1
$\begingroup$

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

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

by The squared-norm of the projection of a Gaussian vector onto an independent $d$-dimensional subspace is a $\chi^2_{d}$. Also,

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

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.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.