Constructing bivariate pdf from polar coordinates I'm working with a radar that gives me range and angle for each of its plots/outputs.  The radar has a fixed range resolution of 20m, and an angular resolution of 2 degrees.  This means that the further away a target is located, the greater the variability.
I want to convert these polar coordinates into a bivariate probability density function, but I am totally stumped as to how this should be done.  I want to keep the calculations as simple as possible (maybe use a rectangle?).
Any advice would be greatly appreciated.
 A: If $X+\mathrm iY=R\mathrm e^{\mathrm i\Theta}$ and $(R,\Theta)$ are independent with respective densities $f_R$ and $f_\Theta$, then the density $f_{(X,Y)}$ of $(X,Y)$ is such that
$$
f_{(X,Y)}(x,y)=\frac1{\sqrt{x^2+y^2}}f_R\left(\sqrt{x^2+y^2}\right)f_\Theta\left(\arctan\frac{y}x\right).
$$
Comparing this to your suggestion, note the missing factor $\frac1{\sqrt{x^2+y^2}}$ in your formula, which arises from the Jacobian of the change of variables $(x,y)\to(r,\theta)$.
In particular, $(X,Y)$ is rarely independent. An exception (the exception, perhaps) is when $f_R(r)\propto r\mathrm e^{-cr^2}$ for some $c\gt0$ and $f_\Theta(\theta)\propto1$, then $(X,Y)$ is i.i.d. centered normal.
A: I'm using the following approach:
(1) Assume r and theta are normal
(2) Set variance = (resolution / 6)^2
p(r,theta)
= p(r).p(theta)
= normal_prob(r).normal_prob(theta)
p(x,y) = normal_prob( sqrt(x^2 + y^2) ) . normal_prob( atan(y/x) )
Does this seem OK?
My next step is to set up a hypothesis between two points from different sensors to see if they are in fact the same points, i.e.:
H(0): X1 - X2 = 0
thus P(X1 - X2 = 0) = alpha

If the above is correct, this is the next hurdle to surmount...
