# If $X$ and $Y$ are zero mean independent Gaussian random variables with different variances, what is the density of $\sqrt{X^2+Y^2}$

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?

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.

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

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.