Geometry of correlated Gaussians

In class we were discussing the geometry of $\rho$-correlated, $n$-dimensional Gaussians; call them $\vec{g}$, $\vec{h}$.

We have that $\parallel \vec{g}{\parallel }^2=\sum _{i=1}^n{g}_i^2$. This is the sum of $n$ very well-behaved independent random variables with mean $1$ and variance $2$, so its distribution value will be $n\pm O\left(\sqrt{n}\right)$ with very high probability. (To be more accurate, say $n\pm O\left(\sqrt{n}\log n\right)$ with probability at least $1-n^{-O\left(1\right)}$.) Hence $\parallel \vec{g}\parallel$ will be $\sqrt{n}(1+O(1/\sqrt{n})$ with high probability. Since $\vec{h}$ is also distributed as a standard $n$-dimensional Gaussian, the same is true of $\vec{h}$.

Now imagine $\vec{g}$ is fixed and we choose $\vec{h}$ to be $\rho$-correlated to $\vec{g}$. Then $\vec{g}\cdot \vec{h}=\sum _{i=1}^n{g}_i{h}_i=\sum _{i=1}^n(\rho g_i^2+\sqrt{1-{\rho }^2}{g}_i{g}_iʹ)$, where the $g_iʹ$ random variables are independent standard normals. I.e., the dot product is $\rho \parallel \vec{g}{\parallel }^2$ plus an independent one-dimensional normal, $\sqrt{1-{\rho }^2}\cdot N(0,\sum _i{g}_i^2)$. Since $\parallel \vec{g}{\parallel }^2=\sum _i{g}_i^2=n\pm O\left(\sqrt{n}\right)$ with high probability, the dot product is $\rho n\pm O\left(\right(\rho +\sqrt{1-{\rho }^2}\left)\sqrt{n}\right)$ with high probability.

So the cosine of the angle between $\vec{g}$ and $\vec{h}$ will be $(\rho n\pm O(\sqrt{n}\left)\right)/(n\pm O(\sqrt{n}\left)\right)=\rho \pm O(1/\sqrt{n})$, and hence the angle will be $\arccos \rho \pm O(1/\sqrt{n})$ with high probability (assuming $\rho$ is treated as a constant in $(0,1)$).

Now this doesn't really yet prove that $\vec{g}$ and $\vec{h}$ are distributed like a random pair on the surface of the $\sqrt{n}$-radius sphere with angle $\arccos \rho$. We just know that their angle will be $\arccos \rho$ and they will both be on the sphere. We should really look at the distribution of $\vec{h}-\rho \vec{g}$. But this is just a scaled $n$-dimensional Gaussian, so its distribution is spherically symmetric. I guess this more or less justifies the overall claim.
Actually, I believe that $\vec{h}-\rho \vec{g}$ will be close to orthogonal to $\rho \vec{g}$ with high probability. The dot product between the two will be distributed like $\sqrt{1-{\rho }^2}\cdot N(0,\sum _i{g}_i^2)$, which will probably be on the order of $\sqrt{1-{\rho }^2}\sqrt{n}$. However the product of the lengths of the two vectors will be like $\rho \sqrt{1-{\rho }^2}n$. So the cosine of the angle between them will be like $1/\left(\rho \sqrt{n}\right)$, and the angle will be close to $90$ degrees.