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@ -18,7 +18,7 @@ The important thing is that $\ell_1 , \ell_2, ...$ is a sequence of \textbf{conv
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\\\\
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In general we define the regret in this way:
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$$
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R_T(u) \ = \ \frac{1}{m} \ \sum_{t=1}^{T} \ell_t(w_t) - \frac{1}{T} \ \sum_{t=1}^{T} \ell_t(u_t)
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R_T(u) \ = \ \sum_{t=1}^{T} \ell_t(w_t) - \sum_{t=1}^{T} \ell_t(u_t)
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$$\\
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The Gradiant descent is one of the simplest algorithm for minimising a convex function. We recall the iteration did by the algorithm:
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$$
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@ -90,6 +90,115 @@ $
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\caption{}
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%\label{fig:}
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\end{figure}\\
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Now we can apply this results to our problem: in particular I rearrange the factors
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$$
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f(w) - f(u) \leq \nabla f(w)^T \, (w-u)
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$$
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This is Ok for $f$ convex and differentiable.
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\\
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I know that: $u-w^T\nabla^2 f(\xi) \, (u-w) \geq 0 $ because f is convex.
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\\
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$$
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\ell_t(w_t) - \ell_t(u) \leq \nabla \ell_t (w_t)^T \, (w_t-u) \qquad \textbf{ Linear Regret}
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$$\\
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How do we proceed? \\
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The first step of the algorithm is : $w'_{t+1} = w_t - \eta_t \nabla \ell_t(w_t) \qquad \eta_t = \frac{\eta}{\sqrt[]{t}}$
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$$
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= - \frac{1}{\eta_t} \, (w'_{t+1} - w_t )^T \, (w_t-u) \ = \
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\frac{1}{\eta_t}\left( \frac{1}{2} \| w_t -u \|^2 - \frac{1}{2} \| w'_{t+1} - u \|^2 + \frac{1}{2} \|w_{t+1} - w_t \|^2 \right) \ \leq
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$$
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$$
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\leq \
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\frac{1}{\eta_t}\left( \frac{1}{2} \| w_t -u \|^2 - \frac{1}{2} \| w_{t+1} - u \|^2 + \frac{1}{2} \|w'_{t+1} - w_t \|^2 \right)
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$$
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$w'$ disappear and add minus sign. I am saying that $\| w_{t+1} -u \| \leq \| w'_{t+1} - u\|$
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.3\linewidth]{../img/lez15-img4.JPG}
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\caption{}
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%\label{fig:}
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\end{figure}\\
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So is telling us that $w_{t+1}$ is closer to $u$ than $w'_{t+1}$
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\\This holds since the ball is convex.
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\\\\
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Now we go back adding and subtracting $\pm \frac{1}{2 \, \eta_{t+1}} \| w_{t+1} -u \|^2$
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min. 38:38
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$$
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= \red{\frac{1}{2 \, \eta_t} \| w_t - u\|^2 -
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\frac{1}{2 \, \eta_{t+1}} \| w_{t+1} - u\|^2}
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-
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\blue{ $
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\frac{1}{2 \, \eta_{t}} \|w_{t+1}-u \|^2
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+
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\frac{1}{2 \, \eta_{t+1}} \|w_{t+1}-u \|^2 $}
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+
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\frac{1}{2 \, \eta_{t}} \|w_{t+1}- w_t \|^2
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$$
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We group the 1,2 and 3,4 elements and sum them up. \\
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$$
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R_T(U) \ = \ \sum_{t=1}^{T} \left( \ell_t(w_t) - \ell_t(u) \right) \leq
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$$
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This is a \textbf{telescopic sum}: $a_1-a_2+a_2-a_3+a_3-a_4+a_t -a_t+1$ and everything in the middle cancel out and remains first and last terms.
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\\
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$$
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\leq \frac{1}{2 \, \eta_t} \|w_1-u\|^2 - \frac{1}{\eta_{T+1}} \| w_{T+1} -u \| ^2 + \frac{1}{2} \sum_{t=1}^T \| w_{t+1} - u \|^2 \left( \frac{1}{\eta_{t+1}} - \frac{1}{\eta_t}\right) + \frac{1}{2} \sum_{t=1}^T \frac{\| w'_{t+1} -w_t \|^2}{\eta_t}
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$$
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where $w_1 = 0 $ \quad and \quad $\| w_{t+1} - u \|^2 \leq 4 \, U^2$ \quad and \quad
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$
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\| w'_{t+1} -w_t \|^2 = \eta_t^2 \| \nabla \ell_t (w_t) \|^2
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$\\
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We know that $\eta_t = \frac{\eta}{\sqrt[]{t}}$ \qquad so $\eta_1 = \frac{\eta}{\sqrt[]{1}} = \eta$
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$$
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R_T(U) \ \leq \ \frac{1}{2 \, \eta} U^2
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\red{
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- \frac{1}{2 \, \eta_{T+1}} \|w_{T+1} - U \|^2
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}
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+2 \, U^2 \, \sum_{t=1}^{T-1} \left(\frac{1}{\eta_t} - \frac{1}{\eta_t}\right)+
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$$
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$$
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\red{
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+ \frac{\| w_{T+1} -U \|^2}{2 \eta_{T+1}} - \frac{\| w_{T} -U \|^2}{\eta_T} }
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+ \frac{1}{2} \sum_{t=1}^T \eta_t \| \nabla \ell_t(w_t) \|^2
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$$
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where \bred{red values} cancel out.
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\\
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I assume that square loss is bounded by some number $G^2$: $ \| \nabla \ell_t(w_t) \|^2 \ \leq \ G^2 $
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\\
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Also, it's a telescopic sum again and all middle terms cancel out.
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\\
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$$
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\max_t \| \nabla_t(w_t) \|^2 \leq G
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$$
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$$
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R_T(U) \leq \red{\frac{1}{2 \, \eta} U^2} + 2\, U^2 \left( \frac{1}{\eta_{T}} - \red{ \frac{1}{\eta_1} }\right) + \frac{G^2}{2} \, \eta \sum_{t=1}^T \frac{1}{\sqrt[]{t}} \qquad \eta_t = \frac{1}{\sqrt[]{t}}
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$$
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where \bred{red values} cancel out.
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\\
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Now how much is this sum $ \sum_{t=1}^T \frac{1}{\sqrt[]{t}}$?
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\\
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It is bounder by the integral
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$
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\leq \int_{1}^{T} \frac{dx}{\sqrt[]{x}} \leq 2 \, \sqrt[]{T}
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$
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$$
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R_T(U) \ \leq \ \frac{2 \, U^2 \sqrt[]{T}}{\eta} + \eta \, G^2 \sqrt[]{T} \ = \ \left( \frac{2\, U^2}{\eta} + \eta \, G^2 \right) \, \sqrt[]{T} $$
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$ \eta = \frac{U}{G} \sqrt[]{2}
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$
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\\
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So finally:
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$$
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\frac{1}{T} \sum_{t=1}^T \ell_t(w_t) \leq \min_{\|U\| \leq U} \frac{1}{T} \sum_{t=1}^T \ell_t(u) + U \, G \ \sqrt[]{\frac{8}{T}}
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$$
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$$
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R_T(U) = \frac{1}{T} \sum_{t=1}^T \left( \ell_t (w_t) - \ell_t(u) \right) \qquad \forall u : \|u\| \leq U : R_T(U) = O \left(\frac{1}{\sqrt[]{T}}\right)
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$$
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Basically my regret is gonna go to 0.
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\\\\
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For $ERM in H$ where $| H| < \infty$, variance error vanishes at rate $\frac{1}{\sqrt[]{m}}$
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\\\\
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The bound $U \, G^2 \, \sqrt[]{\frac{8}{T}}$ on regret holds for any sequence $\ell_1, \ell_2, ...$ of convex and affordable losses, If $\ell_t(w) = \ell(w^T \, x_t, y_t)$ then the bound holds for any sequence of data points $(x_1,y_1), (x_2, y_2)..$
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\\
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This is not a statistical assumption but mathematical so stronger.
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\end{document}
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