Derivation of the Rayleigh Distribution Equation

The Rayleigh Distribution makes the following assumptions:

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

The PDF can be derived in the following two methods.

Simplification of the Bivariate Normal Distribution

Using the assumptions above, the Rayleigh Distribution 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 from First Principles

Given the assumptions in the starting section we first translate the Cartesian coordinates to \((\mu_h, \mu_v)\), and substitute \(\sigma\) for both \(\sigma_h\) and \(\sigma_v\). This simplifies the distributions of \(h\) and \(v\) to:

$$X, Y \sim N(0, \sigma)$$

I.e., we designate the (unknown) true POI as the origin (0, 0). Now we take some number n of "sighter" shots and calculate their center. The distance of this sample center from the true POI is

$$R(n) = \sqrt{\overline{x_i}^2 + \overline{y_i}^2}$$

What is the distribution of Weapon Zero?

Thanks to Alecos Papadopoulos for the following solution:

Given the preceding definitions, the sample means are also mean-zero normal random variables with $$\bar X, \bar Y \sim N(0,\sigma^2/n)$$

Define random variables $$Z_x = \left(\frac{\bar X}{\sigma/\sqrt n}\right)^2 = \frac n{\sigma^2}\bar X^2 \sim \chi^2(1),\;\;Z_y = \left(\frac{\bar Y}{\sigma/\sqrt n}\right)^2 =\frac n{\sigma^2}\bar Y^2\sim \chi^2(1),$$

Then the random variable $$W = Z_x + Z_y =\frac n{\sigma^2}\left(\bar X^2+\bar Y^2\right)\sim \chi^2(2)$$

By the properties of a chi-square random variable, we have $$W_n=\frac {\sigma^2}nW \sim \text {Gamma}(k=1, \theta = 2\sigma^2/n) = \text{Exp}(2\sigma^2/n)$$ i.e. $$f_{W_n}(w_n) = \frac {n}{2\sigma^2}\cdot \exp\Big \{-\frac {n}{2\sigma^2} w_n\Big\}$$

Define \(R_n = \sqrt {W_n}\). By the change-of-variable formula we have

$$W_n = R_n^2 \Rightarrow \frac {dW_n}{dR_n} = 2R_n$$ and so

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

So \(R_n\) follows a Rayleigh distribution with parameter \(\alpha = \sigma / \sqrt{n}\).