Derivation of the Rayleigh Distribution Equation

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The Rayleigh Distribution makes the following assumptions:

  • \(h \sim \mathcal{N}(\mu_h,\sigma_h^2)\), and \(\bar{v} \sim \mathcal{N}(\mu_v,\sigma_v^2)\)
  • 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:
    \(PDF(r) = \frac{r}{\sigma^2 } \exp\left( - \frac{r^2}{2\sigma^2} \right) \)

Accuracy of \(n\) 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 center.

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

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


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

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.

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