Circular Error Probability in polar error expression

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Main Question or Discussion Point

After doing various searching through the google, most of the circular error probability I found are expressed interm of [tex]\sigma_x[/tex] and [tex]\sigma_y[/tex], which the CEP(circular error probability) usually looks like (assume normal distribution):

[tex]P(r) = \int\int \exp^{-.5(\frac{x^2}{\sigma_x^2}+\frac{y^2}{\sigma_y^2})} dxdy[/tex]

or if we integrate the equation in term of polar coordinate

[tex]P(r) = \int\int \exp^{-.5(\frac{x^2}{\sigma_x^2}+\frac{y^2}{\sigma_y^2})} rd\theta dr[/tex]

However, these equations are based on no correlation between [tex]\sigma_x[/tex] and [tex]\sigma_y[/tex]. And also, both [tex]\sigma_x[/tex] and [tex]\sigma_y[/tex] remains constant disregarding the change of [tex]r[/tex] and [tex]\theta[/tex]. Although we usually can express both [tex]x[/tex] and [tex]y[/tex] in term of [tex]r[/tex] and [tex]\theta[/tex],

[tex]x=r\cos(\theta)[/tex] and [tex]y=r\sin(t\theta)[/tex]

But right now, I would like to have the CEP to express in [tex]r[/tex] and [tex]\theta[/tex] only, since the error I will have are [tex]\sigma_\theta[/tex] and [tex]\sigma_r[/tex], and I would like to avoid the correlation issue. So, I'm just not know that if this equation will make sense or not,

[tex]P(r) = \int\int \exp^{-.5(\frac{r^2}{\sigma_r^2}+\frac{\theta^2}{\sigma_\theta^2})} rd\theta dr[/tex]

Anyone has any input/idea about this? One of the other problem is that I need to expand the CEP into spherical error probability (SEP), which is in the 3dimensional. Although I have a paper to show somewhat a close form solution for this problem, however they still consider [tex]\sigma_x[/tex] and [tex]\sigma_y[/tex] with correlation as their error instead of [tex]\sigma_\theta[/tex] and [tex]\sigma_r[/tex]. But the complexity just increases way too high if I'm following this method.
 

Answers and Replies

  • #2
mathman
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Your y expression has a t in it - seems to have come from nowhere. Your final P(r) is just wrong. Plug in the expressions for x and y as functions of r and θ to get the correct integral in polar coordinates.
 
  • #3
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it was a typo for y expression, it should be,

[tex]y=r\sin(\theta)[/tex]

For the final P(r), I'm trying to find the other way that involve both [tex]\sigma_r[/tex] and [tex]\sigma_\theta[/tex]. So it seems like it is impossible unless i substitute them into the following expression?

[tex]P(r) = \int\int \exp^{-.5(\frac{x^2}{\sigma_x^2}+\frac{y^2}{\sigma_y^2})} rd\theta dr[/tex]
 
  • #4
mathman
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Have you tried calculating the r and θ variances in terms of the x and y variances? Off hand it looks messy.
 
  • #5
mathman
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If you consider the case where the x and y variances are the same, the resultant polar coordinates have a distribution where the angle is uniform over a circle and r2 has an exponential distribution.
 

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