math> and updating the integration interval accordingly yields:
:<math>
\int_{a}^{b} f(u | \theta) du = 1
</math>
because <math>f(x | \theta)</math> is a p.d.f. by hypothesis. <math>g(x | \theta, x_0) \ge 0</math> follows from <math>g</math> sharing the same image of <math>f</math>, which is a p.d.f. so its range is contained in <math>[0, 1]</math>.