Difference between revisions of "Derivation of the Rayleigh Distribution Equation"
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Define random variables <math>Z_h</math> and <math>Z_v</math> 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. | Define random variables <math>Z_h</math> and <math>Z_v</math> 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. | ||
| − | <math> | + | <math>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)</math><br /> |
| − | <math> | + | <math>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)</math><br /> |
Define random the variable <math>W</math> which will have a Chi-Squared Distribution with two degrees of freedom like:<br /> | Define random the variable <math>W</math> which will have a Chi-Squared Distribution with two degrees of freedom like:<br /> | ||
<math>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)</math> | <math>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)</math> | ||
| + | |||
| + | Rescale the variable <math>W</math> by <math>\frac {\sigma^2}{n}</math> and denote the new variable <math>w_n</math>:<br /> | ||
| + | |||
| + | <math>w_n=\frac {\sigma^2}nW</math> | ||
By the properties of a chi-square random variable, we have | 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)$$ |
i.e. | i.e. | ||
$$f_{W_n}(w_n) = \frac {n}{2\sigma^2}\cdot \exp\Big \{-\frac {n}{2\sigma^2} w_n\Big\}$$ | $$f_{W_n}(w_n) = \frac {n}{2\sigma^2}\cdot \exp\Big \{-\frac {n}{2\sigma^2} w_n\Big\}$$ | ||
| − | But from above <math>r_n = \sqrt { | + | But from above <math>r_n = \sqrt {w_n}</math>. By the change-of-variable formula we have |
| − | <math> | + | <math>w_n = r_n^2 \Rightarrow \frac {dw_n}{dr_n} = 2r_n</math><br /> |
and so:<br /> | and so:<br /> | ||
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<math> 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</math> | <math> 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</math> | ||
| − | So for any number of shots <math>n, r_n</math> follows a Rayleigh distribution with parameter <math>\alpha = \sigma / \sqrt{n}</math>. | + | So for any number of shots <math>n</math>, the mean distance <math>r_n</math> follows a Rayleigh distribution with parameter <math>\alpha = \sigma / \sqrt{n}</math> where <math>\sigma</math> is the Rayleigh shape factor for one shot. |
'''Thanks to Alecos Papadopoulos for the solution.''' | '''Thanks to Alecos Papadopoulos for the solution.''' | ||
Revision as of 23:58, 2 June 2015
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:
\(PDF(r) = \frac{r}{\sigma^2 }
\exp\left(
- \frac{r^2}{2\sigma^2}
\right)
\)
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 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 like:
\(W = Z_x + Z_y =\frac n{\sigma^2}\left((\bar{v}-\mu_v)^2+(\bar{v}-\mu_v)^2\right)\sim \chi^2(2)\)
Rescale the variable \(W\) by \(\frac {\sigma^2}{n}\) and denote the new variable \(w_n\):
\(w_n=\frac {\sigma^2}nW\)
By the properties of a chi-square random variable, we have $$w_n \sim \text {Gamma}(k=1, \theta = 2\sigma^2/n) = \text{Exp}(2\sigma^2/n)$$ i.e. $$f_{W_n}(w_n) = \frac {n}{2\sigma^2}\cdot \exp\Big \{-\frac {n}{2\sigma^2} w_n\Big\}$$
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 mean distance \(r_n\) 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.
