Covariance Matrix Polar to Cartesian

Question

Let $R$ and $\Theta$ be zero mean independent Gaussian random variables with variances $\sigma_R^2$ and $\sigma_\Theta^2$ respectively. Hence their covariance matrix would be diagonal give by $$C_{R,\Theta}=\begin{bmatrix}\sigma_R^2 & 0 \\ 0 & \sigma_\Theta^2\end{bmatrix}$$ Now, let's define transformations given by $X = R\cos (\Theta)$ and $Y = R \sin (\Theta)$.

The question is, what would be $C_{X,Y}$, the covariance matrix of transformed random variable $X$ and $Y$?

Thank you in advance.

Answer 1

The covariance matrix is given by

$$C_{X,Y}=\begin{bmatrix}E(XX)& E(XY) \\ E(YX )& E(YY) \end{bmatrix}$$

This can be written as below:

$$C_{X,Y}=\begin{bmatrix}E(R^2cos^2(\Theta) )& E(Rcos(\Theta)Rsin(\Theta)) \\ E(Rsin(\Theta)Rcos(\Theta) )& E(R^2sin^2(\Theta)) \end{bmatrix}$$

Since $R$ and $\Theta$ are independent the expectation will distribute. Now the task is to calculate expectation of $Cos^2(\Theta)$ , $Sin^2(\Theta)$ and $Cos(\Theta)Sin(\Theta)$ or equivalently $Sin(2\Theta)$. These can be found by by calculating the expectations of each of these functions for ex: for $Cos^2(\Theta)$ this will be $$E(cos^2(\Theta)) = \int_{ - \infty}^{\infty}\, Cos^2(\theta) f_\Theta(\theta) \, d\theta$$

Where $f_{\Theta}(\theta)$ is the PDF of random varible $\Theta$

Hint: you can also utilize symmetries while calculating these integrals, $Sin(2\Theta)$ is an odd function and it multiples a Gaussian with zero mean which is an even function. Similarly $Cos^2(\Theta)$ and $Sin^2(\Theta)$ are even functions

EDIT, Solving the integral:

$$\cos^2\theta=0.5+0.5\cos 2\theta=0.5+0.25(e^{2i\theta}+e^{-2i\theta})$$

$$I=\int_{-\infty}^{\infty} \cos^{2}(\theta) f_{\Theta}(\theta) d\theta=0.5+0.25\int_{-\infty}^\infty {1\over \sqrt{2\pi}} e^{-\theta^2\over 2}(e^{2i\theta}+e^{-2i\theta})d\theta$$

$$\int_{-\infty}^\infty e^{-\theta^2\over 2}e^{ki\theta}d\theta {= \int_{-\infty}^\infty e^{-\theta^2\over 2}e^{ki\theta}e^{-{(ki)^2\over 2}}e^{{(ki)^2\over 2}}d\theta \\= \int_{-\infty}^\infty e^{-(\theta-ki)^2\over 2}e^{-{k^2\over 2}}d\theta \\= \int_{-\infty}^\infty e^{-\theta^2\over 2}e^{-{k^2\over 2}}d\theta \\= \sqrt{2\pi}e^{-{k^2\over 2}} }$$ hence $$ I=0.5+0.5e^{-2}$$

Original answer to the integral https://math.stackexchange.com/questions/3650453/integral-of-int-infty-infty-cos2-theta-f-theta-theta-d/3650494#3650494

Answer 2

Your covariance matrix should now look like

$$C_{X,Y}=\begin{bmatrix}\sigma^2_{(X,X)} & \sigma^2_{(X,Y)} \\ \sigma^2_{(Y,X)} & \sigma^2_{(Y,Y)} \end{bmatrix}$$

Since the covariance matrix formula is given by

$$C_{X, Y} = \frac{1}{n-1} \sum^{n}_{i=1}{(x_i-\bar{x})(y_i-\bar{y})}$$

and using the fact that both $\bar{x}$ and $\bar{y}$ are zero (zero mean), you should arrive at something like (provided that I did the math correctly)

$$C_{X,Y}=\begin{bmatrix}\sigma^2(R^2*cos^2(\Theta)) & \sigma^2(R^2*cos(\Theta)*sin(\Theta)) \\ \sigma^2(R^2*sin(\Theta)*cos(\Theta)) & \sigma^2(R^2*sin^2(\Theta)) \end{bmatrix}$$

These are still unnormalized values, you should divide by n-1 (n-number of samples) to get normalized values (if you were to apply this to a set of n samples/observations).