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3

"Modern" Guarantees for Feed-Forward Neural Networks My answer will complement nbro's above, which gave a very nice overview of universal approximation theorems for different types of commonly used architectures, by focusing on recent developments specifically for feed-forward networks. I'll try an emphasis depth over breadth (sometimes called ...

1

I will refer to $\mathcal T^{\pi}$as $\mathcal T$ and $P^{\pi}$ as $P$ for notational simplicity \begin{align} (\mathcal{T})^{n+1} Q &= \mathcal{T}(\mathcal{T}(...(\mathcal{T}(Q))))\\ &= r + \gamma P(r + \gamma P(...(r + \gamma P Q)))\\ &= r + r\sum_{i=1}^{n} \gamma^i P^i + \gamma^{n+1} P^{n+1} Q \end{align} \begin{align} \mathcal{T}_{\lambda}Q &...

1

Your bias formula is a bit incorrect. You should subtract a true value, not an estimator. $\mathrm{b}(\hat{\theta}) \stackrel{\text { def }}{=} \mathbb{E}[\hat{\theta}]-\theta$ $\mathrm{b}\left(\max _{a} Q\right)=\mathbb{E}\left[\max _{a} Q\right]-\max _{a} q=\mathbb{E}\left[\max _{a} Q\right]-\max _{a} \mathbb{E}[Q]$ $\mathbb{E}\left[\max _{a} Q\right] \... 2 This is very easy to prove. Let's first prove that, if$\hat{y}_k = y_k$, then the$E = 0\$. I will leave all steps, so that it's super clear. \begin{align} E &=\frac{1}{2}\sum_k(\hat{y}_k - y_k)^2 \\ &=\frac{1}{2}\sum_k(y_k - y_k)^2\\ &=\frac{1}{2}\sum_k(0)^2\\ &=\frac{1}{2}\sum_k 0\\ &=\frac{1}{2} 0\\ &=0\\ \end{align} To prove the ...

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