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The conventional model for probabilistic principal component analysis has a standard normal latent $\vec{y}$ and a loading matrix $\Lambda$:

$P(\vec{y}) \sim N(\vec{0}, I)$, $P(\vec{x}|\vec{y}) \sim N(\Lambda \vec{y}, \psi I)$

An alternative would be to draw $\vec{y}$ from a normal with diagonal covariance (say $\Sigma$, and then restrict $\Lambda$ to have orthonormal columns:

$P(\vec{y}) \sim N(\vec{0}, \Gamma), P(\vec{x}|\vec{y}) \sim N(\Lambda \vec{y}, \psi I)$ with $\Gamma_{ij}=0$ for $i \neq j$ and $\Lambda^T \Lambda = I$

where $\Gamma_{ij} =\Gamma_{ji} =0$ and $\Lambda^T \Lambda=0$.

Question: Show that this alternative model is equivalent to the standard one in the sense that it can model exactly the same set of possible marginal distributions.

I really have no idea what I am being asked to do here. I know that there is some equivalence of PCA models under rotation and scaling - should I be working in that direction, or am I supposed to integrate out the latent variable and show I get the same distribution? Or something else entirely? Help please.

lead to exactly the same marginal distributions.

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Here are some hints. I will rename $\Lambda$ in the first model to $A$ to avoid confusion with the one that is to have orthonormal columns.

As you seem to have deduced, the distributions we get for $x$ are $N(0,\psi I + \Lambda \Gamma \Lambda^T)$ in one model and $N(0,\psi I + AA^T)$ in the other model.

Therefore, it suffices to show that matrices of the form $\Lambda \Gamma \Lambda^T$ (with $\Lambda$ having orthonormal columns and $\Gamma$ diagonal) and $AA^T$ are equivalent (either can be written in the other's form).

To help show that $\Lambda \Gamma \Lambda^T$ can be written as $AA^T$, utilise the square root of $\Gamma$. To help show that $AA^T$ can be written as $\Lambda \Gamma \Lambda^T$, utilise the spectral theorem and drop eigenvectors corresponding to $0$ eigenvalues (and drop these eigenvalues from the diagonal matrix from the spectral theorem).

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