# Derivation of the Rayleigh Distribution Equation

# Mathematical Formulas and Derivations

# Bivariate Normal Distribution

Starting only with the assumptions that the horzontial and vertical measurements are normally distributed as notated by:

\(h \sim \mathcal{N}(\mu_h,\sigma_h^2)\), and \(v \sim \mathcal{N}(\mu_v,\sigma_v^2)\)

then the horizontal and vertical measures follow the general bivariate normal distribution which is given by the following equation:

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

Simplification of the Bivariate Normal Distribution to the Hoyt Distribution

# Correction Factors

The following three correction factors will be used throughout this statistical inference and deduction.

Note that all of these correction factors are > 1, are significant for very small *n*, and converge towards 1 as \(n \to \infty\). Their values are listed for *n* up to 100 in Media:Sigma1ShotStatistics.ods. File:SymmetricBivariate.c uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.

## Bessel correction factor

The Bessel correction removes bias in sample variance.

- \(c_{B}(n) = \frac{n}{n-1}\)

## Gaussian correction factor

The Gaussian correction (sometimes called \(c_4\)) removes bias introduced by taking the square root of variance.

- \(\frac{1}{c_{G}(n)} = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})\)

The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation: \(c_{G}(n)\) `=1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))`

## Rayleigh correction factor

The unbiased estimator for the Rayleigh distribution is also for \(\sigma^2\). The following corrects for the concavity introduced by taking the square root to get *σ*.

- \(c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}\)
^{[1]}

To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: `=EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))`

# The Hoyt Distribution

## Derivation From the Bivariate Normal distribution

Given the Bivariate Normal distribution as follows:

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

a simple translation of the Cartesian Coordinate System converts the Bivariate Normal distribution to the Hoyt distribution. This translation will not affect measurements about COI, but it would of course affect measurements which are measured about POA.

Given a translation to point \((\mu_h, \mu_v)\) then let:

\(h_* = h - \mu_h\) and \(v_* = v - \mu_v\)

Since the derivative of \(h_*\) with respect to \((h - \mu_h)\) is 1, (and similarity for \(v_*\)) then no change results to the integration constant of the function. Thus \(h_*\) can be substituted for \((h - \mu_h)\) and \(v_*\) for \((v - \mu_v)\). At this point the asterisk subscript is superfluous and will be dropped, giving the Hoyt distribution. \( f(h,v) = \frac{1}{2 \pi \sigma_h \sigma_v \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{h^2}{\sigma_h^2} + \frac{v^2}{\sigma_v^2} - \frac{2\rho h v }{\sigma_h \sigma_v} \right] \right) \)

# The Rayleigh distribution

## Derivations

### Derivation OF Single Shot PDF From the Bivariate Normal distribution

The Rayleigh Distribution makes the following simplifying assumptions to the general bivariate normal distribution:

- Horizontal and vertical dispersion are independent.
- \(\sigma_h = \sigma_v\) (realistically \(\sigma_h \approx \sigma_v\))
- \(\rho = 0\)
- No Fliers

for which the PDF for any shot, \(i\), around the horizontal and vertical point \((\mu_h, \mu_v)\) is given by:

\(PDF(r; \sigma_{\Re}) = \frac{r}{\sigma_{\Re}^2 }
\exp\left(
- \frac{r^2}{2\sigma_{\Re}^2}
\right)
\)

- where \(\sigma_{\Re} = \sigma_h = \sigma_v\) and \(r = \sqrt{h_i - \mu_h)^2 + sqrt(v_i - \mu_v)^2}\)

**PROOF**

Using the assumptions in the first section, the distribution of an individual shot 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:

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

### Derivation OF Single Shot CDF from the PDF

Given the single shot Rayleigh distribution, calculate the single shot Cumulative Distribution Function (CDF) for the Rayleigh distribution.

### Calculate the Mode of the Rayleigh distribution from its PDF

Given the Rayleigh distribution, calculate the mode for the Rayleigh distribution.

### Calculate the Median of the Rayleigh distribution from its PDF

Given the Rayleigh distribution, calculate the mean for the Rayleigh distribution.

### Calculate the Mean Radius of the Rayleigh distribution from its PDF

Given the Rayleigh distribution, calculate the mean for the Rayleigh distribution.

## Experimental Uses

### Accuracy of \(n\) Sighting Shots

Given the assumptions in the starting section we again substitute \(\sigma\) for both \(\sigma_h\) and \(\sigma_v\). This simplifies the distributions of \(h\) and \(v\) to:

\(h \sim \mathcal{N}(\mu_h,\sigma^2)\), and \(v \sim \mathcal{N}(\mu_v,\sigma^2)\)

Now we take some number \(n\) of shots \(( n \geq 1)\)and calculate their centers \(\bar{h}\) and \(\bar{v}\) which will be normal distributions as well.

\(\bar{h} \sim \mathcal{N}(\mu_h,\sigma^2/n)\), and \(\bar{v} \sim \mathcal{N}(\mu_v,\sigma^2/n)\)

Let \(r_n\) be the distance of this sample center \((\bar{h}, \bar{v})\) from the true distribution center \((\mu_h, \mu_v)\) as:

\(r_n = \sqrt{(\bar{h}-\mu_h)^2 + (\bar{v}-\mu_v)^2}\)

Define random variables \(Z_h\) and \(Z_v\) as the squared *Studentized* horizontal and vertical errors by dividing by the respective standard deviations. Each of these variables with have a Chi-Squared Distribution with one degree of freedom.

\(Z_h = \left(\frac{(\bar{h}-\mu_h)}{\sigma/\sqrt n}\right)^2 = \frac n{\sigma^2}(\bar{h}-\mu_h)^2 \sim \chi^2(1)\)

\(Z_v = \left(\frac{(\bar{v}-\mu_v)}{\sigma/\sqrt n}\right)^2 = \frac n{\sigma^2}(\bar{v}-\mu_v)^2\sim \chi^2(1)\)

Define random the variable \(W\) which will have a Chi-Squared Distribution with two degrees of freedom as:

\(W = Z_x + Z_y =\frac n{\sigma^2}\left((\bar{v}-\mu_v)^2+(\bar{v}-\mu_v)^2\right)\sim \chi^2(2)\)

Rescale the variable \(W\) by \(\frac {\sigma^2}{n}\) and denote the new variable \(w_n\):

\(w_n=\frac {\sigma^2}nW\) and note that \(w_n=r_n^2\)

By the properties of a chi-square random variable, we have:

\(w_n \sim \text {Gamma}(k=1, \theta = 2\sigma^2/n) = \text{Exp}(2\sigma^2/n)\)

so:

\(PDF(w_n) = \frac {n}{2\sigma^2}\cdot \exp\Big \{-\frac {n}{2\sigma^2} w_n\Big\}\)

But from above \(r_n = \sqrt {w_n}\). By the change-of-variable formula we have

\(w_n = r_n^2 \Rightarrow \frac {dw_n}{dr_n} = 2r_n\)

and so:

\( PDF(r_n) = 2r_n\frac {n}{2\sigma^2}\cdot \exp\Big \{-\frac {n}{2\sigma^2} r_n^2\Big\} = \frac {r_n}{\alpha^2} \exp\Big \{-\frac {r_n^2}{2\alpha^2} \Big\},\;\;\alpha \equiv \sigma/\sqrt n\)

So for any number of shots \(n\), the expected accuracy is given by \(r_n\) follows a Rayleigh distribution with parameter \(\alpha = \sigma / \sqrt{n}\) where \(\sigma\) is the Rayleigh shape factor for one shot.

**Thanks to Alecos Papadopoulos for the solution.**