Distribution of norm of fixed vector projected onto a Gaussian subspace

Let $$\Sigma \in \mathbb{R}^{m \times m}$$, $$\Theta_0 \in \mathbb{R}^{k \times m}$$, $$v = \Theta_0 \beta \in \mathbb{R}^k$$ with $$\| v \| = 1$$ and $$\Theta \sim \mathcal{N}(\Theta_0, \mathrm{Id}_k \otimes \Omega)$$. That is, $$\Theta - \Theta_0 \in \mathbb{R}^{k \times m}$$ consists of i.i.d. draws from a centered multivariate Gaussian with covariance $$\Sigma$$. What can be said about the distribution of

$$X := v^T \Theta (\Theta^T \Theta)^{-1} \Theta^T v = v^T P_\Theta v = \| P_\Theta v \|^2$$

?

One can show that if $$v \in \mathbb{R}^k$$ and $$\Theta_0 = 0$$ (or $$v \perp \Theta_0$$), then $$X \sim \mathrm{Beta}(m/2, (k-m)/2)$$.

Proof:

Let $$Q$$ be uniform in $$O(k)$$, the group of orthogonal (orthonormal) matrices in $$\mathbb{R}^{m \times m}$$. Let $$t^2 \sim \chi^2(k)$$. Then, $$t \cdot Qv \sim \mathcal{N}(0, \mathrm{Id})$$ is independent of $$Q \Theta$$ (independence needs $$\Theta_0=0$$).

Calculate

$$t^2 \cdot v^T P_\Theta v = (t \cdot Qv)^T P_{Q \Theta} (t \cdot Qv) \sim \chi^2(m),$$

$$t^2 = t^2 \cdot \|v \|^2 = t^2 \cdot ( v^T P_\Theta v + v^T M_\Theta v) = (t \cdot Qv)^T P_{Q \Theta} (t \cdot Qv) + (t \cdot Qv)^T M_{Q \Theta} (t \cdot Qv) ,$$

where $$M_\Theta = \mathrm{Id} - P_\Theta$$ and $$v^T M_\Theta v = (t \cdot Qv)^T M_{Q \Theta} (t \cdot Qv) \sim \chi^2(k-m)$$ independent of $$(t \cdot Qv)^T P_{Q \Theta} (t \cdot Qv)$$ by the same argument as above.

Thus,

$$v^T P_\Theta v = \frac{(t \cdot Qv)^T P_{Q \Theta} (t \cdot Qv)}{(t \cdot Qv)^T P_{Q \Theta} (t \cdot Qv) + (t \cdot Qv)^T M_{Q \Theta} (t \cdot Qv)} \sim \frac{a}{a + b}$$

for $$a \sim \chi^2(m)$$ and $$b \sim \chi^2(k-m)$$ independent.