Uniform distribution on the unit sphere rotated by a random orthogonal matrix

Question

Question 1. Let $u\in \mathbb{R}^n $ be a random vector uniformly distributed on $\mathbb{S}^{n-1}$, and $T\in \mathbb{R}^{n\times n}$ be a random orthogonal matrix. If $u$ and $T$ are independent, is $Tu$ uniformly distributed on $S^{n-1}$ and statistically independent of $T$?

I get this question when thinking about the following one.

Question 2. Let $u, v\in\mathbb{R}^n$ be two independent random vectors, and $u$ be uniformly distributed on $\mathbb{S}^{n-1}$. Consider the inner product $u^\top v$. Take a constant vector $v_0\in \mathbb{S}^{n-1}$. It seems that

1. $u^\top v$ has the same distribution as $u^\top v_0$;
2. $u^\top v$ and $v$ are statistically independent.

Is this true?

Towards a positive answer to Question 2, we let $T$ be the Housholder matrix such that $Tv = v_0$. Note that $T$ is independent of $u$. Then $$ u^\top v = u^\top (T^\top v_0) = (Tu)^\top v_0. $$ If the answer to Question 1 is yes, then $Tu$ and $u$ are identically distributed, and $Tu$ is independent of $v$, and hence the answer to Question 2 is yes. (BTW, are Question 1 and Question 2 equivalent?)

Any comments or criticism will be appreciated. Thank you.

(A different but related question: $X$ is independent of $\mathcal{G}$, $f(X ,Y)$ is independent of $Y$, $Y$ is $\mathcal{G}$-measurable, then $f(X,Y)$ is independent of $\mathcal{G}$?)

Answer 1

Let $X$ be a standard Gaussian in $\mathbb{R}^n$. In spherical coordinates, this may be expressed as $(R,u)$ where $R \in [0,\infty)$ and $u$ are independent, and $u$ is uniform on $\mathbb{S}^{n-1}.$

Let $\mathbf T$ is a random orthogonal matrix independent of $X$, with law $\mu$. Notice that $|\det(T)| = 1$ for any orthogonal matrix. So, using Fubini's theorem a couple of times, and a change of coordinates $$P(\mathbf{T}X \in A) = \int_T\int_{x \in T^{-1}(A)} e^{-\|Tx\|^2/2}\mathrm{d}x\,\mu(\mathrm{d}T) = \int_T\int_{y \in A} e^{-\|y\|^2/2}\,\mathrm{d}y\,\mu(\mathrm{d}T)\\ = \int_{x \in A} e^{-\|x\|^2/2}\mathrm{d}x = P(X \in A),$$ and so $\mathbf{T}X$ is also a standard Gaussian. But $\mathbf{T} (R,u) = (R, \mathbf{T} u),$ since $\mathbf{T}$ is norm preserving, so it follows that $\mathbf{T}u$ has the same law as $u$.

Finally, notice that the above calculation was completely agnostic to the law of $T$. This implies that if I replaced the law of $T$ with something else - for instance a conditional law given that $T$ lies in some set, I'll get the same calculation out. Of course this means that $$ P(\mathbf{T} X \in A| \mathbf{T} \in \tau) = P(X \in A) = P(\mathbf{T}X \in A).$$ It follows that $\mathbf{T}X,$ and thus $\mathbf{T}u,$ are indepdendent of $\mathbf{T}$.

Answer 2

The answer to question #1 is yes. Here's the proof.

Let $\mu$ be the uniform distribution on $S^{n-1}$, and let $\nu$ be the distribution for $T$ on $O(n)$. Given any measurable set $A\subseteq S^{n-1}$, let $$ A' = \{(u,T)\in S^{n-1}\times O(n) \mid T(u)\in A\} $$ and let $I_{A'}$ be the indicator function for $A'$. Then $$ P(v\in A) = (\mu\times \nu)(A') = \int_{S^{n-1}\times O(n)} I_{A'} d(\mu\times \nu) = \int_{O(n)} \int_{S^{n-1}} I_{A'}(u,T)\,\mu(du)\,\nu(dT) $$ But for any $T\in O(n)$, we have $$ \int_{S^{n-1}} I_{A'}(u,T)\,\mu(du) = \mu(T^{-1}(A)) = \mu(A) $$ where the last equality follows from the fact that $\mu$ is invariant under the action of $O(n)$. We conclude that $$ P(v\in A) = \int_{O(n)} \mu(A)\,\nu(dT) = \mu(A). $$ This holds for every measurable set $A\subseteq S^{n-1}$, so the distribution for $v$ is also $\mu$.

For independence of $v$ and $T$, let $A$ be any measurable subset of $S^{n-1}$, let $A'$ be as above, let $B$ be any measurable subset of $O(n)$, and let $B'=S^{n-1}\times B$. Then \begin{align*} P(v\in A\text{ and }T\in B) &= \int_{S^{n-1}\times O(n)} I_{A'} I_{B'}\, d(\mu\times \nu) \\[6pt] &= \int_{O(n)}\int_{S^{n-1}} I_{A'}(u,T)\, I_{B'}(u,T) \,\mu(du)\,\nu(dT) \\[6pt] &= \int_{O(n)}\int_{S^{n-1}} I_{A'}(u,T)\, I_{B}(T) \,\mu(du)\,\nu(dT) \\[6pt] &= \int_{O(n)} I_B(T) \int_{S^{n-1}} I_{A'}(u,T) \,\mu(du)\,\nu(dT) \\[6pt] &= \int_{O(n)} I_B(T) \,\mu(A)\,\nu(dT) \\[6pt] &= \mu(A)\, \nu(B) \\[6pt] &= P(v\in A)\,P(T\in B). \end{align*}