Answers.Body: <p><span class="math-container">$\nabla_{\theta_{k-1}} \theta_k$</span> is gradient of <span class="math-container">$\theta_k$</span> with respect to <span class="math-container">$\theta_{k-1}$</span>, it follows chain rule as noted in the side comment in the image. <span class="math-container">$\nabla_{\theta} \mathcal L(\theta_k)$</span> is also not a Hessian but a gradient vector.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-01-13T20:36:18.397Z
Answers.LastActivityDate: 2021-01-13T20:36:18.397Z
Answers.UserDisplayName: Brale
Body: <p>I am attempting to fully understand the explicit derivation and computation of the Hessian and how it is used in MAML.  I came across this blog: <a href="https://lilianweng.github.io/lil-log/2018/11/30/meta-learning.html" rel="nofollow noreferrer">https://lilianweng.github.io/lil-log/2018/11/30/meta-learning.html</a>.</p>
  <p>Specifically, could someone help to clarify this for me: is this term in the red box literally interpreted as the gradient at <span class="math-container">$\theta_{k-1}$</span> multiplied by the <span class="math-container">$\theta_k$</span>?</p>
  <p><a href="https://i.sstatic.net/j6BjW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/j6BjW.png" alt="enter image description here" /></a></p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2021-01-14T00:26:23.597Z
Comments.Text: [Here](https://ai.stackexchange.com/q/18516/2444) is a related question.
Comments.UserDisplayName: nbro
Tags: <math><notation><meta-learning><gradient><model-agnostic-meta-learning>
Title: What is $ \nabla_{\theta_{k-1}} \theta_{k}$ in the context of MAML?
UserDisplayName: Blake Camp
ThreadText: # What is $ \nabla_{\theta_{k-1}} \theta_{k}$ in the context of MAML?
  
  (Asked by: username_0 on 2021-01-13)
  
  I am attempting to fully understand the explicit derivation and computation of the Hessian and how it is used in MAML. I came across this blog: <https://lilianweng.github.io/lil-log/2018/11/30/meta-learning.html>.
  
  Specifically, could someone help to clarify this for me: is this term in the red box literally interpreted as the gradient at $\theta\_{k-1}$ multiplied by the $\theta\_k$?
  
  [![enter image description here](https://i.sstatic.net/j6BjW.png)](https://i.sstatic.net/j6BjW.png)
  
  **Tags:** math, notation, meta-learning, gradient, model-agnostic-meta-learning
  
  > **Comments:**
  > **username_1**: [Here](https://ai.stackexchange.com/q/18516/2444) is a related question.  
  
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  ## 1 answers
  
  ### Answer by username_2 on 2021-01-13. Score: 0 | Accepted answer
  
  $\nabla\_{\theta\_{k-1}} \theta\_k$ is gradient of $\theta\_k$ with respect to $\theta\_{k-1}$, it follows chain rule as noted in the side comment in the image. $\nabla\_{\theta} \mathcal L(\theta\_k)$ is also not a Hessian but a gradient vector.
  
  
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