# Difference between revisions of "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.