Given P(X|z) why is the NLL P(X) proportional to the squared Euclidean distance between f(z) and X

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In the paper Tutorial on Variational Autoencoders, the authors state that

Since $P(X\mid z)$ is an isotropic Gaussian, the negative log probability of $X$ is proportional squared Euclidean distance between $f(z)$ and $X$.

According to the paper, $$P(X) = \int P(X\mid z; \theta)P(z)\,\mathrm dz,$$ and $$P(X\mid z; \theta) = \mathcal N (X\mid f(z; \theta), \sigma^2 ∗ I).$$

I don't see the connection here in proving this.