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activatedgeek t1_jbfjafv wrote

From (1), it looks like V(x_t) is the conditional variance of x_t given x_{t-1} (for the forward process defined by q).

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[deleted] t1_jbhjv7j wrote

I don’t get the last question when V(x_t) = 1 means that beta_t = 1

Why the confusion?

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eugene129 OP t1_jbwdfl2 wrote

As fas as I know, N(Xt ; ... , BtI) means that the V(Xt) = Bt, But if it is so, the equation two equation in the picture seems to be contradictory.

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rkstgr t1_jbia52h wrote

First of all, beta_t is just some predefined variance schedule (in literature often linear interpolated between 1e-2 and 1e-4) and it defines the variance of the noise that is added at step t. What you have in (1) is the variance of sample x_t which does not have to be beta_t.

What does hold for large t is var(x_t)=1 as our sample converges to ~ Normal Gaussian with mean 0 and var 1.

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eugene129 OP t1_jbkfrsr wrote

Hello, thanks for your reply. So... N(Xt ; ... , BtI) doesn't mean that the V(Xt) = Bt ?

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rkstgr t1_jc9sf5c wrote

Exactly, because you have x_t-1 with some (unknown, data/normalisation dependent) variance and you add noise with variance \beta to get x_t.

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[deleted] t1_jblfd4j wrote

[removed]

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eugene129 OP t1_jbwd8i7 wrote

So... N(Xt ; ... , BtI) doesn't mean that the V(Xt) = Bt ?

1