Exercise 12.10 - Efficiently evaluating the PPCA density

Answers

We firstly complete the proof of Theorem 12.2.2, this can be done by plugging (12.61) back into (12.60). Recall that the MLE is where 𝐖 satisfies (using 𝐀 1 𝐀 = 𝐀 1 𝐀 1 ):

( 𝐈 𝐒 𝐂 1 ) 𝐂 1 𝐖 = 0 .

With:

𝐖 = 𝐐 ( λ 1 σ 2 λ L σ 2 0 0 ) ,

we have:

𝐂 1 = 𝐐 ( λ 1 1 0 0 0 λ L 1 0 0 0 0 1 σ 2 0 0 1 σ 2 ) 𝐐 T .

In which 𝐐 is the sigular vectors/eigen vectors for 𝐒 . This would reduce ( 𝐈 𝐒 𝐂 1 ) 𝐂 1 𝐖 to zero, hence complete the proof. For MLE of the σ 2 , using the trace trick.

Now for p ( 𝐱 | 𝐖 ~ , σ ~ 2 ) , it is a normal distribution with zero mean and covariance whose last L D components are uniformly σ ~ 2 .

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2021-03-24 13:42
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