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Let $X\sim\mathcal{N}(0,\sigma_X^2)$ and $Y\sim\mathcal{N}(0,\sigma_Y^2)$ be independent Gaussian random variables. What will be PDF of $Z=\sqrt{X^2+Y^2}$ and $W=\arctan{\left(\frac{Y}{X}\right)}$. Will they still be Rayleigh and uniform distributions respectively?

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According to your description, $x = Z \cdot \cos(W),\ y = Z \cdot \sin(W)$. Follow this answer for the derivation of joint PDF of $(Z, W)$ : Complex Gaussian Magnitude and Phase Joint PDF Derivation

You will reach the following expression after Methid of transformation: $$f_{Z,W}(z,w) = |\mathbf J|.f_{X,Y}(z \cdot \cos(w), z \cdot \sin(w)), \ where \ \mathbf J \ is \ Jacobian$$ Since the computation of $\mathbf J$ will not change hence: $$\mathbf J= z$$

Now, when you put the joint PDF of 2 independent Gaussian random varaibles $X, Y$ in the above expression you get the following : $$f_{Z,W}(z,w) = |\mathbf J|\cdot f_{X,Y}(z\cdot \cos(w), z\cdot \sin(w)) $$ $$= z \cdot \frac{1}{\sqrt{2\pi \sigma_x^2}}\exp(-\frac{1}{2\sigma_x^2}(z\cdot \cos(w))^2)\cdot \frac{1}{\sqrt{2\pi \sigma_y^2}}\exp(-\frac{1}{2\sigma_y^2}(z\cdot \sin(w))^2)$$

Combine this and find the marginal PDFs of $Z$ and $W$ from the above expression. You will find that integrating from $-\pi$ to $\pi$ w.r.t. $w$ will not give you Rayleigh Distribution anymore. And, similarly, integrating from $0$ to $\infty$ w.r.t. $z$ will not give you $\frac{1}{2\pi}$, and hence not Uniform Distribution anymore.

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By definition, $X$ and $Y$ must have a same variance $\sigma^2$ to make $Z$ being Rayleigh.

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Z follows a Hoyt distribution. The shape parameter is $q=max(\sigma_x,\sigma_y)/min(\sigma_x,\sigma_y)$, and the spread parameter is $w=\sigma_x^2+\sigma_y^2$.

The PDF of W is given in On the envelope and phase distributions for correlated gaussian quadratures. But to my knowledge, there are not closed-form expressions for the mean and the variance.

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