I see in this document on variational inference that (p.5):

$L_k = \int q(z_k)E_{-k}[\log(p|z_{-k},x)]dz_k - \int q(z_k)\log q(z_k)dz_k$

It is stated that taking the derivative with respect to $q(z_k)$ results in:

$\frac{dL_k}{dq(z_k)}=E_{-k}[\log(p|z_{-k},x)] - \log q(z_k) - 1$

It looks to me like it has been argued that:

$\frac{d}{dq(z_k)}\int q(z_k)E_{-k}[\log(p|z_{-k},x)]dz_k = \frac{d}{dq(z_k)}[q(z_k)E_{-k}[\log(p|z_{-k},x)]] = E_{-k}[\log(p|z_{-k},x)]$


$\frac{d}{dq(z_k)}[- \int q(z_k)\log q(z_k)dz_k]=-\frac{d}{dq(z_k)}[q(z_k)\log q(z_k)]= -\log q(z_k) - 1 $

I can't see why this would work - it looks like there's been a bonus differentiation by $dz_k$. Can someone fill me in on why this is valid?


It works because these are functional derivatives.


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