# Difference between revisions of "Talk:Derivation of the Rayleigh Distribution Equation"

(Created page with "''' 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. Shoul...") |
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+ | This is topic that I think bears further investigation... | ||

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+ | is it <math>\sigma_h = \sigma_v</math> or <math>\sigma_h^2 = \sigma_v^2</math> ?? | ||

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+ | Of course if they are equal then both equations are true. The problem is in pooling the values if:<br /> | ||

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+ | <math>\sigma_h \approx \sigma_v</math> | ||

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+ | which equation do we use to pool the values? | ||

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+ | * <math>(\sigma_h + \sigma_v)/2</math> | ||

+ | * <math>\sqrt{\sigma_h^2 + \sigma_v^2}</math> | ||

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

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+ | In general it would seem that the ratio <math>(\sigma_h / \sigma_v)</math> could be useful as a guide to stay out of trouble. Obviously the ratio should depend on sample size, ''n''. Think this is sort of the idea, but it doesn't right (two limits should converge) ... | ||

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+ | <math> .33\frac{(\sigma_h + \sigma_v)}{\sqrt{n}} \leq (\sigma_h / \sigma_v) \leq 0.75\frac{(\sigma_h + \sigma_v)}{\sqrt{n}} </math> | ||

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+ | ---- | ||

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

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+ | ---- | ||

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

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+ | ----- | ||

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+ | 4.1 Derivation From the Bivariate Normal distribution | ||

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+ | ... being more repetitive: ''"a simple translation of the Cartesian Coordinate System converts the Bivariate Normal distribution to the Hoyt distribution"'' is misleading: The bivariate normal distribution does not become the (univariate) Hoyt distribution. After converting (h,v) coordinates to polar (r,azimuth) coordinates, the r-coordinate (distance to true COI) follows a Hoyt distribution. | ||

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+ | 5.1 Derivation OF Single Shot PDF From the Bivariate Normal distribution | ||

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+ | ''"The Rayleigh Distribution makes the following simplifying assumptions to the general bivariate normal distribution:"'' is a bit weird because a distribution does not make assumptions. Making the assumption that shots (in the sense of (h,v) coordinates) follow a restricted (circular) bivariate normal distribution implies that the distance to the true COI follows a Rayleigh distribution. | ||

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+ | :: You're right, the wording is sloppy.<br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 20:13, 24 June 2015 (EDT) |

## Latest revision as of 19:13, 24 June 2015

** 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 ratio should depend on sample size, *n*. Think this is sort of the idea, but it doesn't right (two limits should converge) ...

\( .33\frac{(\sigma_h + \sigma_v)}{\sqrt{n}} \leq (\sigma_h / \sigma_v) \leq 0.75\frac{(\sigma_h + \sigma_v)}{\sqrt{n}} \)

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.

4.1 Derivation From the Bivariate Normal distribution

... being more repetitive: *"a simple translation of the Cartesian Coordinate System converts the Bivariate Normal distribution to the Hoyt distribution"* is misleading: The bivariate normal distribution does not become the (univariate) Hoyt distribution. After converting (h,v) coordinates to polar (r,azimuth) coordinates, the r-coordinate (distance to true COI) follows a Hoyt distribution.

5.1 Derivation OF Single Shot PDF From the Bivariate Normal distribution

*"The Rayleigh Distribution makes the following simplifying assumptions to the general bivariate normal distribution:"* is a bit weird because a distribution does not make assumptions. Making the assumption that shots (in the sense of (h,v) coordinates) follow a restricted (circular) bivariate normal distribution implies that the distance to the true COI follows a Rayleigh distribution.