Derivation of the Rayleigh Distribution Equation

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:

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

The Rayleigh Distribution makes the following simplying 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:
    \(PDF(r) = \frac{r}{\sigma^2 } \exp\left( - \frac{r^2}{2\sigma^2} \right) \)

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

Distribution of An Individual Shot

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

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.