Derivation of the Rayleigh Distribution Equation

The Rayleigh Distribution makes the following assumptions:

• \(\bar{h} \sim \mathcal{N}(\mu_h,\sigma_h^2), \bar{v} \sim \mathcal{N}(\mu_v,\sigma_v^2)\)
• Horizontal and vertical dispersion are independent.
• \(\sigma_h = \sigma_v\) (realistically \(\sigma_h \approx \sigma_v\))
• \(\rho = 0\)
• No Fliers

for which the PDF is given by:
    \(PDF(r) = \frac{r}{\sigma^2 } \exp\left( - \frac{r^2}{2\sigma^2} \right) \)

The PDF can be derived in the following two methods.

Simplification of the Bivariate Normal Distribution

Using the assumptions above, the Rayleigh Distribution is easily simplified from the Bivariate Normal Distribution which has the equation:

    \( f(h,v) = \frac{1}{2 \pi \sigma_h \sigma_v \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{(h-\mu_h)^2}{\sigma_h^2} + \frac{(v-\mu_v)^2}{\sigma_v^2} - \frac{2\rho(h-\mu_h)(v-\mu_v)}{\sigma_h \sigma_v} \right] \right) \)

By substituting \(\rho = 0\) the equation reduces to:

    \( f(h,v) = \frac{1}{2 \pi \sigma_h \sigma_v } \exp\left( -\frac{1}{2}\left[ \frac{(h-\mu_h)^2}{\sigma_h^2} + \frac{(v-\mu_v)^2}{\sigma_v^2} \right] \right) \)

Since \(\sigma_h\) and \(\sigma_v\) are equal, substitute \(\sigma\) for each, then collect terms in the exponential, after which the equation reduces to:

    \( f(h,v) = \frac{1}{2 \pi \sigma^2 } \exp\left( -\left[ \frac{(h-\mu_h)^2 + (v-\mu_h)^2}{2\sigma^2} \right] \right) \)

Letting \(r^2 = (h-\mu_h)^2 + (v-\mu_v)^2\) the equation becomes:

    \( f(h,v) = \frac{1}{2 \pi \sigma^2 } \exp\left( - \frac{r^2}{2\sigma^2} \right) \)

Now transforming to the polar coordinate system:

** ok, here I'm lost **

and finally:
    \( f(r) = \frac{r}{\sigma^2 } \exp\left( - \frac{r^2}{2\sigma^2} \right) \)