Polar Coordinates for Bivariate Random Vectors

Why Polar Coordinates Change the Density

For a bivariate random vector, rectangular coordinates describe location through horizontal and vertical components. Polar coordinates instead describe distance from the origin and direction. This can make a density easier to understand when its support or its values depend on distance from the origin. The change is not merely a relabelling: equal increments of radius and angle do not cover equal areas everywhere. The area represented by an angular increment grows with the radius, and that fact determines the Jacobian.

We use an angle in $[0,2\pi)$, measured counterclockwise from the positive horizontal axis. The choice of a single interval avoids counting a direction twice. Since the origin has no defined direction, we assume throughout that $(X,Y)$ has a joint density; in particular, it has probability zero of being exactly the origin. We may assign any angle there without changing its distribution.

Definition: For a bivariate random vector $(X,Y)$, its radius and angle are $R=\sqrt{X^2+Y^2}$ and $\Theta=\operatorname{atan2}(Y,X)$, with $\Theta\in[0,2\pi)$. For $r>0$, the inverse coordinate map is $x=r\cos\theta,\ y=r\sin\theta$.

The Polar Jacobian and Density

Write $T(r,\theta)=(r\cos\theta,r\sin\theta)$. Its derivative matrix and determinant are

DT(r,\theta)= \begin{pmatrix} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{pmatrix}, \qquad \det DT(r,\theta)=r(\cos^2\theta+\sin^2\theta)=r.

Thus the absolute Jacobian is $r$. Geometrically, a small change in radius contributes a length of about $dr$, while an angular change $d\theta$ at radius $r$ contributes an arc length of about $r\,d\theta$. The corresponding area is approximately $r\,dr\,d\theta$.

Theorem: Suppose $(X,Y)$ has joint density $f_{X,Y}$ with respect to planar area, and let $S$ be a measurable set outside which that density is zero. For $r>0$ and $0\leq\theta<2\pi$, define $A=\{(r,\theta): (r\cos\theta,r\sin\theta)\in S\}$. A joint density of $(R,\Theta)$ is $$ f_{R,\Theta}(r,\theta) = r f_{X,Y}(r\cos\theta,r\sin\theta)\,\mathbf{1}_{A}(r,\theta). $$ The marginal densities, for almost every $r>0$ and $0\leq\theta<2\pi$, are $$ f_R(r)=r\int_0^{2\pi} f_{X,Y}(r\cos\theta,r\sin\theta)\,d\theta, \qquad f_\Theta(\theta)=\int_0^\infty r f_{X,Y}(r\cos\theta,r\sin\theta)\,dr, $$ where the density is taken to be zero outside its support.

Proof. Apart from the origin and the positive horizontal axis, which are sets of planar area zero, the polar map is one-to-one on $r>0,\ 0\leq\theta<2\pi$. Its inverse is differentiable there, and the absolute Jacobian of the forward map is $r$. The change-of-variables formula therefore gives, for any measurable set $B$ in polar-coordinate space,

P((R,\Theta)\in B) = \int_B r f_{X,Y}(r\cos\theta,r\sin\theta) \,\mathbf{1}_A(r,\theta)\,dr\,d\theta.

This identifies the joint density. Integrating it over $\theta$ gives the density of $R$, and integrating it over $r$ gives the density of $\Theta$. These marginal integrations are valid by Tonelli’s theorem because the integrand is measurable and nonnegative. The support indicator can be omitted in the displayed marginal formulas when $f_{X,Y}$ is defined as zero outside $S$. $\square$

Worked Examples

Worked Example: A Uniform Distribution on a Disk

Suppose $(X,Y)$ is uniform on the disk $x^2+y^2\leq a^2$, where $a>0$. The disk has area $\pi a^2$, so the joint density is $1/(\pi a^2)$ on that disk and zero elsewhere. In polar coordinates the support is $0\leq r\leq a,\ 0\leq\theta<2\pi$. The transformed density is

f_{R,\Theta}(r,\theta)= \frac{r}{\pi a^2} \quad (0<r<a,\ 0\leq\theta<2\pi).

Integrating over the full angular range gives $f_R(r)=2r/a^2$ for $0<r<a$, and zero otherwise. Integrating over the radial range gives $f_\Theta(\theta)=1/(2\pi)$ for $0\leq\theta<2\pi$. Both densities normalize:

\int_0^a \frac{2r}{a^2}\,dr=1, \qquad \int_0^{2\pi}\frac{1}{2\pi}\,d\theta=1.

Moreover, the joint density factors as $(2r/a^2)(1/(2\pi))$ on the product support. Thus $R$ and $\Theta$ are independent in this example. The factor $r$ in the transformed density is essential: even though the original density is constant over the disk, points farther from the center correspond to more area per unit radial increment.

Worked Example: A Uniform Distribution on the Unit Square

Let $(X,Y)$ be uniform on $0\leq x\leq1,\ 0\leq y\leq1$. The square has area one, so its density is one there. Its polar support lies in the first quadrant. For a fixed angle in that quadrant, the constraints on $r$ are $r\cos\theta\leq1$ and $r\sin\theta\leq1$. Therefore $0<r\leq \min(\sec\theta,\csc\theta)$. The maximum radius is $\sqrt{2}$, the distance to the far corner. The joint density is

f_{R,\Theta}(r,\theta) = r\,\mathbf{1}\left\{ 0<\theta<\frac{\pi}{2},\ 0<r<\min(\sec\theta,\csc\theta) \right\}.

For a fixed $r$, if $0<r\leq1$, every first-quadrant angle is allowed. If $1<r\leq\sqrt{2}$, both coordinates must remain at most one, so $\arccos(1/r)\leq\theta\leq\arcsin(1/r)$. The angular length of this interval is $\pi/2-2\arccos(1/r)$. Integrating the joint density over the allowed angles yields

f_R(r)= \begin{cases} \frac{\pi r}{2},&0<r\leq1,\\ r\left(\frac{\pi}{2}-2\arccos(1/r)\right),&1<r\leq\sqrt{2},\\ 0,&\text{otherwise}. \end{cases}

For a fixed angle, the largest permissible radius is $\sec\theta$ on $0<\theta\leq\pi/4$, and $\csc\theta$ on $\pi/4<\theta<\pi/2$. Integrating $r$ from zero to that bound gives

f_\Theta(\theta)= \begin{cases} \frac{1}{2}\sec^2\theta,&0<\theta\leq\frac{\pi}{4},\\ \frac{1}{2}\csc^2\theta,&\frac{\pi}{4}<\theta<\frac{\pi}{2},\\ 0,&\text{otherwise}. \end{cases}

As a normalization check, the two angular integrals are $\frac12[\tan\theta]_0^{\pi/4}=1/2$ and $\frac12[-\cot\theta]_{\pi/4}^{\pi/2}=1/2$, which sum to one. Radius and angle are not independent: the radial upper limit depends on angle, so their joint support is not the product of their marginal supports.

Worked Example: A Standard Bivariate Normal Distribution

Suppose the joint density is $f_{X,Y}(x,y)=(1/(2\pi))\exp(-(x^2+y^2)/2)$ over the whole plane. Substituting $x=r\cos\theta,\ y=r\sin\theta$ gives $x^2+y^2=r^2$, and the polar Jacobian contributes $r$. Hence

f_{R,\Theta}(r,\theta) = \frac{r}{2\pi}e^{-r^2/2}, \qquad r>0,\quad 0\leq\theta<2\pi.

The support is the product of all positive radii and all angles in the chosen interval, and the density factors into a radial density and an angular density:

f_R(r)=r e^{-r^2/2}, \qquad f_\Theta(\theta)=\frac{1}{2\pi}.

The radial density integrates to one because $\int_0^\infty r e^{-r^2/2}\,dr=1$, using $u=r^2/2$. The angle density also integrates to one. Their product is the joint density, so $R$ and $\Theta$ are independent. Integrating the radial density from zero to $t\geq0$ gives $P(R\leq t)=1-e^{-t^2/2}$.

Support Determines the Marginals

The examples illustrate a general procedure: first transform the support, then integrate over the resulting allowable values of the other coordinate. For a fixed radius, the angular integral is only over angles that place the point inside the original support. For a fixed angle, the radial integral is only over radii that remain inside it. The square example required different limits for the two integrations; using the full circle for every radius, or the same radial limit for every direction, would give incorrect marginals.

Independence requires more than having simple-looking marginal densities. For continuous variables, a joint density that factors as $f_R(r)f_\Theta(\theta)$ almost everywhere on a product support implies independence. A support whose allowable radius depends on angle, as in the square, generally prevents such a factorization. Conversely, the disk and the isotropic normal example have a product support and a density that factors, so independence follows directly.

Common Mistakes

Dropping the Jacobian. Replacing the original coordinates in the density but leaving out the factor $r$ does not preserve probability. The transformed density must include the absolute Jacobian.
Assuming a uniform planar density gives a uniform radius. In a disk, the radial density is $2r/a^2$, not a constant: outer rings have more area than inner rings of the same thickness.
Using the full angular range without checking support. A first-quadrant square only permits angles in the first quadrant, and its maximum radius varies with angle.
Assuming radius and angle are always independent. Independence must be justified by the joint density and support; it is not a consequence of changing coordinates.

Key takeaway: The polar-coordinate density is the original density evaluated at $(r\cos\theta,r\sin\theta)$, multiplied by $r$, and restricted to the transformed support. Marginal radial and angular laws come from integrating this density over the allowable values of the other coordinate.

Check Your Understanding

Use the transformed density and support to answer each question.

Why does the polar-coordinate change of variables contribute a factor of $r$?
For a uniform distribution on a disk of radius $a$, what are the density of the radius and the density of the angle?
For a uniform unit square, why does the maximum allowable radius depend on the angle?
What support and density factorization establish independence of radius and angle in the standard bivariate normal example?
What is the error in treating the radius as uniform for a point uniformly distributed over a disk?