# Talk:Derivation of the Rayleigh Distribution Equation

rev 13:28, 3 June 2015‎ Herb

Ok, name of page needs a bit of fixing.

more verbosity in text

I goofed in accuracy section. "PDF" functions are set up wrong. Should be something like "PDF of X as x"

Been about 45 years since I took calculus in college. I'll have to look up conversion from Cartesian to Polar coordinates, but I know the conversion will swizzle to the right answer...

This is topic that I think bears further investigation...

is it $$\sigma_h = \sigma_v$$ or $$\sigma_h^2 = \sigma_v^2$$ ??

Of course if they are equal then both equations are true. The problem is in pooling the values if:

$$\sigma_h \approx \sigma_v$$

which equation do we use to pool the values?

• $$(\sigma_h + \sigma_v)/2$$
• $$\sqrt{\sigma_h^2 + \sigma_v^2}$$

I think $$\sqrt{\sigma_h^2 + \sigma_v^2}$$ should be corrected by factor $$\frac{1}{\sqrt{2}}$$ since $$(\sigma + \sigma)/2 = \sigma$$ but $$\sqrt{\sigma^2 + \sigma^2} = \sigma \sqrt{2}$$

In general it would seem that the ratio $$(\sigma_h / \sigma_v)$$ could be useful as a guide to stay out of trouble. Obviously the ration should depend on sample size, n. but something like $$.33(\sigma_h + \sigma_v) \leq (\sigma_h / \sigma_v) \leq 0.75(\sigma_h + \sigma_v)$$ is the idea...

I also really don't like using a $$\sigma$$ as a factor in the equation. If you think about radial values then there is a $$\sigma$$ which can be calculated from the $$r_i$$ values. The two $$\sigma$$'s aren't equal.

The !@#\$%^&* literature really messes up calculation of radial standard deviation. needs $$\frac{1}{\sqrt{2}}$$ correction.