MetaLearning-Standford-Lecture3

MetaLearning Learning Note - 2

Stanford CS330: Multi-Task and Meta-Learning, 2019 | Lecture 3 - Optimization-Based Meta-Learning

• Recap the probabilistic formulation of meta-learning
• general recipe of meta-learning algorithm
• black box adaption appraoches
• optimization-based meta-learning algorithm

Recap

meta-learning to learn the $\theta$ then to learn $\phi$ in train dataset.

We will use omniglot dataset. 1623 characters 50 different alphabets.

In her's introduction to the meta-supervised learning:

Input is the $D_{tr}, x_{test}, \theta$

The steps to design a meta-learning algorithm

1. choosing a form of $p(\phi | D^{TR}_i,\theta)$
2. choosing how to optimizae $\theta$

Treat $p(\phi | D^{TR}_i,\theta)$ as a inference problem?

Black-Box adaption

we ca see we use $D^{tr}_i$ to train data and generate the results parameters $\phi$

In my opinion, they train $\theta$ to generate $\phi_i$ then use the use $\phi_i$ to generate $y^{test}$ we need to let this loss to get minimized!

Yes the step of black box training:

1. sample task $T_i$
2. sample disjoint dataset $D^{tr}_i,D^{test}_i$
3. Compute $\phi \leftarrow f_{\theta}{D^{tr}_i}$
4. Update $\theta$ using $\Delta_{\theta}\mathbb{L}(\phi,D^{test})$

Challenge: Output all neural net parameters doest not seem scalable.

We only ouput sufficient statistics (SNAIL!) Oh I read this paper but I don't understand. Let me try to review this paper tommorrow.

The first Homework is omnigenet.

The proble of Black-Box adaption need a large number of meta-learning dataset. - data inefficent.

Optimiation Based Inference

Motivation: in the above problem, to let the $\theta$ become scalable, we only genereate several fixed parameter from our meta-learner. How can we genereate all parameters?

First we get the formula:

$$max_{\phi_i}logP(D_i^{tr}|\phi_i)+logp(\phi_i|\theta)$$

whye $\theta$ in the behind of the formula?

meta parameter are server as prior parameter. ( Recap: Fine-tune your parameters in the test data set!)

Pre-training can be taken as art more than others.

Our goal:

$$min_\theta\sum_{task\ i}$$

and MAML:

key idea: acquire $\phi_i through optimization$

1. Sample task $T_i$
2. Sample disjoint datasets $D^{tr}_i$,$D^{test}_i$ from $D_i$
3. Optimize $\phi_i \leftarrow \theta - \alpha \nabla_\theta L(\theta,D^{tr}_i)$
4. use $\phi$ from step 3's trainging to update $\theta$

Specificly, the phi is theta its self. And MAML summarized alll task's gradient to do a 'the whole gradient steps' to find the best value.

Shall we care the second order derivative?

Follow her formula:

$$\phi = u(\theta, D^{tr})\ d: total\ derivative\ and \bigtriangledown : partial derivative$$

$$min_\theta L(\phi,D^{test}) = min_\theta(u(\theta,D^{tr}),D^{test})$$

Let us see the GD process of this formula.

$$\frac{d}{d_\theta}L(\phi,D^{test}) = \bigtriangledown_\phi * L(\phi,D^{test}) \circ d_{theta} u(\theta,D^tr) _{\phi = u(\theta,D^tr)}$$

I consider the first derivative for $\phi$ is second derivative for $\theta$?

Let $u(\theta,D^tr) = \theta - \alpha * d_\theta * L(\theta, D^tr)$

$d_\theta u(\theta, D^tr) = I - \alpha * d_\theta^2 * L(\theta, D^tr)$ is a hessian matrix for this problem.

I am a little confused about the $\phi \& \theta$ who is the meta-parameters?

So it can repeate computing the higher-order derivations.