SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Here, the inequality comes from the left inequality of Equation 4 and the fact that  $c \leq \frac{1}{\min_k w_k} \eta_t \zeta$ . If we also have that  $\eta_t \zeta \leq (1 - \alpha)^2 (w_i^t)^\alpha$ , then the last claim of Lemma A.1 gives us that
$$
\left(w _ {i} ^ {t + 1}\right) ^ {\alpha - 1} = - h _ {i} + c \leq \left(w _ {i} ^ {t}\right) ^ {\alpha - 1} + \eta_ {t} \zeta + m \eta_ {t} \zeta = \left(w _ {i} ^ {t}\right) ^ {\alpha - 1} + (m + 1) \eta_ {t} \zeta .
$$
$$
(w _ {i} ^ {t + 1}) ^ {\alpha - 1} = - h _ {i} + c \geq (w _ {i} ^ {t}) ^ {\alpha - 1} - \frac {\eta_ {t} \zeta}{w _ {i} ^ {t}} - \eta_ {t} \zeta = (w _ {i} ^ {t}) ^ {\alpha - 1} - \left(\frac {1}{w _ {i} ^ {t}} + 1\right) \eta_ {t} \zeta .
$$
Lemma A.1. Let  $-1 < a < 0$  and  $\mathbf{g} \in \mathbb{R}^m$ . There is a unique  $c \in \mathbb{R}$  such that  $\sum_{i}(g_i + c)^{1/a} = 1$ . Furthermore, let  $\mathbf{v} \in \Delta^m$  and  $\kappa \geq 0$ . Then:
- If  $g_{i} \leq v_{i}^{a} + \kappa$  for all  $i$ , then  $c \geq -\kappa$ .
- If  $g_{i} \geq v_{i}^{a} - \frac{\kappa}{v_{i}}$  for all  $i$ , then  $c \leq \frac{\kappa}{\min_{i} v_{i}}$ .
- And if, furthermore,  $\kappa \leq a^{2}v_{i}^{a + 1}$  for all  $i$ , then  $c \leq m\kappa$ .
Proof. Observe that  $\sum_{i}(g_{i} + c)^{1 / a}$  is a continuous, monotone decreasing function on  $c\in (-\min_i g_i,\infty)$ ; the range of the function on this interval is  $(0,\infty)$ . Therefore, there is a unique  $c\in (-\min_i g_i,\infty)$  such that the sum equals 1.
We now prove the first bullet. Since  $x^{1/a}$  decreases in  $x$  and  $g_i \leq v_i^a + \kappa$ , we have that
$$
1 = \sum_ {i} (g _ {i} + c) ^ {1 / a} \geq \sum_ {i} (v _ {i} ^ {a} + \kappa + c) ^ {1 / a}.
$$
Suppose for contradiction that  $c < -\kappa$ . Then  $v_{i}^{a} + \kappa + c < v_{i}^{a}$  for all  $i$ , so
$$
\sum_ {i} \left(v _ {i} ^ {a} + \kappa + c\right) ^ {1 / a} > \sum_ {i} \left(v _ {i} ^ {a}\right) ^ {1 / a} = \sum_ {i} v _ {i} = 1.
$$
This is a contradiction, so in fact  $c \geq -\kappa$ .
The first claim of the second bullet is analogous. Since  $x^{1/a}$  decreases in  $x$  and  $g_i \geq v_i^a - \kappa v_i$ , we have that
$$
1 = \sum_ {i} \left(g _ {i} + c\right) ^ {1 / a} \leq \sum_ {i} \left(v _ {i} ^ {a} - \frac {\kappa}{v _ {i}} + c\right) ^ {1 / a}. \tag {5}
$$
Suppose for contradiction that  $c > \frac{\kappa}{v_i}$  for every  $i$ . Then  $v_i^a - \frac{\kappa}{v_i} + c > v_i^a$  for all  $i$ , so
$$
\sum_ {i} \left(v _ {i} ^ {a} - \frac {\kappa}{v _ {i}} + c\right) ^ {1 / a} <   \sum_ {i} \left(v _ {i} ^ {a}\right) ^ {1 / a} = \sum_ {i} v _ {i} = 1.
$$
This is a contradiction, so in fact  $c \leq \frac{\kappa}{\min_i v_i}$ .
We now prove the second claim of the second bullet. To do so, we note the following technical lemma (proof below).
Lemma A.2. For  $-1 < a < 0$  and  $\kappa, c \geq 0$ , the function  $f(x) = (x^a - \frac{\kappa}{x} + c)^{1/a}$  is defined and concave at any value of  $x > 0$  such that  $a^2 x^{a+1} \geq \kappa$ .
Since for a general concave function  $f$  it holds that  $\frac{1}{m}\sum_{i = 1}^{m}f(x_i)\leq f\left(\frac{1}{m}\sum_{i = 1}^{m}x_i\right)$ , the following inequality follows from Lemma A.2:
$$
\sum_ {i} \left(v _ {i} ^ {a} - \frac {\kappa}{v _ {i}} + c\right) ^ {1 / a} \leq m \left(\left(\frac {1}{m}\right) ^ {a} - \kappa m + c\right) ^ {1 / a}.
$$
(Here we are using the fact that  $\sum_{i}v_{i} = 1$ .) Now, combining this fact with Equation 5, we have that
$$
m \left(\left(\frac {1}{m}\right) ^ {a} - \kappa m + c\right) ^ {1 / a} \geq 1
$$
$$
\left(\frac {1}{m}\right) ^ {a} - \kappa m + c \leq \left(\frac {1}{m}\right) ^ {a}
$$
so  $c\leq m\kappa$  ,as desired.
Proof of Lemma A.2. To show that  $f$  is defined for any  $x$  such that  $a^2 x^{a + 1} \geq \kappa$ , we need to show that  $x^a - \frac{\kappa}{x} + c > 0$  for such values of  $x$ . This is indeed the case:
$$
x ^ {a} - \frac {\kappa}{x} + c \geq x ^ {a} - a ^ {2} x ^ {a} + c = (1 - a ^ {2}) x ^ {a} + c > c \geq 0.
$$
Now we show concavity. We have
$$
f ^ {\prime \prime} (x) = \frac {1}{- a} \left(\left(1 + \frac {1}{- a}\right) \left(x ^ {a} - \frac {\kappa}{x} + c\right) ^ {1 / a - 2} \left(a x ^ {a - 1} + \frac {\kappa}{x ^ {2}}\right) ^ {2} - \left(x ^ {a} - \frac {\kappa}{x} + c\right) ^ {1 / a - 1} \left(a (a - 1) x ^ {a - 2} - \frac {2 \kappa}{x ^ {3}}\right)\right)
$$
so we wish to show that
$$
\left(1 + \frac {1}{- a}\right) \left(x ^ {a} - \frac {\kappa}{x} + c\right) ^ {1 / a - 2} \left(a x ^ {a - 1} + \frac {\kappa}{x ^ {2}}\right) ^ {2} \leq \left(x ^ {a} - \frac {\kappa}{x} + c\right) ^ {1 / a - 1} \left(a (a - 1) x ^ {a - 2} - \frac {2 \kappa}{x ^ {3}}\right)
$$
for every  $x$  such that  $a^2 x^{a + 1} \geq \kappa$ . Fix any such  $x$ , and let  $d = \frac{\kappa}{x^{a + 1}}$  (so  $0 \leq d \leq a^2$ ). We have
$$
d \leq a ^ {2}
$$
$$
(1 + a) (a ^ {2} - d) d \geq 0
$$
$$
(1 - a) (a + d) ^ {2} \leq - a (1 - d) (a (a - 1) - 2 d) \quad (\text {r e a r r a n g e t e r m s})
$$
$$
\left(1 - \frac {1}{a}\right) (a + d) ^ {2} x ^ {a} \leq ((1 - d) x ^ {a}) (a (a - 1) - 2 d) \quad (\text {m u l t i p l y b y} \frac {x ^ {a}}{- a})
$$
$$
\left(1 - \frac {1}{a}\right) (a + d) ^ {2} x ^ {a} \leq ((1 - d) x ^ {a} + c) (a (a - 1) - 2 d) \quad (c (a (a - 1) - 2 d) \geq 0)
$$
$$
\left(1 - \frac {1}{a}\right) \left((a + d) x ^ {a - 1}\right) ^ {2} \leq \left((1 - d) x ^ {a} + c\right) (a (a - 1) - 2 d) x ^ {a - 2} \quad \text {(m u l t i p l y b y} x ^ {a - 2})
$$
$$
\left(1 - \frac {1}{a}\right) \left(a x ^ {a - 1} + \frac {\kappa}{x ^ {2}}\right) ^ {2} \leq \left(x ^ {a} - \frac {\kappa}{x} + c\right) \left(a (a - 1) x ^ {a - 2} - \frac {2 \kappa}{x ^ {3}}\right) \quad (\text {s u b s t i t u t e} d = \kappa x ^ {- a - 1}).
$$
Note that the fifth line is justified by the fact that  $c \geq 0$  and  $a(a - 1) \geq 2d$  (because  $a^2 \geq d$  and  $-a > a^2 \geq d$ ). Now, multiplying both sides by  $\left(x^{a} - \frac{\kappa}{x} + c\right)^{1 / a - 2}$  completes the proof.
Proof of Corollary 4.5. Note that  $\eta \gamma = \frac{1}{\sqrt{T}m^{(1 + \alpha) / 2}}$  and also that  $\eta_t \leq \eta$  for all  $t$ ; we will be using these facts.
To prove (#1), we proceed by induction on  $t$ . In the case of  $t = 1$ , all weights are  $1/m$ , so the claim holds for sufficiently large  $T$ . Now assume that the claim holds for a generic  $t < T$ ; we show it for  $t + 1$ .
By the small gradient assumption, we may use Lemma 4.4 with  $\zeta = \gamma$ . By the inductive hypothesis (and the fact that  $\eta_t \leq \eta$ ), we may apply the second part of Lemma 4.4:
$$
\begin{array}{l} \left(w _ {i} ^ {t + 1}\right) ^ {\alpha - 1} \leq \left(w _ {i} ^ {t}\right) ^ {\alpha - 1} + (m + 1) \eta \gamma \leq \dots \leq (1 / m) ^ {\alpha - 1} + t (m + 1) \eta \gamma . \\ \leq (1 / m) ^ {\alpha - 1} + \frac {(T - 1) (m + 1)}{m ^ {(1 + \alpha) / 2} \sqrt {T}} \leq 3 m ^ {(1 - \alpha) / 2} \sqrt {T}. \\ \end{array}
$$
Since  $\frac{-1}{2} < \alpha - 1 < 0$ , this means that  $w_{i}^{t} \geq \frac{1}{10\sqrt{m}} T^{1 / (2(\alpha - 1))}$ .
$$
\left(w _ {i} ^ {t + 1}\right) ^ {\alpha} \geq \frac {1}{(1 0 \sqrt {m}) ^ {\alpha}} T ^ {\alpha / (2 (\alpha - 1))} \geq \frac {4}{m ^ {(1 + \alpha) / 2}} T ^ {- 1 / 2} = 4 \eta \gamma
$$
for  $T$  sufficiently large, since  $\frac{\alpha}{2(\alpha - 1)} > \frac{-1}{2}$ . This completes the inductive step, and thus the proof of (#1).
To prove (#2), we use the following technical lemma (see below for the proof).
Lemma A.3. Fix  $x > 0$  and  $-1 < a < 0$ . Let  $f(y) = (x^a + y)^{1/a}$ . Then for all  $y > -x^a$ , we have
$$
x - f (y) \leq \frac {- 1}{a} x ^ {1 - a} y \tag {6}
$$
and for all  $-1 < c \leq 0$ , for all  $cx^a \leq y \leq 0$ , we have
$$
f (y) - x \leq \frac {1}{a} (1 + c) ^ {1 / a - 1} x ^ {1 - a} y. \tag {7}
$$
We apply Equation 6 to  $x = w_{i}^{t}$ ,  $y = (m + 1)\eta \gamma$ , and  $a = \alpha - 1$ . This tells us that
$$
w _ {i} ^ {t} - w _ {i} ^ {t + 1} \leq w _ {i} ^ {t} - ((w _ {i} ^ {t}) ^ {\alpha - 1} + (m + 1) \eta \gamma) ^ {1 / (\alpha - 1)} \leq 2 (w _ {i} ^ {t}) ^ {2 - \alpha} (m + 1) \eta \gamma .
$$
The first step follows by the second part of Lemma 4.4 and the fact that  $\eta_t \leq \eta$ . The second step follows from Equation 6 and uses the fact that  $\frac{1}{1 - \alpha} > 2$ .
For the other side of (#2), we observe that since by (#1) we have  $(w_i^t)^\alpha \geq 4\eta \gamma$ , it follows that  $\frac{1}{2} (w_i^t)^\alpha \geq (w_i^t + 1)\eta \gamma$ , and so  $\left(\frac{1}{w_i^t} + 1\right)\eta \gamma \leq \frac{1}{2} (w_i^t)^{\alpha - 1}$ . Therefore, we can apply Equation 7 to  $x = w_i^t$ ,  $y = -\left(\frac{1}{w_i^t} + 1\right)\eta \gamma$ ,  $a = \alpha - 1$ , and  $c = -\frac{1}{2}$ . This tells us that
$$
\begin{array}{l} w _ {i} ^ {t + 1} - w _ {i} ^ {t} \leq \left((w _ {i} ^ {t}) ^ {\alpha - 1} - \left(\frac {1}{w _ {i} ^ {t}} + 1\right) \eta \gamma\right) ^ {1 / (\alpha - 1)} - w _ {i} ^ {t} \leq 1 6 (w _ {i} ^ {t}) ^ {2 - \alpha} \left(\frac {1}{w _ {i} ^ {t}} + 1\right) \eta \gamma \\ \leq 3 2 (w _ {i} ^ {t}) ^ {1 - \alpha} \eta \gamma . \\ \end{array}
$$
Proof of Lemma A.3. For all  $y > -x^a$ , we have
$$
f ^ {\prime} (y) = \frac {1}{a} (x ^ {a} + y) ^ {1 / a - 1}
$$
$$
f ^ {\prime \prime} (y) = \frac {1}{a} \left(\frac {1}{a} - 1\right) (x ^ {a} + y) ^ {1 / a - 2} > 0,
$$
so  $f'$  is increasing. Thus, for positive values of  $y$  we have
$$
f ^ {\prime} (0) \leq \frac {f (y) - f (0)}{y} = \frac {f (y) - x}{y} \leq f ^ {\prime} (y)
$$
and for negative values of  $y$  we have
$$
f ^ {\prime} (y) \leq \frac {f (y) - f (0)}{y} = \frac {f (y) - x}{y} \leq f ^ {\prime} (0).
$$
Regardless of whether  $y$  is positive or negative, this means that  $x - f(y) \leq -yf'(0) = \frac{-1}{a} x^{1 - a}y$ . Now, let  $-1 < c \leq 0$  and suppose that  $cx^a \leq y \leq 0$ . Since  $f'$  is increasing, we have that
$$
f ^ {\prime} (y) \geq f ^ {\prime} (c x ^ {a}) = \frac {1}{a} ((1 + c) x ^ {a}) ^ {1 / a - 1} = \frac {1}{a} (1 + c) ^ {1 / a - 1} x ^ {1 - a},
$$
$$
f (y) - x \leq y f ^ {\prime} (y) \leq \frac {1}{a} (1 + c) ^ {1 / a - 1} x ^ {1 - a} y.
$$
Proof of Lemma 4.8. We first derive an expression for  $\partial_i L(\mathbf{w})$  given expert reports  $\mathbf{p}^1, \ldots, \mathbf{p}^m$ , where  $L(\mathbf{w})$  is the log loss of the logarithmic pool  $\mathbf{p}^*(\mathbf{w})$  of  $\mathbf{p}^1, \ldots, \mathbf{p}^m$  with weights  $\mathbf{w}$ , and  $j$  is the realized outcome. We have
$$
\begin{array}{l} \partial_ {i} L (\mathbf {w}) = - \partial_ {i} \ln \frac {\prod_ {k = 1} ^ {m} (p _ {j} ^ {k}) ^ {w _ {k}}}{\sum_ {\ell = 1} ^ {n} \prod_ {k = 1} ^ {m} (p _ {\ell} ^ {k}) ^ {w _ {k}}} = \partial_ {i} \ln \left(\sum_ {\ell = 1} ^ {n} \prod_ {k = 1} ^ {m} (p _ {\ell} ^ {k}) ^ {w _ {k}}\right) - \partial_ {i} \ln \left(\prod_ {k = 1} ^ {m} (p _ {j} ^ {k}) ^ {w _ {k}}\right) \\ = \frac {\sum_ {\ell = 1} ^ {n} \ln p _ {\ell} ^ {i} \cdot \prod_ {k = 1} ^ {m} \left(p _ {\ell} ^ {k}\right) ^ {w _ {k}}}{\sum_ {\ell = 1} ^ {n} \prod_ {k = 1} ^ {m} \left(p _ {\ell} ^ {k}\right) ^ {w _ {k}}} - \ln p _ {j} ^ {i} = \sum_ {\ell = 1} ^ {n} p _ {\ell} ^ {*} (\mathbf {w}) \ln p _ {\ell} ^ {i} - \ln p _ {j} ^ {i}. \tag {8} \\ \end{array}
$$
It should be noted that  $\nabla L(\mathbf{w})$  is most naturally thought of as living in  $\mathbb{R}^m /T(\mathbf{1}_m)$ , i.e.  $m$ -dimensional space modulo translation by the all-ones vector, since  $\mathbf{w}$  lives in a place that is orthogonal to the all-ones vector. As an arbitrary but convenient convention, we define  $\partial_iL(\mathbf{w})$  to be the specific value derived below, and define the small gradient assumption accordingly.
Equation (2) now follows fairly straightforwardly. Equation 8 tells us that  $\partial_i L(\mathbf{w}) \leq -\ln p_J^i$ , where  $J$  is the random variable corresponding to the realized outcome. Therefore, we have
where the last step follows by the calibration property. This proves Equation (2).
We now prove Equation (3). The proof has a similar idea, but is somewhat more technical. We begin by proving the following lemma; we again use the calibration property in the proof.
Lemma A.4. For all  $q$ , we have
$$
\mathbb {P} \left[ \forall j \exists i: p _ {j} ^ {i} \leq q \right] \leq m n q.
$$
Proof. Let  $J$  be the random variable corresponding to the index of the outcome that ends up happening. We have
$$
\begin{array}{l} \mathbb {P} \left[ \forall j \exists i: p _ {j} ^ {i} \leq q \right] \leq \mathbb {P} \left[ \exists i: p _ {J} ^ {i} \leq q \right] = \sum_ {j \in [ n ]} \mathbb {P} \left[ J = j \& \exists i: p _ {j} ^ {i} \leq q \right] \\ \leq \sum_ {j \in [ n ]} \sum_ {i \in [ m ]} \mathbb {P} \left[ J = j \& p _ {j} ^ {i} \leq q \right] \\ = \sum_ {j \in [ n ]} \sum_ {i \in [ m ]} \mathbb {P} \left[ p _ {j} ^ {i} \leq q \right] \mathbb {P} \left[ J = j \mid p _ {j} ^ {i} \leq q \right] \leq \sum_ {j \in [ n ]} \sum_ {i \in [ m ]} 1 \cdot q = m n q, \\ \end{array}
$$
where the fact that  $\mathbb{P}\left[J = j\mid p_j^i\leq q\right]\leq q$  follows by the calibration property.
Corollary A.5. For any reports  $\mathbf{p}^1, \ldots, \mathbf{p}^m$ , weight vector  $\mathbf{w}$ ,  $i \in [m]$ , and  $j \in [n]$ , we have
$$
\mathbb {P} \left[ p _ {j} ^ {*} (\mathbf {w}) \geq \frac {\left(p _ {j} ^ {i}\right) ^ {w _ {i}}}{q} \right] \leq m n q.
$$
$$
p _ {j} ^ {*} (\mathbf {w}) = \frac {\prod_ {k = 1} ^ {m} \left(p _ {j} ^ {k}\right) ^ {w _ {k}}}{\sum_ {\ell = 1} ^ {n} \prod_ {k = 1} ^ {m} \left(p _ {\ell} ^ {k}\right) ^ {w _ {k}}} \leq \frac {\left(p _ {j} ^ {i}\right) ^ {w _ {i}}}{\sum_ {\ell = 1} ^ {n} \prod_ {k = 1} ^ {m} \left(p _ {\ell} ^ {k}\right) ^ {w _ {k}}}.
$$
Now, assuming that there is an  $\ell$  such that for every  $k$  we have  $p_{\ell}^{k} > q$ , the denominator is greater than  $q$ , in which case we have  $p_{j}^{*}(\mathbf{w}) < \frac{(p_{j}^{i})^{w_{i}}}{q}$ . Therefore, if  $p_{j}^{*}(\mathbf{w}) \geq \frac{(p_{j}^{i})^{w_{i}}}{q}$ , it follows that for every  $\ell$  there is a  $k$  such that  $p_{\ell}^{k} \leq q$ . By Lemma A.4, this happens with probability at most  $mnq$ .
We now use Corollary A.5 to prove Equation (3). Note that the equation is trivial for  $\zeta < n$ , so we assume that  $\zeta \geq n$ . By setting  $q := e^{\frac{-\zeta}{n}}$ , we may restate Equation (3) as follows: for any  $q \leq \frac{1}{e}$ , any  $i \in [m]$ , and any weight vector  $\mathbf{w}$ ,
$$
\mathbb {P} \left[ \partial_ {i} L (\mathbf {w}) \leq - \frac {n \ln 1 / q}{w _ {i}} \right] \leq m n ^ {2} q.
$$
(Note that the condition  $q \leq \frac{1}{e}$  is equivalent to  $\zeta \geq n$ .) We prove this result.
From Equation 8, we have
$$
\partial_ {i} L (\mathbf {w}) = \sum_ {j = 1} ^ {n} p _ {j} ^ {*} (\mathbf {w}) \ln p _ {j} ^ {i} - \ln p _ {j} ^ {i} \geq \sum_ {j = 1} ^ {n} p _ {j} (\mathbf {w}) \ln p _ {j} ^ {i}.
$$
Now, it suffices to show that for each  $j \in [n]$ , the probability that  $p_j(\mathbf{w}) \ln p_j^i \leq -\frac{\ln 1 / q}{w_i} = \frac{\ln q}{w_i}$  is at most  $mnq$ ; the desired result will then follow by the union bound. By Corollary A.5, for each  $j$  we have that
$$
\mathbb {P} \left[ p _ {j} (\mathbf {w}) \ln p _ {j} ^ {i} \leq \frac {\left(p _ {j} ^ {i}\right) ^ {w _ {i}}}{q} \ln p _ {j} ^ {i} \right] \leq m n q.
$$
Additionally, we know for a fact that  $p_j(\mathbf{w}) \ln p_j^i \geq \ln p_j^i$  (since  $p_j(\mathbf{w}) \leq 1$ ), so in fact
$$
\mathbb {P} \left[ p _ {j} (\mathbf {w}) \ln p _ {j} ^ {i} \leq \max  \left(\frac {\left(p _ {j} ^ {i}\right) ^ {w _ {i}}}{q} \ln p _ {j} ^ {i}, \ln p _ {j} ^ {i}\right) \right] \leq m n q.
$$
It remains only to show that  $\max \left( \frac{(p_j^i)^{w_i}}{q} \ln p_j^i, \ln p_j^i \right) \geq \frac{\ln q}{w_i}$ . If  $p_j^i \geq q^{1/w_i}$  then this is clearly true, since in that case  $\ln p_j^i \geq \frac{\ln q}{w_i}$ . Now suppose that  $p_j^i < q^{1/w_i}$ . Observe that  $\frac{x^{w_i}}{q} \ln x$  decreases on  $(0, e^{-1/w_i})$ , and that (since  $q \leq \frac{1}{e}$ ) we have  $q^{1/w_i} \leq e^{-1/w_i}$ . Therefore,
$$
\frac {\left(p _ {j} ^ {i}\right) ^ {w _ {i}}}{q} \ln p _ {j} ^ {i} \leq \frac {\left(q ^ {1 / w _ {i}}\right) ^ {w _ {i}}}{q} \ln q ^ {1 / w _ {i}} = \frac {\ln q}{w _ {i}}.
$$
This completes the proof of Equation (3), and thus of Lemma 4.8.
The following lemma lower bounds the regret of Algorithm 1 as a function of  $\zeta$ .
Lemma A.6. Consider a run of Algorithm 1. Let  $\zeta$  be such that  $-\frac{\zeta}{w_i^t} \leq \partial_i L^t(\mathbf{w}^t) \leq \zeta$  for all  $i, t$ . The total regret is at most
$$
O \left(\zeta^ {2 (2 - \alpha) / (1 - \alpha)} T ^ {(5 - \alpha) / (1 - \alpha)}\right).
$$
Proof of Lemma A.6. We first bound  $w_{i}^{t}$  for all  $i, t$ . From Lemma 4.4, we have that
$$
(w _ {i} ^ {t + 1}) ^ {\alpha - 1} \leq (w _ {i} ^ {t}) ^ {\alpha - 1} + \left(\frac {1}{\min _ {i} w _ {i} ^ {t}} + 1\right) \eta_ {t} \zeta \leq (w _ {i} ^ {t}) ^ {\alpha - 1} + 2 \zeta .
$$
Here we use that  $\frac{1}{\min_i w_i} + 1 \leq \frac{2}{\min_i w_i}$  and that  $\eta_t \leq \min_i w_i$ . Therefore, we have that
$$
\left(w _ {i} ^ {t}\right) ^ {\alpha - 1} \leq \left(w _ {i} ^ {t - 1}\right) ^ {\alpha - 1} + 2 \zeta \leq \dots \leq m ^ {1 - \alpha} + 2 \zeta (t - 1) \leq m ^ {1 - \alpha} + 2 \zeta T.
$$
Thus,  $w_{i}^{t}\geq (m^{1 - \alpha} + 2\zeta T)^{1 / (\alpha -1)}\geq \Omega ((\zeta T)^{1 / (\alpha -1)})$  for all  $i,t$
We now use the standard regret bound for online mirror descent, see e.g. (Orabona, 2021, Theorem 6.8):
$$
\operatorname {R e g r e t} \leq \max  _ {t} \frac {B _ {R} (\mathbf {u} ; \mathbf {w} ^ {t})}{\eta_ {T}} + \frac {1}{2 \lambda} \sum_ {t = 1} ^ {T} \eta_ {t} \| \nabla L ^ {t} (\mathbf {w} ^ {t}) \| _ {*} ^ {2} \tag {9}
$$
where  $B_{R}(\cdot ;\cdot)$  is the Bregman divergence of with respect to  $R$ ,  $\mathbf{u}$  is the optimal (overall loss-minimizing) point,  $\lambda$  is a constant such that  $R$  is  $\lambda$ -strongly convex with respect to a norm of our choice over  $\Delta^m$ , and  $\| \cdot \| _*$  is the dual norm of the aforementioned norm.
Note that for any  $\mathbf{x} \in \Delta^m$ , we have
$$
\max  _ {\mathbf {v} \in \Delta^ {m}} B _ {R} (\mathbf {v}; \mathbf {x}) = \max  _ {\mathbf {v} \in \Delta^ {m}} R (\mathbf {v}) - R (\mathbf {x}) - \nabla R (\mathbf {x}) \cdot (\mathbf {v} - \mathbf {x}) \leq \frac {m ^ {1 - \alpha}}{\alpha} + \left(\min  _ {i} x _ {i}\right) ^ {\alpha - 1}.
$$
In the last step, we use the fact that  $-\nabla R(\mathbf{x}) = (x_1^{\alpha - 1}, \ldots, x_m^{\alpha - 1})$  (all of these coordinates are positive), so  $-\nabla R(\mathbf{x}) \cdot (\mathbf{v} - \mathbf{x}) \leq (x_1^{\alpha - 1}, \ldots, x_m^{\alpha - 1}) \cdot \mathbf{v}$ , and that all coordinates of  $\mathbf{v}$  are non-negative and add to 1.
Therefore, given our bound on  $w_{i}^{t}$ , we have that this first component of our regret bound (9) is at most
$$
\frac {1}{\eta_ {T}} \left(\frac {m ^ {1 - \alpha}}{\alpha} + m ^ {1 - \alpha} + 2 \zeta T\right) \leq O \left(\frac {\zeta T}{\eta_ {T}}\right) \leq O \left(\frac {\zeta T}{(\zeta T) ^ {1 / (\alpha - 1)}}\right) = O \left((\zeta T) ^ {(2 - \alpha) / (1 - \alpha)}\right).
$$
To bound the second term, we choose to work with the  $\ell_1$  norm. To show that  $R$  is  $\lambda$ -convex it suffices to show that for all  $\mathbf{x}, \mathbf{y} \in \Delta^m$  we have  $(\nabla^2 R(\mathbf{x})\mathbf{y}) \cdot \mathbf{y} \geq \lambda \| \mathbf{y}\|^2$ , where  $\nabla^2 R$  is the Hessian of  $R$  (Shalev-Shwartz, 2007, Lemma 14) (see also (Orabona, 2021, Theorem 4.3)). Equivalently, we wish to find a  $\lambda$  such that
$$
(1 - \alpha) \sum_ {i} x _ {i} ^ {\alpha - 2} y _ {i} ^ {2} \geq \lambda .
$$
Since  $x_{i}^{\alpha - 2} \geq 1$  for all  $i$ , the left-hand side is at least  $(1 - \alpha) \sum_{i} y_{i}^{2} \geq \frac{1 - \alpha}{m}$ , so  $\lambda = \frac{1 - \alpha}{m}$  suffices.
Now, given  $\theta \in \mathbb{R}^m$ , we have  $\| \theta \|_* = \max_{\mathbf{x} : \| \mathbf{x} \| \leq 1} \theta \cdot \mathbf{x}$ . In the case of the  $\ell_1$  primal norm, the dual norm is the largest absolute component of  $\theta$ . Thus, we have
$$
\left\| \nabla L ^ {t} (\mathbf {x} ^ {t}) \right\| _ {*} \leq \frac {\zeta}{w _ {i} ^ {t}} \leq O \left(\zeta (\zeta T) ^ {1 / (1 - \alpha)}\right) = O \left(\zeta^ {(2 - \alpha) / (1 - \alpha)} T ^ {1 / (1 - \alpha)}\right).
$$
Since  $\eta_t \leq O(T^{-1/2})$ , we have that the second component of our regret bound (9) is at most
$$
O \left(T \cdot T ^ {- 1 / 2} \cdot \zeta^ {2 (2 - \alpha) / (1 - \alpha)} T ^ {2 / (1 - \alpha)}\right) \leq O \left(\zeta^ {2 (2 - \alpha) / (1 - \alpha)} T ^ {(5 - \alpha) / (1 - \alpha)}\right).
$$
This component dominates our bound on the regret of the first component, in both  $\zeta$  and  $T$ . This concludes the proof.
Proof of Corollary 4.9. Let  $Z$  be the minimum value of  $\zeta$  such that  $-\frac{\zeta}{w_i^t} \leq \partial_i L^t(\mathbf{w}^t) \leq \zeta$  for all  $i, t$ . Note that by Lemma 4.8, we have that
$$
\mathbb {P} \left[ Z \geq x \right] \leq \sum_ {i = 1} ^ {m} \sum_ {t = 1} ^ {T} \left(m n ^ {2} e ^ {- \frac {x}{n}} + n e ^ {- x}\right) \leq 2 m ^ {2} n ^ {2} T e ^ {- \frac {x}{n}}.
$$
Let  $\mu$  be the constant hidden in the big-O of Lemma A.6, i.e. a constant (dependent on  $m$ ,  $n$ , and  $\alpha$ ) such that
$$
\operatorname {R e g r e t} \leq \mu Z ^ {2 (2 - \alpha) / (1 - \alpha)} T ^ {(5 - \alpha) / (1 - \alpha)}.
$$
Let  $r(Z, T)$  be the expression on the right-hand side. The small gradient assumption not holding is equivalent to  $Z > 12n\ln T$ , or equivalently,  $r(Z, T) > r(12n\ln T, T)$ . The expected regret of our algorithm conditional on the small gradient assumption not holding, times the probability of this event, is therefore at most the expected value of  $r(Z, T)$  conditional on the value being greater than  $r(12n\ln T, T)$ , times this probability. This is equal to
$$
\begin{array}{l} r (1 2 n \ln T, T) \cdot \mathbb {P} [ Z > 1 2 n \ln T ] + \int_ {x = r (1 2 n \ln T, T)} ^ {\infty} \mathbb {P} [ r (Z, T) \geq x ] d x \\ \leq \sum_ {k = 1 1} ^ {\infty} r ((k + 1) n \ln T, T) \cdot \mathbb {P} [ Z \geq k n \ln T ] \\ \leq \sum_ {k = 1 1} ^ {\infty} \mu \cdot ((k + 1) n \ln T) ^ {2 (2 - \alpha) / (1 - \alpha)} T ^ {(5 - \alpha) / (1 - \alpha)} \cdot 2 m ^ {2} n ^ {2} T \cdot T ^ {- k} \\ \leq \sum_ {k = 1 1} ^ {\infty} \tilde {O} (T ^ {1 + (5 - \alpha) / (1 - \alpha) - k}) = \tilde {O} (T ^ {(5 - \alpha) / (1 - \alpha) - 1 0}), \\ \end{array}
$$
as desired. (The first inequality follows by matching the first term with the  $k = 11$  summand and upper-bounding the integral with subsequent summands, noting that  $r((k + 1)n\ln T,T)\geq 1.$
Note that  $\frac{5 - \alpha}{1 - \alpha} - 10 \leq \frac{5 - 1/2}{1 - 1/2} - 10 = -1$ . Therefore, the contribution to expected regret from the case that the small gradient assumption does not hold is  $\tilde{O}(T^{-1})$ , which is negligible. Together with Corollary 4.7 (which bounds regret under the small gradient assumption), this proves Theorem 3.2.
We now extend Theorem 3.2 by showing that the theorem holds even if experts are only approximately calibrated.
Definition A.7. For  $\tau \geq 1$ , we say that expert  $i$  is  $\tau$ -calibrated if for all  $\mathbf{p} \in \Delta^n$  and  $j \in [n]$ , we have that  $\mathbb{P}\left[J = j \mid \mathbf{p}^j = \mathbf{p}\right] \leq \tau p_j$ . We say that  $\mathbb{P}$  satisfies the  $\tau$ -approximate calibration property if every expert is  $\tau$ -calibrated.
Corollary A.8. For any  $\tau$ , Theorem 3.2 holds even if the calibration property is replaced with the  $\tau$ -approximate calibration property.
(Note that the  $\tau$  is subsumed by the big- $O$  notation in Theorem 3.2; Corollary A.8 does not allow experts to be arbitrarily miscalibrated.)
Technically, Corollary A.8 is a corollary of the proof of Theorem 3.2, rather than a corollary of the theorem itself. $^{10}$