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.