Circular Error Probability in polar error expression

[dbeeo]

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

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

or if we integrate the equation in term of polar coordinate

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

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

x=r\cos(\theta) and y=r\sin(t\theta)

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

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

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 \sigma_x and \sigma_y with correlation as their error instead of \sigma_\theta and \sigma_r. But the complexity just increases way too high if I'm following this method.

 

[mathman]

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.

 

[dbeeo]

it was a typo for y expression, it should be,

y=r\sin(\theta)

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

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

 

[mathman]

Have you tried calculating the r and θ variances in terms of the x and y variances? Off hand it looks messy.

 

[mathman]

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.