SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\mathbb {E} \left[ M _ {1} \left(\boldsymbol {f} _ {1}, \boldsymbol {\lambda} _ {1}\right) \right] = \mathbb {E} \left[ \mathbb {E} \left[ M _ {1} \left(\boldsymbol {f} _ {1}, \boldsymbol {\lambda} _ {1}\right) \mid \mathcal {H} _ {0} \right] \right] = 1.
$$
Now, we verify the martingale property. For any  $t \geq 2$ , we have
$$
\begin{array}{l} \mathbb {E} \left[ M _ {t} (\pmb {f} _ {t}, \pmb {\lambda} _ {t}) | \mathcal {H} _ {t - 1} \right] = \mathbb {E} \left[ \exp \left(\sum_ {k = 1} ^ {t} \lambda_ {k} Z _ {k} (f _ {k}) - \psi_ {k} (f _ {k}, \lambda_ {k})\right) \Bigg | \mathcal {H} _ {t - 1} \right] \\ = \exp \left(\sum_ {k = 1} ^ {t - 1} \lambda_ {k} Z _ {k} (f _ {k}) - \psi_ {k} (f _ {k}, \lambda_ {k})\right) \mathbb {E} \left[ \exp \left(\lambda_ {t} Z _ {t} (f _ {t}) - \psi_ {t} (f _ {t}, \lambda_ {t})\right) \mid \mathcal {H} _ {t - 1} \right] \\ = \exp \left(\sum_ {k = 1} ^ {t - 1} \lambda_ {k} Z _ {k} (f _ {k}) - \psi_ {k} (f _ {k}, \lambda_ {k})\right) \\ = M _ {t - 1} \left(\boldsymbol {f} _ {t - 1}, \boldsymbol {\lambda} _ {t - 1}\right). \\ \end{array}
$$
Using the tower rule again, we have for any  $t \geq 2$
$$
\mathbb {E} \left[ M _ {t} \left(\boldsymbol {f} _ {t}, \boldsymbol {\lambda} _ {t}\right) \right] = \mathbb {E} \left[ \mathbb {E} \left[ M _ {t} \left(\boldsymbol {f} _ {t}, \boldsymbol {\lambda} _ {t}\right) \mid \mathcal {H} _ {t - 1} \right] \right] = \mathbb {E} \left[ M _ {t - 1} \left(\boldsymbol {f} _ {t - 1}, \boldsymbol {\lambda} _ {t - 1}\right) \right].
$$
$$
\mathbb {E} \left[ M _ {t} \left(\boldsymbol {f} _ {t}, \boldsymbol {\lambda} _ {t}\right) \right] = \mathbb {E} \left[ M _ {t - 1} \left(\boldsymbol {f} _ {t - 1}, \boldsymbol {\lambda} _ {t - 1}\right) \right] = \dots = \mathbb {E} \left[ M _ {1} \left(\boldsymbol {f} _ {1}, \boldsymbol {\lambda} _ {1}\right) \right] = 1.
$$
Lemma A.2. For any adaptive sequence of mixture distributions  $(P_t|t \in \mathbb{N})$ ,  $(\mathbb{E}_{\boldsymbol{f}_t \sim P_t}[M_t(\boldsymbol{f}_t, \boldsymbol{\lambda}_t)]|t \in \mathbb{N})$  is a martingale and  $\mathbb{E}[\mathbb{E}_{\boldsymbol{f}_t \sim P_t}[M_t(\boldsymbol{f}_t, \boldsymbol{\lambda}_t)]] = 1$  for all  $t \in \mathbb{N}$ .
Proof. For any  $t \geq 1$ , since  $M_{t}(\pmb{f}_{t},\pmb{\lambda}_{t})$  is non-negative and  $P_{t}$  is  $\mathcal{H}_{t - 1}$ -measurable, Tonelli's theorem implies
$$
\mathbb {E} \left[ \mathbb {E} _ {\boldsymbol {f} _ {t} \sim P _ {t}} \left[ M _ {t} \left(\boldsymbol {f} _ {t}, \boldsymbol {\lambda} _ {t}\right) \right] \mid \mathcal {H} _ {t - 1} \right] = \mathbb {E} _ {\boldsymbol {f} _ {t} \sim P _ {t}} \left[ \mathbb {E} \left[ M _ {t} \left(\boldsymbol {f} _ {t}, \boldsymbol {\lambda} _ {t}\right) \mid \mathcal {H} _ {t - 1} \right] \right].
$$
The requirement that the distributions  $P_{1}, P_{2}, \ldots$  have coinciding marginals, i.e.  $\int P_t(\pmb{f}_t) \, \mathrm{d}f_t = P_{t-1}(\pmb{f}_{t-1})$ , means that for all  $t \geq 2$
$$
\mathbb {E} _ {\boldsymbol {f} _ {t} \sim P _ {t}} [ M _ {t - 1} (\boldsymbol {f} _ {t - 1}, \boldsymbol {\lambda} _ {t - 1}) ] = \mathbb {E} _ {\boldsymbol {f} _ {t - 1} \sim P _ {t - 1}} [ M _ {t - 1} (\boldsymbol {f} _ {t - 1}, \boldsymbol {\lambda} _ {t - 1}) ].
$$
Using these two results, and the fact that  $(M_t(\pmb{f}_t, \pmb{\lambda}_t) | t \in \mathbb{N})$  is a martingale for any sequence  $(f_t | t \in \mathbb{N})$ , we now verify that the martingale mixture  $(\mathbb{E}_{\pmb{f}_t \sim P_t} [M_t(\pmb{f}_t, \pmb{\lambda}_t)] | t \in \mathbb{N})$  is a martingale with expected value 1. For  $t = 1$ , we have
$$
\begin{array}{l} \mathbb {E} \left[ \mathbb {E} _ {\boldsymbol {f} _ {1} \sim P _ {1}} \left[ M _ {1} \left(\boldsymbol {f} _ {1}, \boldsymbol {\lambda} _ {1}\right) \right] | \mathcal {H} _ {0} \right] = \mathbb {E} _ {\boldsymbol {f} _ {1} \sim P _ {1}} \left[ \mathbb {E} \left[ M _ {1} \left(\boldsymbol {f} _ {1}, \boldsymbol {\lambda} _ {1}\right) | \mathcal {H} _ {0} \right] \right] \\ = \mathbb {E} _ {\boldsymbol {f} _ {1} \sim P _ {1}} [ 1 ] \\ = 1. \\ \end{array}
$$
Using the tower rule as before, this also means that  $\mathbb{E}\left[\mathbb{E}_{\pmb{f}_1\sim P_1}[M_1(\pmb {f}_1,\pmb {\lambda}_1)]\right] = 1$  . For any  $t\geq 2$  we have
$$
\begin{array}{l} \mathbb {E} \left[ \mathbb {E} _ {\boldsymbol {f} _ {t} \sim P _ {t}} \left[ M _ {t} \left(\boldsymbol {f} _ {t}, \boldsymbol {\lambda} _ {t}\right) \right] | \mathcal {H} _ {t - 1} \right] = \mathbb {E} _ {\boldsymbol {f} _ {t} \sim P _ {t}} \left[ \mathbb {E} \left[ M _ {t} \left(\boldsymbol {f} _ {t}, \boldsymbol {\lambda} _ {t}\right) | \mathcal {H} _ {t - 1} \right] \right] \\ = \mathbb {E} _ {\boldsymbol {f} _ {t} \sim P _ {t}} \left[ M _ {t - 1} (\boldsymbol {f} _ {t - 1}, \boldsymbol {\lambda} _ {t - 1}) \right] \\ = \mathbb {E} _ {\boldsymbol {f} _ {t - 1} \sim P _ {t - 1}} \left[ M _ {t - 1} \left(\boldsymbol {f} _ {t - 1}, \boldsymbol {\lambda} _ {t - 1}\right) \right]. \\ \end{array}
$$
Using the tower rule one more time, we have for any  $t \geq 2$
$$
\mathbb {E} \left[ \mathbb {E} _ {\boldsymbol {f} _ {t} \sim P _ {t}} \left[ M _ {t} \left(\boldsymbol {f} _ {t}, \boldsymbol {\lambda} _ {t}\right) \right] \right] = \mathbb {E} \left[ \mathbb {E} _ {\boldsymbol {f} _ {t} \sim P _ {t}} \left[ M _ {t} \left(\boldsymbol {f} _ {t}, \boldsymbol {\lambda} _ {t}\right) \right] \mid \mathcal {H} _ {t - 1} \right] = \mathbb {E} \left[ \mathbb {E} _ {\boldsymbol {f} _ {t - 1} \sim P _ {t - 1}} \left[ M _ {t - 1} \left(\boldsymbol {f} _ {t - 1}, \boldsymbol {\lambda} _ {t - 1}\right) \right] \right].
$$
$$
\mathbb {E} \left[ \mathbb {E} _ {\boldsymbol {f} _ {t} \sim P _ {t}} \left[ M _ {t} \left(\boldsymbol {f} _ {t}, \boldsymbol {\lambda} _ {t}\right) \right] \right] = \mathbb {E} \left[ \mathbb {E} _ {\boldsymbol {f} _ {1} \sim P _ {1}} \left[ M _ {1} \left(\boldsymbol {f} _ {1}, \boldsymbol {\lambda} _ {1}\right) \right] \right] = 1.
$$
To prove Thm. 5.1, we use Ville's inequality for non-negative supermartingales (Ville, 1939), which can be thought of as a time-uniform version of Markov's inequality. Instead of the martingale property  $\mathbb{E}[M_t|\mathcal{H}_{t - 1}] = M_{t - 1}$ , a supermartingale satisfies  $\mathbb{E}[M_t|\mathcal{H}_{t - 1}]\leq M_{t - 1}$ , which means that any martingale is also a supermartingale. Therefore, Ville's inequality for non-negative supermartingales also holds for the non-negative martingale in Lemma A.2.
Lemma A.3 (Ville's inequality for non-negative supermartingales (Ville, 1939)). Let  $(M_t|t \in \mathbb{N})$  be a non-negative supermartingale with respect to the filtration  $(\mathcal{H}_t|t \in \mathbb{N})$ , which satisfies  $M_0 = 1$ . For any  $\delta \in (0,1]$ , it holds with probability at least  $1 - \delta$ :
$$
\forall t \geq 1: \quad M _ {t} \leq 1 / \delta .
$$
Proof of Thm. 5.1. We choose an arbitrary  $\delta \in (0,1]$ . From Lemma A.2, for any adaptive sequence of mixture distributions  $(P_{t}|t\in \mathbb{N})$ ,  $(\mathbb{E}_{\pmb{f}_t\sim P_t}[M_t(\pmb {f}_t,\pmb {\lambda}_t)]|t\in \mathbb{N})$  is a martingale and  $\mathbb{E}[\mathbb{E}_{\pmb {f}_t\sim P_t}[M_t(\pmb {f}_t,\pmb {\lambda}_t)]] = 1$ . In addition,  $(\mathbb{E}_{\pmb {f}_t\sim P_t}[M_t(\pmb {f}_t,\pmb {\lambda}_t)]|t\in \mathbb{N})$  is clearly non-negative. Therefore, using Lemma A.3, with probability at least  $1 - \delta$
$$
\forall t \geq 1, \quad \mathbb {E} _ {\boldsymbol {f} _ {t} \sim P _ {t}} \left[ M _ {t} \left(\boldsymbol {f} _ {t}, \boldsymbol {\lambda} _ {t}\right) \right] \leq 1 / \delta .
$$
Taking the logarithm of both sides yields the statement of Thm. 5.1.
B Closed-Form Gaussian Integration
Here, we calculate the integral in the inequality (see beginning of Sec. 5.2):
$$
\underset {\boldsymbol {f} _ {t} \sim \mathcal {N} \left(\boldsymbol {\mu} _ {t}, \boldsymbol {T} _ {t}\right)} {\mathbb {E}} \left[ \exp \left\{\sum_ {k = 1} ^ {t} \lambda_ {k} \left(f _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*}\right) \left(r _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*}\right) - \frac {\sigma^ {2}}{2} \lambda_ {k} ^ {2} \left(f _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*}\right) ^ {2} \right\} \right] \leq \frac {1}{\delta}. \tag {12}
$$
First, we rearrange the integrand into a more convenient form. For every  $k$ , using  $r_k = \phi(a_k)^\top \pmb{\theta}^* + \epsilon_k$ , we have
$$
\begin{array}{l} \left(\phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*} - r _ {k}\right) ^ {2} - \left(f _ {k} - r _ {k}\right) ^ {2} = \epsilon_ {k} ^ {2} - \left(f _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*} - \epsilon_ {k}\right) ^ {2} \\ = \epsilon_ {k} ^ {2} - \left(f _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*}\right) ^ {2} + 2 \left(f _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*}\right) \epsilon_ {k} - \epsilon_ {k} ^ {2} \\ = 2 \left(f _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*}\right) \left(r _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*}\right) - \left(f _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*}\right) ^ {2}. \\ \end{array}
$$
$$
\begin{array}{l} \lambda_ {k} \left(f _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*}\right) \left(r _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*}\right) - \frac {\sigma^ {2}}{2} \lambda_ {k} ^ {2} \left(f _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*}\right) ^ {2} \\ = \frac {\lambda_ {k}}{2} (\phi (a _ {k}) ^ {\top} \pmb {\theta} ^ {*} - r _ {k}) ^ {2} - \frac {\lambda_ {k}}{2} (f _ {k} - r _ {k}) ^ {2} + \frac {1}{2} (\lambda_ {k} - \sigma^ {2} \lambda_ {k} ^ {2}) (f _ {k} - \phi (a _ {k}) ^ {\top} \pmb {\theta} ^ {*}) ^ {2}. \\ \end{array}
$$
Equation (12) can now be re-written as
$$
\underset {\boldsymbol {f} _ {t} \sim \mathcal {N} \left(\boldsymbol {\mu} _ {t}, \boldsymbol {T} _ {t}\right)} {\mathbb {E}} \left[ \exp \left\{\sum_ {k = 1} ^ {t} \frac {\lambda_ {k}}{2} \left(\phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*} - r _ {k}\right) ^ {2} - \frac {\lambda_ {k}}{2} \left(f _ {k} - r _ {k}\right) ^ {2} + \frac {1}{2} \left(\lambda_ {k} - \sigma^ {2} \lambda_ {k} ^ {2}\right) \left(f _ {k} - \phi \left(a _ {k}\right) ^ {\top} \boldsymbol {\theta} ^ {*}\right) ^ {2} \right\} \right] \leq \frac {1}{\delta}. \tag {13}
$$
In the special case where  $\lambda_t \equiv 1 / \sigma^2$ , we have  $\lambda_k - \sigma^2 \lambda_k^2 = 0$ , which means that  $\frac{1}{2} (\lambda_k - \sigma^2 \lambda_k^2)(f_k - \phi(a_k)^\top \pmb{\theta}^*)^2$  disappears. In addition,  $(\lambda_k / 2)(\phi(a_k)^\top \pmb{\theta}^* - r_k)^2$  does not depend on  $f_k$ , so it can be moved outside the integral.
B.1 General  $\lambda_{t}$
Let  $\Lambda_{t}$  be the  $t\times t$  diagonal matrix with diagonal elements  $\lambda_1,\lambda_2,\ldots ,\lambda_t$ . Starting from (13), taking the logarithm of both sides, rearranging terms and then writing everything in matrix notation, we arrive at
$$
\begin{array}{l} \left(\Phi_ {t} \boldsymbol {\theta} ^ {*} - \boldsymbol {r} _ {t}\right) \boldsymbol {\Lambda} _ {t} \left(\Phi_ {t} \boldsymbol {\theta} ^ {*} - \boldsymbol {r} _ {t}\right) \leq 2 \ln (1 / \delta) \tag {14} \\ \left. \right.\left. - 2 \ln \left(\underset {\boldsymbol {f} _ {t} \sim \mathcal {N} \left(\boldsymbol {\mu} _ {t}, \boldsymbol {T} _ {t}\right)} {\mathbb {E}} \left[ \exp \left(- \frac {1}{2} (\boldsymbol {f} _ {t} - \boldsymbol {r} _ {t}) ^ {\top} \boldsymbol {\Lambda} _ {t} (\boldsymbol {f} _ {t} - \boldsymbol {r} _ {t}) + \frac {1}{2} (\boldsymbol {f} _ {t} - \Phi_ {t} \boldsymbol {\theta} ^ {*}) ^ {\top} \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) (\boldsymbol {f} _ {t} - \Phi_ {t} \boldsymbol {\theta} ^ {*})\right)\right]\right)\right) \\ \end{array}
$$
The expected value inside the logarithm can be re-written as
$$
\begin{array}{l} \frac {1}{\sqrt {(2 \pi) ^ {t} \det (\boldsymbol {T} _ {t})}} \int \exp \left(- \frac {1}{2} (\boldsymbol {f} _ {t} - \boldsymbol {\mu} _ {t}) ^ {\top} \boldsymbol {T} _ {t} ^ {- 1} (\boldsymbol {f} _ {t} - \boldsymbol {\mu} _ {t}) - \frac {1}{2} (\boldsymbol {f} _ {t} - \boldsymbol {r} _ {t}) ^ {\top} \boldsymbol {\Lambda} _ {t} (\boldsymbol {f} _ {t} - \boldsymbol {r} _ {t}) \right. \tag {15} \\ + \frac {1}{2} (\pmb {f} _ {t} - \Phi_ {t} \pmb {\theta} ^ {*}) ^ {\top} \left(\pmb {\Lambda} _ {t} - \sigma^ {2} \pmb {\Lambda} _ {t} ^ {2}\right) (\pmb {f} _ {t} - \Phi_ {t} \pmb {\theta} ^ {*}) \Biggr) \mathrm {d} \pmb {f} _ {t}. \\ \end{array}
$$
We will calculate the integral by "completing the square", i.e. rewriting the exponent in the form  $-\frac{1}{2} (\pmb{f}_t - \pmb{b})^\top \pmb{A}(\pmb{f}_t - \pmb{b}) + c$ , to recover the integral of a Gaussian density function. For a symmetric matrix  $\pmb{A}$ , we have
$$
- \frac {1}{2} (\boldsymbol {f} _ {t} - \boldsymbol {b}) ^ {\top} \boldsymbol {A} (\boldsymbol {f} _ {t} - \boldsymbol {b}) + c = - \frac {1}{2} \boldsymbol {f} _ {t} ^ {\top} \boldsymbol {A} \boldsymbol {f} _ {t} + \boldsymbol {b} ^ {\top} \boldsymbol {A} \boldsymbol {f} _ {t} - \frac {1}{2} \boldsymbol {b} ^ {\top} \boldsymbol {A} \boldsymbol {b} + c.
$$
$$
\begin{array}{l} - \frac {1}{2} \left(\boldsymbol {f} _ {t} - \boldsymbol {\mu} _ {t}\right) ^ {\top} \boldsymbol {T} _ {t} ^ {- 1} \left(\boldsymbol {f} _ {t} - \boldsymbol {\mu} _ {t}\right) = - \frac {1}{2} \boldsymbol {f} _ {t} ^ {\top} \boldsymbol {T} _ {t} ^ {- 1} \boldsymbol {f} _ {t} + \boldsymbol {\mu} _ {t} ^ {\top} \boldsymbol {T} _ {t} ^ {- 1} \boldsymbol {f} _ {t} - \frac {1}{2} \boldsymbol {\mu} _ {t} ^ {\top} \boldsymbol {T} _ {t} ^ {- 1} \boldsymbol {\mu} _ {t} \\ - \frac {1}{2} (\boldsymbol {f} _ {t} - \boldsymbol {r} _ {t}) ^ {\top} \boldsymbol {\Lambda} _ {t} (\boldsymbol {f} _ {t} - \boldsymbol {r} _ {t}) = - \frac {1}{2} \boldsymbol {f} _ {t} ^ {\top} \boldsymbol {\Lambda} _ {t} \boldsymbol {f} _ {t} + \boldsymbol {r} _ {t} ^ {\top} \boldsymbol {\Lambda} _ {t} \boldsymbol {f} _ {t} - \frac {1}{2} \boldsymbol {r} _ {t} ^ {\top} \boldsymbol {\Lambda} _ {t} \boldsymbol {r} _ {t} \\ \end{array}
$$
$$
\begin{array}{l} \frac {1}{2} \left(\boldsymbol {f} _ {t} - \Phi_ {t} \boldsymbol {\theta} ^ {*}\right) ^ {\top} \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \left(\boldsymbol {f} _ {t} - \Phi_ {t} \boldsymbol {\theta} ^ {*}\right) = \frac {1}{2} \boldsymbol {f} _ {t} ^ {\top} \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \boldsymbol {f} _ {t} - \boldsymbol {\theta} ^ {* \top} \Phi_ {t} ^ {\top} \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \boldsymbol {f} _ {t} \\ + \frac {1}{2} \boldsymbol {\theta} ^ {* \top} \Phi_ {t} ^ {\top} \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \Phi_ {t} \boldsymbol {\theta} ^ {*}. \\ \end{array}
$$
We now equate coefficients to find  $A$ ,  $b$  and  $c$ . We find that  $A$  is
$$
\boldsymbol {A} = \boldsymbol {T} _ {t} ^ {- 1} + \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}.
$$
Note that  $\mathbf{A}$  is symmetric. We find that  $\mathbf{b}$  is
$$
\begin{array}{l} \boldsymbol {b} ^ {\top} \boldsymbol {A} = \boldsymbol {\mu} _ {t} ^ {\top} \boldsymbol {T} _ {t} ^ {- 1} + \boldsymbol {r} _ {t} ^ {\top} \boldsymbol {\Lambda} _ {t} - \boldsymbol {\theta} ^ {* \top} \Phi_ {t} ^ {\top} \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \\ \Rightarrow \boldsymbol {A} \boldsymbol {b} = \boldsymbol {T} _ {t} ^ {- 1} \boldsymbol {\mu} _ {t} + \boldsymbol {\Lambda} _ {t} \boldsymbol {r} _ {t} - \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \boldsymbol {\Phi} _ {t} \boldsymbol {\theta} ^ {*} \\ \Rightarrow \boldsymbol {b} = \left(\boldsymbol {T} _ {t} ^ {- 1} + \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) ^ {- 1} \left(\boldsymbol {T} _ {t} ^ {- 1} \boldsymbol {\mu} _ {t} + \boldsymbol {\Lambda} _ {t} \boldsymbol {r} _ {t} - \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \boldsymbol {\Phi} _ {t} \boldsymbol {\theta} ^ {*}\right). \\ \end{array}
$$
Finally, we find that  $c$  is
$$
\begin{array}{l} c = \frac {1}{2} \boldsymbol {b} ^ {\top} \boldsymbol {A} \boldsymbol {b} - \frac {1}{2} \boldsymbol {\mu} _ {t} ^ {\top} \boldsymbol {T} _ {t} ^ {- 1} \boldsymbol {\mu} _ {t} - \frac {1}{2} \boldsymbol {r} _ {t} ^ {\top} \boldsymbol {\Lambda} _ {t} \boldsymbol {r} _ {t} + \frac {1}{2} \boldsymbol {\theta} ^ {* \top} \boldsymbol {\Phi} _ {t} ^ {\top} \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \boldsymbol {\Phi} _ {t} \boldsymbol {\theta} ^ {*} \\ = \frac {1}{2} \left(\boldsymbol {T} _ {t} ^ {- 1} \boldsymbol {\mu} _ {t} + \boldsymbol {\Lambda} _ {t} \boldsymbol {r} _ {t} - \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \Phi_ {t} \boldsymbol {\theta} ^ {*}\right) ^ {\top} \left(\boldsymbol {T} _ {t} ^ {- 1} + \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) ^ {- 1} \left(\boldsymbol {T} _ {t} ^ {- 1} \boldsymbol {\mu} _ {t} + \boldsymbol {\Lambda} _ {t} \boldsymbol {r} _ {t} - \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \Phi_ {t} \boldsymbol {\theta} ^ {*\prime}\right) \\ - \frac {1}{2} \boldsymbol {\mu} _ {t} ^ {\top} \boldsymbol {T} _ {t} ^ {- 1} \boldsymbol {\mu} _ {t} - \frac {1}{2} \boldsymbol {r} _ {t} ^ {\top} \boldsymbol {\Lambda} _ {t} \boldsymbol {r} _ {t} + \frac {1}{2} \boldsymbol {\theta} ^ {* \top} \Phi_ {t} ^ {\top} \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \Phi_ {t} \boldsymbol {\theta} ^ {*} \\ \end{array}
$$
Now, we can rewrite and calculate the integral in (15) as
$$
\begin{array}{l} \frac {\exp (c)}{\sqrt {(2 \pi) ^ {t} \det (\boldsymbol {T} _ {t})}} \int \exp \left(- \frac {1}{2} (\boldsymbol {f} _ {t} - \boldsymbol {b}) ^ {\top} \boldsymbol {A} (\boldsymbol {f} _ {t} - \boldsymbol {b})\right) \mathrm {d} \boldsymbol {f} _ {t} = \frac {\exp (c) \sqrt {(2 \pi) ^ {t} \det (\boldsymbol {A} ^ {- 1})}}{\sqrt {(2 \pi) ^ {t} \det (\boldsymbol {T} _ {t})}} \\ = \exp (c) \sqrt {\frac {\det  (\boldsymbol {A} ^ {- 1})}{\det  (\boldsymbol {T} _ {t})}} \\ \end{array}
$$
Substituting this into (14), we obtain the constraint
$$
\begin{array}{l} \left(\Phi_ {t} \boldsymbol {\theta} ^ {*} - \boldsymbol {r} _ {t}\right) ^ {\top} \boldsymbol {\Lambda} _ {t} \left(\Phi_ {t} \boldsymbol {\theta} ^ {*} - \boldsymbol {r} _ {t}\right) \leq - 2 \ln \left(\exp (c) \sqrt {\frac {\det  (\boldsymbol {A} ^ {- 1})}{\det  (\boldsymbol {T} _ {t})}}\right) + 2 \ln (1 / \delta) \\ = - 2 c + \ln (\det (\boldsymbol {A} \boldsymbol {T} _ {t})) + 2 \ln (1 / \delta) \\ = - \left(\boldsymbol {T} _ {t} ^ {- 1} \boldsymbol {\mu} _ {t} + \boldsymbol {\Lambda} _ {t} \boldsymbol {r} _ {t} - \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \Phi_ {t} \boldsymbol {\theta} ^ {*}\right) ^ {\top} \left(\boldsymbol {T} _ {t} ^ {- 1} + \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) ^ {- 1} \left( \right.\boldsymbol {T} _ {t} ^ {- 1} \boldsymbol {\mu} _ {t} + \boldsymbol {\Lambda} _ {t} \boldsymbol {r} _ {t} - \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \Phi_ {t} \boldsymbol {\theta} ^ {*}\) \\ + \boldsymbol {\mu} _ {t} ^ {\top} \boldsymbol {T} _ {t} ^ {- 1} \boldsymbol {\mu} _ {t} + \boldsymbol {r} _ {t} ^ {\top} \boldsymbol {\Lambda} _ {t} \boldsymbol {r} _ {t} - \boldsymbol {\theta} ^ {* \top} \Phi_ {t} ^ {\top} \left(\boldsymbol {\Lambda} _ {t} - \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2}\right) \Phi_ {t} \boldsymbol {\theta} ^ {*} + \ln \left(\det (\mathbb {1} + \sigma^ {2} \boldsymbol {\Lambda} _ {t} ^ {2} \boldsymbol {T} _ {t})\right) + 2 \ln (1 / \delta) \\ \end{array}
$$
Note that  $\pmb{\theta}^{*}$  appears on both the left-hand-side and right-hand-side of this inequality. However, when  $\Lambda_{t} - \sigma^{2}\Lambda_{t}^{2}$  is the zero matrix (e.g. when  $\lambda_t\equiv 1 / \sigma^2$ ), all the  $\pmb{\theta}^{*}$ -dependent terms on the right-hand-side disappear.
B.2 The Special Case  $\lambda_t \equiv 1 / \sigma^2$
Starting from (13), choosing  $\lambda_t \equiv 1 / \sigma^2$ , taking the logarithm of both sides and then rearranging terms, we arrive at
$$
\left. \right.\left\| \Phi_ {t} \boldsymbol {\theta} ^ {*} - \boldsymbol {r} _ {t} \right\| _ {2} ^ {2} \leq - 2 \sigma^ {2} \ln \left(\underset {\boldsymbol {f} _ {t} \sim \mathcal {N} \left(\boldsymbol {\mu} _ {t}, \boldsymbol {T} _ {t}\right)} {\mathbb {E}} \left[ \exp \left(- \frac {1}{2 \sigma^ {2}} \left(\boldsymbol {f} _ {t} - \boldsymbol {r} _ {t}\right) ^ {\top} \left(\boldsymbol {f} _ {t} - \boldsymbol {r} _ {t}\right)\right)\right]\right) + 2 \sigma^ {2} \ln (1 / \delta). \tag {16}
$$
For any  $t \times t$  covariance matrix  $\pmb{T}$ , let  $Z(\pmb{T})$  denote the normalising constant of a Gaussian distribution with covariance  $\pmb{T}$ , so
$$
Z (\boldsymbol {T}) = \sqrt {(2 \pi) ^ {t} \det (\boldsymbol {T})}.
$$
For any  $t$ -dimensional vectors  $\pmb{x}$  and  $\pmb{\mu}$ , and any  $t\times t$  covariance matrix  $\pmb{T}$ , let  $p(\pmb{x}|\pmb{\mu},\pmb{T})$  denote the density function of a Gaussian distribution with mean  $\pmb{\mu}$  and covariance  $\pmb{T}$ , evaluated at  $\pmb{x}$ . This means that
$$
p (\pmb {x} | \pmb {\mu}, \pmb {T}) = \frac {1}{Z (\pmb {T})} \mathrm {e x p} \left(- \frac {1}{2} (\pmb {x} - \pmb {\mu}) ^ {\top} \pmb {T} ^ {- 1} (\pmb {x} - \pmb {\mu})\right).
$$
We will use the product of Gaussians trick from (Petersen et al., 2008) (Section 8.1.8, Equation 371), which states
$$
p (\boldsymbol {x} | \boldsymbol {\mu} _ {1}, \boldsymbol {\Sigma} _ {1}) p (\boldsymbol {x} | \boldsymbol {\mu} _ {2}, \boldsymbol {\Sigma} _ {2}) = p (\boldsymbol {\mu} _ {1} | \boldsymbol {\mu} _ {2}, \boldsymbol {\Sigma} _ {1} + \boldsymbol {\Sigma} _ {2}) p (\boldsymbol {x} | \boldsymbol {\mu} _ {c}, \boldsymbol {\Sigma} _ {c}), \tag {17}
$$
$$
\boldsymbol {\mu} _ {c} = \left(\boldsymbol {\Sigma} _ {1} ^ {- 1} + \boldsymbol {\Sigma} _ {2} ^ {- 1}\right) ^ {- 1} \left(\boldsymbol {\Sigma} _ {1} ^ {- 1} \boldsymbol {\mu} _ {1} + \boldsymbol {\Sigma} _ {2} ^ {- 1} \boldsymbol {\mu} _ {2}\right), \qquad \boldsymbol {\Sigma} _ {c} = \left(\boldsymbol {\Sigma} _ {1} ^ {- 1} + \boldsymbol {\Sigma} _ {2} ^ {- 1}\right) ^ {- 1}.
$$
$$
\begin{array}{l} \underset {\boldsymbol {f} _ {t} \sim \mathcal {N} (\boldsymbol {\mu} _ {t}, \boldsymbol {T} _ {t})} {\mathbb {E}} \left[ \exp \left(- \frac {1}{2 \sigma^ {2}} \left(\boldsymbol {f} _ {t} - \boldsymbol {r} _ {t}\right) ^ {\top} \left(\boldsymbol {f} _ {t} - \boldsymbol {r} _ {t}\right)\right) \right] = \underset {\boldsymbol {f} _ {t} \sim \mathcal {N} (\boldsymbol {\mu} _ {t}, \boldsymbol {T} _ {t})} {\mathbb {E}} \left[ Z (\sigma^ {2} \mathbb {1}) p \left(\boldsymbol {f} _ {t} | \boldsymbol {r} _ {t}, \sigma^ {2} \mathbb {1}\right) \right] \\ = Z (\sigma^ {2} \mathbb {1}) \int_ {\mathbb {R} ^ {t}} p (\pmb {f} _ {t} | \pmb {\mu} _ {t}, \pmb {T} _ {t}) p (\pmb {f} _ {t} | \pmb {r} _ {t}, \sigma^ {2} \mathbb {1}) \mathrm {d} \pmb {f} _ {t} \\ = Z \left(\sigma^ {2} \mathbb {1}\right) \int_ {\mathbb {R} ^ {t}} p \left(\boldsymbol {\mu} _ {t} \mid \boldsymbol {r} _ {t}, \boldsymbol {T} _ {t} + \sigma^ {2} \mathbb {1}\right) p \left(\boldsymbol {f} _ {t} \mid \boldsymbol {\mu} _ {c}, \boldsymbol {\Sigma} _ {c}\right) \mathrm {d} \boldsymbol {f} _ {t} \\ = Z \left(\sigma^ {2} \mathbb {1}\right) p \left(\boldsymbol {\mu} _ {t} \mid \boldsymbol {r} _ {t}, \boldsymbol {T} _ {t} + \sigma^ {2} \mathbb {1}\right) \\ = \sqrt {\frac {\operatorname * {d e t} (\sigma^ {2} \mathbb {1})}{\operatorname * {d e t} (\boldsymbol {T} _ {t} + \sigma^ {2} \mathbb {1})}} \exp \left(- \frac {1}{2} (\boldsymbol {\mu} _ {t} - \boldsymbol {r} _ {t}) ^ {\top} (\boldsymbol {T} _ {t} + \sigma^ {2} \mathbb {1}) ^ {- 1} (\boldsymbol {\mu} _ {t} - \boldsymbol {r} _ {t})\right) \\ \end{array}
$$
Substituting this into (16), the constraint becomes