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I'm reading the paper "Optimizing Millions of Hyperparameters by Implicit Differentiation". The key contribution of the paper is to show that you can replace optimizing through the optimization process/path by using implicit gradients to effectively optimize hyper-parameters.

Problem:

I don't quite understand how the optimization path fits into the derivation. For example, per their derivation, let's assume that we want to optimize some hyper-paramaeters $\lambda$, where $w^{*}(\lambda)$ is the locally optimal base-parameters using some value of $\lambda$. Furthermore, let us assume we are using some vanilla gradient descent style optimization process for both the inner and outer optimization.

\begin{align} \lambda_{new} &= \lambda - \frac{\partial}{\partial \lambda}\mathcal{L}(w^{*}(\lambda)) \\ &= \lambda - \frac{\partial}{\partial w^{*}(\lambda)} \mathcal{L}(w^{*}(\lambda)) \cdot \frac{\partial}{\partial \lambda} w^{*}(\lambda) \end{align}

Where $\frac{\partial}{\partial\lambda}w^{*}(\lambda)$ can be replaced with some closed-form solution as follows using the implicit gradients theorm:

\begin{align} \frac{\partial}{\partial\lambda}\bigg[\frac{\partial}{\partial w} \mathcal{L}(w(\lambda), \lambda) \bigg] & = 0 \\ \frac{\partial}{\partial w}\bigg[\frac{\partial}{\partial\lambda} \mathcal{L}(w(\lambda), \lambda) \bigg] & = 0 \\ \frac{\partial}{\partial w}\bigg[\frac{\partial\mathcal{L}}{\partial w} \cdot \frac{\partial w}{\partial\lambda} + \frac{\partial\mathcal{L}}{\partial\lambda} \bigg] & = 0 \\ \frac{\partial^{2}\mathcal{L}}{\partial w \partial w^{T}} \cdot \frac{\partial^{2} w}{\partial\lambda\partial w^{T}} + \frac{\partial^{2}\mathcal{L}}{\partial\lambda\partial w^{T}} & = 0 \\ \frac{\partial^{2}\mathcal{L}}{\partial w \partial w^{T}} \cdot \frac{\partial w}{\partial\lambda} + \frac{\partial^{2}\mathcal{L}}{\partial\lambda\partial w^{T}} & = 0 \\ \frac{\partial^{2}\mathcal{L}}{\partial w \partial w^{T}} \cdot \frac{\partial w}{\partial\lambda} & = - \frac{\partial^{2}\mathcal{L}}{\partial\lambda\partial w^{T}} \\ \frac{\partial w}{\partial\lambda} & = - \bigg[\frac{\partial^{2}\mathcal{L}}{\partial w \partial w^{T}}\bigg]^{-1} \cdot \frac{\partial^{2}\mathcal{L}}{\partial\lambda\partial w^{T}} \end{align}

However, I don't understand how $w^{*}(\lambda)$ gets transformed into the following:

\begin{align} \frac{\partial}{\partial \lambda} w^{*}(\lambda) &= \cdots \\ &= \cdots \\ &= \frac{\partial}{\partial\lambda}\bigg[\frac{\partial}{\partial w} \mathcal{L}(w(\lambda), \lambda) \bigg] \end{align}

Intuitively, $w^{*}(\lambda)$ is the result of the optimization path taking $k$ gradient steps:

\begin{align} w_{0} &= \cdots\\ w_{1} &= w_{0} - \frac{\partial}{\partial w_{0}} \mathcal{L}(w_{0}(\lambda)) \\ w_{2} &= w_{1} - \frac{\partial}{\partial w_{1}} \mathcal{L}(w_{1}(\lambda)) \\ \vdots \\ w^{*} &= w_{k-1} - \frac{\partial}{\partial w_{k-1}} \mathcal{L}(w_{k-1}(\lambda)) \\ \end{align}

Which we can substitute back into our equation as follows

$$\frac{\partial}{\partial \lambda} w^{*}(\lambda) = \frac{\partial}{\partial \lambda} \bigg[w_0 - \sum^{k}_{i=1}\frac{\partial}{\partial w_{k-1}} \mathcal{L}(w_{k-1}(\lambda))\bigg]$$

However, this isn't the quantity we want since

$$\frac{\partial}{\partial\lambda}\bigg[\frac{\partial}{\partial w} \mathcal{L}(w(\lambda), \lambda) \bigg] \neq \frac{\partial}{\partial \lambda} \bigg[w_0 - \sum^{k}_{i=1}\frac{\partial}{\partial w_{k-1}} \mathcal{L}(w_{k-1}(\lambda))\bigg]$$

Can anyone help resolve this issue for me?

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  • $\begingroup$ Could you please put your specific question in the title? "Differentiating through Optimization Paths" is quite general. It seems that your question is related to a discrepancy between their derivation and your derivation. So, your title should probably ask about that. $\endgroup$
    – nbro
    May 10, 2022 at 7:31
  • $\begingroup$ Thanks for the suggestion, I have updated the title to better reflect the contents of the question. $\endgroup$
    – Decadz
    May 11, 2022 at 11:47

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