Covariance matrix of rotation and translation applied to a point

Question

I'm trying to calculate a 2x2 covariance matrix in Cartesian coordinates that represents the amount of uncertainty when rotating and translating a point in 2D space,

$\Sigma = \begin{pmatrix} \sigma_{xx}&\sigma_{xy}\\ \sigma_{yx}&\sigma_{yy} \end{pmatrix}$

The position of a point in 2D space is represented by a vector,

$p= \begin{pmatrix} x\\y \end{pmatrix}$

I then rotate and translate this point to a new position as follows,

$p'=Rp + t$

where $R$ is a 2D rotation matrix given by,

$R= \begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}$

and t is a translation vector given by,

$t= \begin{pmatrix} t_x\\t_y \end{pmatrix}$

If I assume independent Gaussian noise on the translations $(t_x, t_y)$ and the rotation angle $\theta$, then I can perform a simple Monte Carlo simulation to see what the true uncertainty should look like, by passing many sampled points through the rotation and translation function.

In the above image, the blue point is the original 2D point, the green point is the rotated and translated 2D point, and the small red dots are the samples with random noise applied to $t_x, t_y$ and $\theta$.

As such, I know that my covariance matrix is a function of the amount of uncertainty in rotation and translation. However, I'm struggling to produce an analytical expression for this covariance matrix (the true distribution is nonlinear due to rotation, but I'm content with a linear representation of the covariance matrix that assumes small amounts of rotation)

So my question is, given that I know the equations to move a point to a new position using rotation and translation, how do I use these equations to derive an analytical expression for the amount of uncertainty in this rotation and translation represented by a 2x2 covariance matrix in 2D Cartesian coordinates?

Answer 1

If you are content with linear small rotation angle approximation, then you have

$$ R \approx \mathbb{I} + d\theta \begin{pmatrix} 0 & -1 \\ 1 &0 \end{pmatrix}$$

so

$$ Rp \approx p + d\theta \begin{pmatrix} -y \\ x \end{pmatrix} .$$

Assuming that $d\theta$ has a normal distribution with variance $\sigma_\theta^2$, then the covariance matrix of $Rp$ is given by

$$ Cov(Rp) = \begin{pmatrix} -y \\ x \end{pmatrix} \sigma_\theta^2 \begin{pmatrix} -y & x \end{pmatrix} = \sigma_\theta^2 \begin{pmatrix} y^2 & -xy \\ -xy & x^2 \end{pmatrix}.$$

If you further assume that $t_x$ and $t_y$ are independent, having normal distribution with variance $\sigma_t^2$, then the covariance matrix of $t$ is $\sigma_t^2$ times the unit matrix. The covariance matrix of the sum is just the sum of covariance matrices:

$$Cov(p')=Cov(Rp)+Cov(t)= \sigma_\theta^2 \begin{pmatrix} y^2 & -xy \\ -xy & x^2 \end{pmatrix} + \sigma_t^2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$