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I am currently trying to understand a paper on bayesian neural networks whereby the authors use a bayes by backprop approach to learn weight uncertainties in the neural networks.

I am trying to understand the derivation for proposition 1 in the paper. Particularly, I am not sure how

$$\frac{\partial}{\partial\theta}\int f(\boldsymbol{w},\theta)q(\epsilon)d\epsilon = E_{q(\epsilon)}[\frac{\partial f(w,\theta)}{\partial w} \frac{\partial w}{\partial \theta} + \frac{\partial f(w,\theta)}{\partial \theta}]$$

I am not sure why there is an additional $\frac{\partial f(w,\theta)}{\partial \theta}$ inside the expectation ? since I thought $\frac{\partial f(w,\theta)}{\partial w} \frac{\partial w}{\partial \theta} = \frac{\partial f(w,\theta)}{\partial w}$

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  • $\begingroup$ isn't the first term inside the expectation quantity on the left hand side equal to the right hand side term by chain rule? $\endgroup$
    – calveeen
    Commented Dec 7, 2020 at 1:36

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Okay I think this is why there is an additional term. $f$ is a function of $\textbf{w}$ and $\theta$. Since $\textbf{w}$ is composed of $\theta$, we apply chain rule to obtain $\frac{\partial f}{\partial w}\frac{\partial w}{\partial \theta}$. Since $f$ is also a function of $\theta$. we also differentiate with respect to $\theta$, yielding $\frac{\partial f}{\partial \theta}$

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  • $\begingroup$ yes (+1), it's the multivariable chain rule. And the reason why you can swap the order of integration and differentiation is the proposition on slide 7 here $\endgroup$
    – Taylor
    Commented Dec 7, 2020 at 19:53

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