Following this question: Can we get the concentration inequality of the inner product of two unit vectors distributed on the sphere?.
Let $u$ and $v$ be two random vectors on $R^n$ that are independent and uniformly distributed on the unit sphere. That means we can represent it as Gaussian random vectors $g\sim N(0, I_n)$, $$u=\frac{g}{\|g\|_2}.
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Why do we have that $u\cdot v$ (may be with some orders?) converges weakly to a standard Gaussian random variables as $n\to \infty$? That means $u\cdot v=O_p(1)$.
Best Answer
Also write $v = h/\lVert h \rVert_2$ for Gaussian $h$. We then have that $$ u\cdot v = \frac{\sum_{i=1}^n g_i h_i}{\sqrt{\sum_{i=1}^n g_i^2 \sum_{i=1}^n h_i^2}} = \frac{1}{\sqrt{n}} \frac{\frac{1}{\sqrt{n}}\sum_{i=1}^n g_i h_i}{\sqrt{\frac{1}{n}\sum_{i=1}^n g_i^2 \frac{1}{n}\sum_{i=1}^n h_i^2}}. $$
By the CLT, the numerator converges in distribution to $\mathcal{N}(0, 1)$, whereas by the LLN, the denominator converges in probability to $1$.
Therefore, Slutsky's theorem yields that $$ \sqrt{n}(u\cdot v) = \frac{\frac{1}{\sqrt{n}}\sum_{i=1}^n g_i h_i}{\sqrt{\frac{1}{n}\sum_{i=1}^n g_i^2 \frac{1}{n}\sum_{i=1}^n h_i^2}} \xrightarrow{\mathrm{d}}\mathcal{N}(0, 1). $$
By the way, the intuition here is that as $n \rightarrow \infty$, the norm of an $n$-dimensional Gaussian random vector concentrates around $\sqrt{n}$, and so such a random vector can be thought of as being very similar to a random vector uniformly distributed on the $(n-1)$-sphere of radius $\sqrt{n}$.