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In online convex optimization, the regret of an algorithm $\mathcal{A}$ as defined in Introduction to Online Convex Optimization (Page 5) is: $$ regret_T(\mathcal{A}) = \sup_{\{f_1,...,f_T\}} \sum_{t=1}^{T}f_t(x_t) − \min_{x \in \mathcal{K}} \sum_{t=1}^{T} f_t (x) $$ where at iteration t, the online player chooses a decision $x_t \in \mathcal{K}$ and $\mathcal{K}$ is a convex set in $\mathbb{R}^n$. Let set $\mathcal{F}$ consists of bounded family of cost functions available to the adversary and $f_t \in \mathcal{F}: \mathcal{K} \rightarrow \mathbb{R}$ is the convex cost function reveled after player chooses decision $x_t$.

As far as I have understood, the second term is the sum of $T$ convex functions, Hence the overall sum is a convex function in $x$ and we set the minimum of this convex function as a baseline for our algorithm $i.e.$ the performance of an algorithm is analyzed with respect to this minimum.

But as far as I can see the regret can still be negative because an algorithm can still play by always choosing the decision $x_t = \min_{x \in \mathcal{K}} f_t (x)$. which would result in a non-positive regret.

Am I mistaken somewhere or Is negative regret allowed in such settings?

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The sup is over the whole expression

$$regret_T(\mathcal{A}) = \sup_{\{f_1,...,f_T\}} \left \{ \sum_{t=1}^{T}f_t(x_t) − \min_{x \in \mathcal{K}} \sum_{t=1}^{T} f_t (x) \right \} $$,

and by definition you have that the second terms is smaller than the first. So the difference between the two terms in brackets will always be positive.

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  • $\begingroup$ Can you elaborate on the claim: "by definition you have that the second terms is smaller than the first"? $\endgroup$ – durdi Jul 1 at 10:24
  • $\begingroup$ The claim stems by the elementary observation that for any function $f$ and any $x \in dom(f)$ you have that $\min_x f(x) < f(x)$. $\endgroup$ – Apprentice Jul 1 at 13:20
  • $\begingroup$ But what you are saying is that, $\min_{x \in \mathcal{K}} \sum_{t=1}^T f_t(x) \leq \sum_{t=1}^T f_t(y)$, for any fixed $y \in \mathcal{K}$. However, in the first sum of the regret definition (i.e. $\sum_{t=1}^T f_t(x_t)$), the online player may choose a different action $x_t$ for each time $t$. $\endgroup$ – durdi Jul 1 at 14:28
  • $\begingroup$ I honestly don't get what you mean. The loss $f_t(x_t)$ for any action $x_t$ that the online player can choose will always be larger than or equal to $min_{x} f_t(x)$. There is not much more to add to this very simple observation. $\endgroup$ – Apprentice Jul 1 at 15:22
  • $\begingroup$ What I am saying is that $\min_{y \in \mathcal{K}} f_t (y) \leq f_t(x_t)$ does not imply that $\min_{x \in \mathcal{K}} \sum_{t=1}^{T}f_t(x) \leq \sum_{t=1}^{T}f_t(x_t)$, for all action sequences $x_1,\dots,x_T$. To see this, define $x^*_t := \text{arg}\min_{x \in \mathcal{K}} f_t(x)$ and $x^* := \text{arg}\min_{x \in \mathcal{K}} \sum_{t=1}^{T} f_t(x)$. By the definition of $x^*_t$, we have that $f_t (x^*) \geq f_t(x^*_t)$, for any $t$. Summing this inequality over $t = 1,\dots,T$, we have $\sum_{t=1}^{T} f_t(x^*) = \min_{x \in \mathcal{K}} \sum_{t=1}^{T}f_t(x) \geq \sum_{t=1}^{T}f_t(x^*_t)$. $\endgroup$ – durdi Jul 1 at 16:32

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