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3.2.1 Transformation of probability densities

Given $p(x)$ and a bijective mapping $y=f(x); \; x=f^{-1}(y)$; then
$\displaystyle \vert p(y) \, dy \vert$ $\textstyle =$ $\displaystyle \vert p(x)\,dx\vert$  

or
$\displaystyle p(y)$ $\textstyle =$ $\displaystyle p(x) \, \left\vert \frac{dx}{dy}\right\vert
= p[f^{-1}(y)] \, \left\vert \frac{df^{-1}(y)}{dy} \right\vert$  



\begin{figure}\includegraphics[height=180pt]{figures/f3tr.ps}
\end{figure}


This relation holds for any kind of density.

EXAMPLE: The spectral density of black body radiation is usually written in terms of the angular frequency $\omega$:
$\displaystyle I(\omega)$ $\textstyle =$ $\displaystyle \frac{\hbar \omega^{3}}{\pi c^{3}} \frac{1}{e^{\hbar \omega/kT}-1}$  

If we prefer to give the spectral density in terms of the wave length $\lambda \equiv 2 \pi c / \omega$, we have
$\displaystyle I(\lambda)$ $\textstyle =$ $\displaystyle I[\omega(\lambda)] \, \left\vert \frac{d \omega}{d \lambda}\right...
...)^{3} \frac{1}{e^{(hc/\lambda)/kT}-1}
\left( \frac{2 \pi c}{\lambda^{2}}\right)$  




EXERCISE: A powder of approximately spherical metallic grains is used for sintering. The diameters of the grains obey a normal distribution with $\langle d \rangle =2 \mu m$ and $\sigma=0.25 \mu m$. Determine the distribution of the grain volumes.



next up previous
Next: 3.2.2 Transformation Method Up: 3.2 Other Distributions Previous: 3.2 Other Distributions
Franz J. Vesely Oct 2005
See also:
"Computational Physics - An Introduction," Kluwer-Plenum 2001