Asymptotic distribution of ratio of two coordinates of random unit-vector Let $n$ be a large positive integer. Let $X=(X_1,X_2,\ldots,X_n)$ be uniform on the unit-sphere in $\mathbb R^n$ and define $R:=X_1/X_2$.

Question. What is the distribution of $R$ ?

Note. I'm really only interested in estimating $\mathbb P(R \le r)$ upto an error which goes to zero with $n$, uniformly in $r$.
Observations

*

*The CDF of $X_j$ converges pointwise to the CDF of $N(0,1/n)$, with a uniform error of $\mathcal O(1/n^\alpha)$.

*If $Z_1$ and $Z_2$ are independent gaussians of zero mean and same variance, then $Z_1/Z_2$ has Cauchy distribution.

Thus my guess is that asymptotic distribution of $R$ is Cauchy.
 A: The sought pdf is obtained by
$$
p(R)=C\int_0^\infty\cdots\int_0^\infty dx_1\cdots dx_n\delta\left(\sum_{i=1}^n x_i^2-1\right)\delta(R-x_1/x_2)\ .
$$
Define the auxiliary object
$$
p(R,t)=C\int_0^\infty\cdots\int_0^\infty dx_1\cdots dx_n\delta\left(\sum_{i=1}^n x_i^2-t\right)\delta(R-x_1/x_2)\ ,
$$
and take its Laplace transform w.r.t. $t$
$$
\hat p(R,s)=\int_0^\infty dt e^{-st}p(R,t)=C\int_0^\infty\cdots\int_0^\infty dx_1\cdots dx_n e^{-s\sum_{i=1}^n x_i^2}\delta(R-x_1/x_2)\ .
$$
Use the delta to kill the $x_1$ integral, as $\delta(R-x_1/x_2)=x_2\delta(x_1-R x_2)$, to obtain
$$
\hat p(R,s)=C\int_0^\infty dx_2~x_2 e^{-x_2^2(s+s R^2)}\left[\int_0^\infty dx e^{-s x^2}\right]^{n-2}=\frac{C}{2s(R^2+1)}(\sqrt{\pi}/2)^{n-2}\frac{1}{s^{(n-2)/2}}\ .$$
All you have to do is to take the inverse Laplace transform of this result (Mathematica can do that), set $t=1$, and compute the normalisation constant $C$, such that $\int_0^\infty dR p(R)=1$. The $R$-dependence is anyway independent of this procedure, and it yields indeed a Cauchy pdf for all $n$.
EDIT: if you allow $x_i$ to be also negative, the procedure above also works with $\int_0^\infty \to\int_{-\infty}^\infty$, and using $\delta(R-x_1/x_2)=|x_2|\delta(x_1-R x_2)$.
A: It turns out this can be done in a more direct way: $R_S:=X_1/X_2$ has the same distribution as $R_G:=Z_1/Z_2$ (where $(Z_1,Z_2) \sim N(0,I_2)$, $X$ has the same distribution as $X=Z/\|Z\|$, where $Z \sim N(0,I_n)$. But $R_G$ can be shown to have Cauchy distribution by elementary calculations!
Don't know why I didn't think of this earlier...
