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

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.

• How would you randomly choose an orthogonal matrix? Feb 9, 2021 at 15:00
• There is an example in Question 2. Here I do not make any assumption on the distribution of the random orthogonal matrix except that it is statistically independent of $u$. Thank you.
– Nuno
Feb 9, 2021 at 15:01

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 $$\mathbf{T}$$.
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*}