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Say $X,Y,Z$ are iid Std Normal RVs. I'm interested in finding the joint distribution of $X/\sqrt{X^2+Y^2+Z^2}$, $Y/\sqrt{X^2+Y^2+Z^2},Z/\sqrt{X^2+Y^2+Z^2}$. Various other questions and answers on tell me that this will follow Uniform distribution on the surface of sphere. To verify this, I tried to first transform to spherical coordinates and then set $r=1$. As we all know, the joint distribution of $X,Y,Z$ is

$$f(X,Y,Z)=\frac1{(2\pi)^{3/2}}e^{-(x^2+y^2+z^2)/2}$$

To verify the claim, I set $x=r\sin\theta\cos\phi, y=r\sin\theta\sin\phi, z=r\cos\theta$. The determinant of the jacobian is $r^2\sin\theta$. Hence $$f(r,\theta,\phi)=\frac{r^2|\sin\theta|}{(2\pi)^{3/2}}e^{-r^2/2}$$ Integrating the $r$ out in order to obtain joint distribution of just $\theta$ and $\phi$, we obtain $f(\theta,\phi)=|\sin(\theta)|/4\pi$. Clearly, this is not the distribution of uniform distribution on the surface of unit sphere. Where am I going wrong?

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  • $\begingroup$ math.stackexchange.com/questions/1864519/… $\endgroup$
    – lmaosome
    Commented Feb 7, 2021 at 19:52
  • $\begingroup$ @lmaosome I had seen that before, my only problem with that answer was the magical "surface area" measure. I wanted to explicitly show all steps. The series of steps I've done above work for 2D case but seem to fail in 3D case $\endgroup$
    – Anvit
    Commented Feb 7, 2021 at 19:57

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There is no mistake in your proof, except that you thought the distribution of angles also have to be uniform for uniform distribution on unit sphere.

Consider the density of uniform distribution of unit sphere: $$f(x,y,z) = \frac1{4\pi}\mathbb I_{[x^2+y^2+z^2=1]}$$ Now perform the same substitution, so that we'll have the same Jacobian, and the density becomes $$f(r,\theta,\phi)=\frac{r^2|\sin\theta|}{4\pi}\mathbb I_{[r=1]}$$ Integrating out $r$, you get the same density as yours.

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