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I have a vector $X=(1,X_2,X_3)$, where $(X_2,X_3)$ is a random vector in $\mathbb{R}^2$. Now consider $Y=g(X)=X/\|X\|$. What is a density function of $Y$ with respect to the uniform spherical measure, if I know the density of $(X_2,X_3)$?

The change of variables formula tells us that if $X=(X_1,\dots,X_n)$ has a density $f_X$ and $g:\mathbb{R}^n\to\mathbb{R}^n$ is continuously differentiable and injective, then $Y=g(X)$ has a density $$f_Y(y)=\begin{cases} f_X(g^{-1}(y))|\det J_{g^{-1}}| & \text{ if $y$ is in the range of $g$} \\ 0 & \text{ otherwise } \end{cases} $$

The transformation $g:\mathbb{R}^2\to\mathbb{R}^3$, $(x_2,x_3)\mapsto\left(\frac{1}{\sqrt{1+x_2^2+x_3^2}},\frac{x_2}{\sqrt{1+x_2^2+x_3^2}},\frac{x_3}{\sqrt{1+x_2^2+x_3^2}}\right) $ is injective with inverse $$g^{-1}:(y_1,y_2,y_3)\mapsto \left(\frac{y_2}{y_1},\frac{y_3}{y_1}\right).$$

But now, I see some difficulties with the standard change of variables formula. First, $g^{-1}$ is a transformation between $\mathbb{R}^3$ and $\mathbb{R}^2$. The Jacobian is a $3\times 2$ matrix, which is not square.

Is there still a way to find a density of $Y$ with respect to the uniform spherical measure?

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    $\begingroup$ Just remove the first function $1/\sqrt{1 + x_2^2 + x_3^2}$. Then you'll have a $2\times2$ matrix. $\endgroup$
    – muaddib
    Commented Aug 4, 2015 at 18:38
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    $\begingroup$ You can't. The three random variables in $Y$ are not jointly continuous and as such do not enjoy a joint density function $f_{Y_1,Y_2,Y_3}(y_1,y_1,y_3)$. On a smaller scale, note that $W$ and $Z=W^2$ are two real random variables that do not possess a joint density because all the probability mass lies on the curve $z=w^2$ in the $w$-$z$ plane and the density (measured in mass per unit area) of the probability mass at points on this curve is unbounded. $\endgroup$ Commented Aug 4, 2015 at 19:47
  • $\begingroup$ @Dilip Sarwate: I forgot to mention, that I'm interested in the joint density with respect to the uniform spherical measure. $\endgroup$
    – Aad
    Commented Aug 4, 2015 at 20:48
  • $\begingroup$ @muaddib: Could you help me with some intuition why am I allowed to do this, please? $\endgroup$
    – Aad
    Commented Aug 5, 2015 at 8:47
  • $\begingroup$ @Aad - I don't understand your addition to the problem: "What is a density function of Y with respect to the uniform spherical measure" $\endgroup$
    – muaddib
    Commented Aug 5, 2015 at 12:48

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You asked for an explanation of why you can restrict to two variables:

As you have found, when given two input random variables $X_2, X_3$ you need two output random variables $Y_2, Y_3$ to use the change of random variables formula. Here $$Y_2 = \frac{X_2}{1 + X_2^2 + X_3^2} \,\,\,\,\,\,\, Y_3 = \frac{X_2}{1 + X_2^2 + X_3^2}$$

I'll use your wording to show how this compares to your partial solution: The transformation $g:\mathbb{R}^2\to\mathbb{R}^2$, $(x_2,x_3)\mapsto\left(\frac{x_2}{\sqrt{1+x_2^2+x_3^2}},\frac{x_3}{\sqrt{1+x_2^2+x_3^2}}\right)$. i.e. $(Y_2, Y_3) = g(X_2, X_3)$. After computing the Jacobian of $g$ we see that is never $0$, hence the function is injective. Now we can apply the change of random variables formula.

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