New answers tagged proofs
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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