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Example: Gauss distribution

Definition 2 (Gauss Distribution)   The function
$\displaystyle \fbox{$ \displaystyle
\rho(x) \equiv \frac{1}{\sqrt{2\pi \sigma^2}}\exp \left(-\frac{(x-x_0)^2}{2\sigma^2}\right)
$}$     (26)

is the probability density of the Gauss distribution with parameters $ x_0$ and $ \sigma>0$.

We have the following important integral (Gauss integral):
$\displaystyle \fbox{\fbox{$ \displaystyle
\int_{-\infty}^{\infty}dx e^{-ax^2 + bx} = \sqrt{\frac{\pi}{a}} \exp \left({\frac{b^2}{4a}}\right).
$}}$     (27)

Proof: First consider $ b=0,a=1$. Calculate
$\displaystyle \left[ \int_{-\infty}^{\infty}dx e^{-x^2} \right] = \int \int dx ...
...\int_{0}^{\infty}dr r e^{-r^2} = 2\pi \int_{0}^{\infty}\frac{dx}{2}e^{-x} = \pi$     (28)

and take the square-root of this equation. Then do the case with general $ a$ and $ b$ by completing the square in the exponential and substitution (exercise!).



Tobias Brandes 2004-02-04