# Maximum likelihood estimate for mixture with components using both cartesian and polar coordinates

I have a set of points (x,y) that were generated from a mixture of two components: one component uses Cartesian coordinates, and the other polar coordinates.

For example, with probability $$\gamma$$ I sample points:

$$X, Y \sim \mathcal{N}(\boldsymbol{\mu}, \Sigma)$$

and with probability $$1 - \gamma$$ I sample:

$$R \sim \mathcal{Gamma}(\alpha, \beta) \\ \Theta \sim \mathcal{U}_{[0, 2\pi)} \\ X = R\cos(\Theta) \\ Y = R\sin(\Theta) \\$$

Where $$R$$ and $$\Theta$$ are independent.

I'd like to estimate the model parameters $$\boldsymbol{\mu}, \Sigma, \alpha, \beta, \gamma$$, where $$\gamma$$ is a mixing proportion.

I have previously used EM to fit a mixture of Gaussians, but I'm having trouble getting started due to the polar coordinate part of this problem.

Below is a generated toy data set as a visual aid. The orange cluster was sampled from the bivariate normal distribution.

• Is $R$ and $\Theta$ independent? Feb 8, 2023 at 21:14
• If your assumption is slightly modified to $R^2 \sim \exp(\lambda)$ plus $R$ and $\Theta$ are independent, you can actually show that the distribution the second component is also Gaussian. Feb 8, 2023 at 21:34
• Yes $R$ and $\Theta$ are independent, but I was looking for a general approach for handling the polar coordinate part of the problem. It could be $R \sim \mathcal{U}_{[a,b]}$ Feb 8, 2023 at 21:38
• You can always get the component density of $(X, Y)$ from the anti-polar transformation. But after that, if the mixture is not Gaussians, the EM algorithm (or other optimization methods) will be ugly. Are you only interested in getting the density or the subsequent numerical steps? If former, I can post an answer. Feb 8, 2023 at 21:43
• I'm interested in the numerical steps, however the density would be useful to understand. Maybe I can use some black-box optimization methods. Feb 8, 2023 at 21:47

By condition, the joint density of $$(R, \Theta)$$ is \begin{align} f(r, \theta) = \frac{\beta^\alpha}{2\pi\Gamma(\alpha)}r^{\alpha - 1}e^{-\beta r}, \quad r > 0, \theta \in [0, 2\pi). \tag{1} \end{align} It is easy to verify that the Jacobian determinant of the transformation: \begin{align} \begin{cases} r = \sqrt{x^2 + y^2}, \\[1em] \theta = \arctan(y/x). \end{cases} \end{align} is $$\frac{1}{\sqrt{x^2 + y^2}}$$, which together with $$(1)$$ imply that the joint density of $$(X, Y)$$ is \begin{align} f_2(x, y) = \frac{\beta^\alpha}{2\pi\Gamma(\alpha)}(x^2 + y^2)^{\frac{\alpha}{2} - 1}e^{-\beta\sqrt{x^2 + y^2}}, \quad (x, y) \in \mathbb{R}^2. \end{align} Let $$f_1(x, y)$$ be the density of $$N(\mu, \Sigma)$$. The density of the mixture distribution is then given by \begin{align} g(x, y) = \gamma f_1(x, y) + (1 - \gamma) f_2(x, y). \tag{2} \end{align}
Given $$(2)$$, if you decided to use the EM algorithm to get the MLE of the parameter $$\Psi = (\gamma, \mu, \Sigma, \alpha, \beta)$$, you can follow the steps outlined in Section 2.8 in Finite Mixture Models by McLachlan G. and Peel D.