SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Note that at each round, we either select an approximate zCDP mechanism or select a Brownian noise reduction, and in the latter case  $\delta_n(x) \equiv 0$  and  $\rho_n(x) \equiv 0$ , which then means  $P_n' \equiv P_n$  and  $Q_n' \equiv Q_n$  at those rounds  $n$ . We will write the distribution  $P_{1:n}$  for any prefix of outcomes from  $A_1(x), \dots, A_n(x)$  and similarly we will write the distribution  $Q_{1:n}$  for the prefix of outcomes from  $A_1(x'), \dots, A_n(x')$ . We can then write these likelihood as a convex combination of likelihoods, using the fact that  $\sum_{n=1}^{\infty} \delta_n(y_{1:n-1}) \leq \delta$  for all  $y_1, y_2, \dots$ .
$$
P _ {1: n} \left(y _ {1}, \dots , y _ {n}\right) = (1 - \delta) \underbrace {\prod_ {\ell = 1} ^ {n} P _ {\ell} ^ {\prime} \left(y _ {\ell} \mid y _ {1 : \ell - 1}\right)} _ {P _ {1: n} ^ {\prime} \left(y _ {1}, \dots , y _ {n}\right)} + \delta P _ {1: n} ^ {\prime \prime} \left(y _ {1}, \dots , y _ {n}\right). \tag {7}
$$
$$
Q _ {1: n} \left(y _ {1}, \dots , y _ {n}\right) = (1 - \delta) \underbrace {\prod_ {\ell = 1} ^ {n} Q _ {\ell} ^ {\prime} \left(y _ {\ell} \mid y _ {1 : \ell - 1}\right)} _ {Q _ {1: n} ^ {\prime} \left(y _ {1}, \dots , y _ {n}\right)} + \delta Q _ {1: n} ^ {\prime \prime} \left(y _ {1}, \dots , y _ {n}\right). \tag {8}
$$
For any fixed  $\lambda \geq 1$  and  $k\geq 1$ , consider the following filtration  $\mathcal{F}' = (\mathcal{F}_n')_{n\geq 1}$  where  $\mathcal{F}_n' = \sigma (Y_1',\dots ,Y_n')$ , with  $Y_{1}^{\prime}\sim P_{1}^{\prime},Y_{2}^{\prime}\mid Y_{1}^{\prime}\sim P_{2}^{\prime}(\cdot \mid Y_{1}^{\prime}),\dots ,Y_{k}^{\prime}\mid Y_{1:k - 1}^{\prime}\sim P_{k}^{\prime}(\cdot \mid Y_{1:k - 1}^{\prime})$ .
We will first consider the Brownian noise reduction mechanisms with time values  $(t_n^{(k)}: k \geq 1)_{n \geq 1}$  to be stopped at fixed rounds  $(k_n)_{n \geq 1}$ , although not every round will have a Brownian noise reduction mechanism selected with Brownian motion  $(B_n^{(t)})_{t > 0}$ . We will write out the privacy loss for the Brownian noise reduction mechanisms stopped at rounds  $(k_n)_{n \geq 1}$  as  $\mathcal{L}_n^{(1:k_n)}$  where from Theorem 3 we have,
$$
\mathcal {L} _ {n} ^ {(1: k _ {n})} = \mathcal {L} _ {n} ^ {(k _ {n})} = \frac {1}{2 t _ {n} ^ {(k _ {n})}} + \frac {B _ {n} ^ {(k _ {n})}}{t _ {n} ^ {(k _ {n})}}.
$$
We will add the noise reduction outcomes to the filtration, so that  $\mathcal{F}_n'' = \sigma(Y_1', \dots, Y_{n-1}', (B_n^{(t)})_{t \geq k_n})$ . From Lemma 6.1, we know for all  $\lambda \in \mathbb{R}$
$$
\begin{array}{l} \mathbb {E} \left[ \exp \left(\lambda \left(\mathcal {L} _ {n} ^ {(1: k _ {n})} - \frac {1}{2 t _ {n} ^ {(k _ {n})}}\right) - \frac {\lambda^ {2}}{2 t _ {n} ^ {(k _ {n})}}\right) \mid \mathcal {F} _ {n - 1} ^ {\prime \prime} \right] \\ = \mathbb {E} \left[ \exp \left(\lambda \mathcal {L} _ {n} ^ {(1: k _ {n})} - \frac {\lambda (\lambda + 1)}{2 t _ {n} ^ {(k _ {n})}}\right) \mid \mathcal {F} _ {n - 1} ^ {\prime \prime} \right] \\ \leq 1. \\ \end{array}
$$
We then replace  $(k_{i})_{i\leq n}$  with an adaptive stopping time functions  $(T_{i})_{i\leq n}$  with corresponding stopping times  $(T_{i}(x))_{i\leq n}$ , rather than fixing them in advance, in which case we rename  $\mathcal{L}_n^{(1:k_n)}$  as  $\mathcal{L}_n^{\mathtt{BM}_n}$  and the same inequality still holds with the filtration  $\mathcal{F}_n^{\prime \prime \prime} = \sigma (Y_1',\dots ,Y_{n - 1}',(B_n^{(t)})_{t\geq T_n(x)})$ . Note that we will call  $Y_{n}^{\prime} = (f(x) + B_{n}^{(t)})_{t\geq T_{n}(x)}$  so that  $\mathcal{F}_n^{\prime \prime \prime} = \mathcal{F}_n'$ .
At rounds  $n$  where a Brownian noise reduction mechanism is not selected, we simply have  $1 / t_n^{(k_n)} = 0$  and  $\mathcal{L}_n^{(1:k_n)} = 0$ . We also consider the modified privacy losses for the approximate zCDP mechanisms  $\mathcal{L}_n'$  where
$$
\mathcal {L} _ {n} ^ {\prime} = \mathcal {L} _ {n} ^ {\prime} (y _ {1: n}) = \log \left(\frac {P _ {n} ^ {\prime} (y _ {n} | y _ {<   n})}{Q _ {n} ^ {\prime} (y _ {n} | y _ {<   n})}\right), \text {w h e r e} y _ {1: n} \sim P _ {1: n} ^ {\prime}.
$$
Due to mechanisms being zCDP, we then have for any  $\lambda \geq 1$
$$
\mathbb {E} \left[ \exp \left((\lambda - 1) \mathcal {L} _ {n} ^ {\prime} - \lambda (\lambda - 1) \rho_ {n} \left(Y _ {1: n - 1} ^ {\prime}\right) / 2\right) \mid \mathcal {F} _ {n - 1} ^ {\prime} \right] \leq 1.
$$
Because at each round  $n$ , the mechanism selected is either approximate zCDP or a Brownian noise reduction with a stopping function, we can write the privacy loss at each round  $i$  as the sum  $\mathcal{L}_i' + \mathcal{L}_i^{\mathrm{BM}}$  so that for all  $\lambda \geq 1$  we have
$$
X _ {n} ^ {(\lambda)} := \sum_ {i \leq n} \left\{\mathcal {L} _ {i} ^ {\prime} + \mathcal {L} _ {i} ^ {\mathrm {B M}} - \lambda \left(\rho_ {i} \left(Y _ {1: n - 1} ^ {\prime}\right) + \frac {1}{t _ {i} ^ {(T _ {i} (x))}}\right) \right\}, \tag {9}
$$
$$
M _ {n} ^ {(\lambda)} := \exp \left((\lambda - 1) X _ {n} ^ {(\lambda)}\right). \tag {10}
$$
We know that  $(M_{n})$  is a supermartingale with respect to  $(\mathcal{F}_n^{\prime})_{n\geq 1}$ . From the optional stopping theorem, we have
$$
\mathbb {E} \left[ M _ {N (x)} ^ {(\lambda)} \right] \leq 1.
$$
This will ensure (4) holds. Although for rounds  $n$  where we select a Brownian noise reduction we have  $P_{n}^{\prime} = P_{n}$  and  $Q_{n}^{\prime} = Q_{n}$ , we still need to show that for rounds  $n$  where approximate zCDP mechanisms were selected the original distributions  $P_{n}$  and  $Q_{n}$  can be written as weighted mixtures including  $P_{n}^{\prime}$  and  $Q_{n}^{\prime}$ , respectively. This follows from the same analysis as in Whitehouse et al. [2023], so that for all outcomes  $y_{1}, y_{2}, \dots$  where  $\sum_{n=1}^{\infty} \delta_{n}(y_{1:n-1}) \leq \delta$  we have
$$
P _ {1: n} \left(y _ {1}, y _ {2}, \dots , y _ {n}\right) \geq (1 - \delta) \prod_ {m = 1} ^ {n} P _ {m} ^ {\prime} \left(y _ {m} \mid y _ {1: m - 1}\right),
$$
and similarly for  $Q_{1:n}$ . As is argued in Whitehouse et al. [2023], it suffices to show that the two likelihoods of the stopped process  $P_{1:N}$  and  $Q_{1:N}$  can be decomposed as weighted mixtures of  $P_{1:N}'$  and  $P_{1:N}''$  as well as  $Q_{1:N}'$  and  $Q_{1:N}''$ , respectively such that the weights on  $P_{1:N}'$  and  $Q_{1:N}'$  are at least  $1 - \delta$ . Note that from our stopping rule, we haven for all  $\lambda > 0$
$$
\mathbb {E} \left[ M _ {N (x)} ^ {(\lambda)} \right] \leq 1 \Rightarrow \mathbb {E} \left[ e ^ {\lambda \sum_ {n = 1} ^ {N (x)} \left(\mathcal {L} _ {n} ^ {\prime} + \mathcal {L} _ {n} ^ {\mathtt {B M}}\right)} \right] \leq e ^ {\lambda (\lambda + 1) \rho}.
$$
We then still need to convert to DP, which we do with the stopping rule  $N(x)$  and the conversion lemma between approximate zCDP and DP in Lemma 2.1.
Figure 3: Data from a Zipf Distribution with parameter  $a = 0.75$  and max value of 300.
B.4 Data Distribution Plot from Section 7
In Section 7 we generate data from a Zipfian distribution, which we present in Figure 3.
C Noise Reduction Mechanisms
We provide a more general approach to handle both the Laplace and Brownian noise reduction mechanisms from Koufogiannis et al. [2017], Ligett et al. [2017], Whitehouse et al. [2022]. Consider a discrete set of times  $\{a^{(j)}\}$  with  $0 < a^{(j)} \leq b$ . We will allow the stopping time and the number of steps to be adaptive. We will handle the univariate case and note that the analysis extends to the multivariate case. We will write time functions  $(\phi^{(k)}, k \geq 1)$  to satisfy the following
$$
\phi^ {(1)} \equiv b > 0,
$$
$$
\phi^ {(k)}: \left\{a ^ {(j)} \right\} \times \mathbb {R} \rightarrow \left\{a ^ {(j)} \right\} \text {s u c h} \phi^ {(k)} (t, z) <   t \text {f o r a l l} (t, z) \in \left\{a ^ {(j)} \right\} \times \mathbb {R} \text {a n d} k \geq 2.
$$
Consider the noise reduction mechanisms where we have  $M^{(1,p)} = t^{(1)} = b$ ,  $M^{(1,q)}(x) = f(x) + Z(t^{(1)})$  where  $Z(t)$  is either a standard Brownian motion or a Laplace process Koufogiannis et al. [2017], and for  $k > 1$  we have
$$
M ^ {(k, p)} (x) = \phi^ {(k)} \left(M ^ {(k - 1, p)} (x), M ^ {(k - 1, q)} (x)\right),
$$
$$
M ^ {(k, q)} (x) = f (x) + Z \left(M ^ {(k, p)} (x)\right).
$$
We will then write
$$
M ^ {(1: k)} (x) = \left(\left(M ^ {(k, p)} (x), M ^ {(k, q)} (x)\right), \dots , \left(M ^ {(1, p)} (x), M ^ {(1, q)} (x)\right)\right).
$$
We then consider the extended mechanism  $M^{*}(x) = (T(x), M^{(1:T(x))}(x))$ , where  $T(x)$  is a stopping function. Hence,  $M^{*}(x)$  takes values in  $\mathbb{Z} \times ((0,b) \times \mathbb{R})^{\infty,*}$  where  $((0,b] \times \mathbb{R})^{\infty,*}$  is a collection of all finite sequences of the type  $((t^{(i)}, z^{(i)}): i \in [k])$  for  $k = 1,2,\dots$  and  $0 < t^{(k)} < \dots$ ,  $t^{(1)} = b$  and  $z^{(i)} \in \mathbb{R}$  for  $i \in [k]$ .
Let  $m^*$  be the probability measure on this space generated by  $M^*(x^*)$  where  $x^*$  is a point where  $f(x^*) = 0$ . Note that  $x^*$  might be a fake value that needs to be added to make the function equal zero. Let  $m(x)$  be the probability measure on this space generated by  $M^*(x)$ . We will then compute densities with respect to  $m^*$ .
Let  $A$  be a measurable set in  $\mathbb{Z} \times ((0, b] \times \mathbb{R})^{\infty, *}$ . Let  $A^{(j)} \subseteq A$  on which the last (smallest)  $p$ -coordinate of every point is equal to  $a^{(j)}$ . Then
$$
\Pr [ M ^ {*} (x) \in A ] = \sum_ {j} \Pr [ M ^ {*} (x) \in A ^ {(j)} ].
$$
By construction,  $M^{*}(x)$  is a function of  $(Z(t) + f(x), 0 \leq t \leq b)$ . Therefore, for each  $j$ , we have
$$
\Pr \left[ M ^ {*} (x) \in A ^ {(j)} \right] = \Pr \left[ \left(Z (t) + f (x), 0 \leq t \leq b\right) \in \left(M ^ {*}\right) ^ {- 1} \left(A ^ {(j)}\right) \right],
$$
where the set of paths is of the following form for a measurable subset  $D^{(j)} \in C([a^{(j)}, b])$  of continuous functions on  $[a^{(j)}, b]$ :
$$
(M ^ {*}) ^ {- 1} (A ^ {(j)}) = \left\{(y (t): 0 \leq t \leq b): (y (t): a ^ {(j)} \leq t \leq b) \in D ^ {(j)} \right\}.
$$
The standard Brownian motion and the Laplace process both have independence of increments, so we have the following with  $p_t$  as the density for  $Z(t)$
$$
\begin{array}{l} \Pr \left[ \left(Z (t) + f (x): 0 \leq t \leq b\right) \in \left\{\left(y (t): 0 \leq t \leq b\right): \left(y (t): a ^ {(j)} \leq t \leq b\right) \in D ^ {(j)} \right\} \right] \\ = \Pr [ (Z (t) + f (x): a ^ {(j)} \leq t \leq b) \in D ^ {(j)} ] \\ = \mathbb {E} \left[ \mathbb {1} \left\{\left(Z (t): a ^ {(j)} \leq t \leq b\right) \in D ^ {(j)} \right\} \frac {p _ {a ^ {(j)}} \left(Z \left(a ^ {(j)} - f (x)\right)\right)}{p _ {a ^ {(j)}} \left(Z \left(a ^ {(j)}\right)\right)} \right] \\ = \mathbb {E} \left[ \mathbb {1} \left\{M ^ {*} (x ^ {*}) \in A ^ {(j)} \right\} \frac {p _ {M ^ {(T (x ^ {*}) , q)} (x ^ {*})} (Z (M ^ {(T (x ^ {*} , q)} (x ^ {*})) - f (x))}{p _ {M ^ {(T (x ^ {*}) , q)} (x ^ {*})} (Z (M ^ {(T (x ^ {*} , q)} (x ^ {*})))} \right], \\ \end{array}
$$
where the second equality follows from the fact that the law of a shifted process with independent increments on  $(a^{(j)},b)$  is equivalent to the law of the non-shifted process and its density is the ratio of the two densities evaluated at its left most point. We then conclude that
$$
\Pr [ M ^ {*} (x) \in A ] = \mathbb {E} \left[ \mathbb {1} \left\{M ^ {*} (x ^ {*}) \in A \right\} \frac {p _ {M ^ {(T (x ^ {*}) , p)} (x ^ {*})} \left(Z \left(M ^ {(T (x ^ {*} , p)} (x ^ {*})\right) - f (x)\right)}{p _ {M ^ {(T (x ^ {*} , p)} (x ^ {*})} \left(Z \left(M ^ {(T (x ^ {*} , p)} (x ^ {*}))\right)\right)}. \right]
$$
Therefore  $m(x)$  is absolutely continuous with respect to  $m^*$ . To write the density (the Radon-Nikodym derivative), recall that the density is evaluated at some point  $\mathbf{s} \in \mathbb{Z} \times (\{a^{(j)}\} \times \mathbb{R})^{\infty, *}$ . Let  $t(\mathbf{s})$  be the smallest  $p$ -value in  $\mathbf{s}$  and  $z(\mathbf{s})$  be the corresponding space value. Then we have
$$
p _ {x} ^ {(1: T (x))} (\mathbf {s}) = \frac {d m (x)}{d m ^ {*}} (\mathbf {s}) = \frac {p _ {t (\mathbf {s})} (z (\mathbf {s}) - f (x))}{p _ {t (\mathbf {s})} (z (\mathbf {s}))}.
$$
Similarly, we consider  $M_{\mathrm{last}}^{*}(x) = (T(x), M^{(T(x))}(x))$  so that the state space is  $\mathbb{Z} \times \{a^{(j)}\} \times \mathbb{R}$ . We will write  $\hat{m}^*$  to be the probability measure on this space generated by  $M_{\mathrm{last}}^{*}(x^{*})$  and  $\hat{m}(x)$  to be the probability measure on this space generated by  $M_{\mathrm{last}}^{*}(x)$ . We then compute densities with respect to  $\hat{m}^*$ .
Let  $A$  now be a measurable subset of  $\mathbb{Z} \times \{a^{(j)}\} \times \mathbb{R}$  and let  $A^{(j)} \subseteq A$  with the  $p$ -coordinate equal to  $a^{(j)}$ . Then we follow a similar argument to what we have above to show that
$$
\Pr \left[ M _ {\text {l a s t}} ^ {*} (x) \in A \right] = \mathbb {E} \left[ \mathbb {1} \left\{M _ {\text {l a s t}} ^ {*} (x ^ {*}) \in A \right\} \frac {p _ {M ^ {(T (x ^ {*}) , p)} (x ^ {*})} \left(Z \left(M ^ {(T (x ^ {*}) , p)} (x ^ {*})\right) - f (x)\right)}{p _ {M ^ {(T (x ^ {*}) , p)} (x ^ {*})} \left(Z \left(M ^ {(T (x ^ {*}) , p)} (x ^ {*})\right)\right)} \right].
$$
Hence, for  $(k,t,z)\in \mathbb{Z}\times \{a^{(j)}\} \times \mathbb{R}$  we have
$$
p _ {x} ^ {(T (x))} (k, t, z) = \frac {d m (x)}{d m ^ {*}} (k, t, z) = \frac {p _ {t} (z - f (x))}{p _ {t} (z)}.
$$
We then consider the privacy loss for both  $M^{*}$ , denoted as  $\mathcal{L}^{(1:T(x))}$ , and  $M_{\mathrm{last}}^{*}$ , denoted as  $\mathcal{L}^{(T(x))}$  under neighboring data  $x$  and  $x'$ . We have
$$
\mathcal {L} ^ {(1: T (x))} = \log \left(\frac {p _ {M ^ {(T (x) , p)} (x)} (Z (M ^ {(T (x) , p)} (x)) - f (x))}{p _ {M ^ {(T (x) , p)} (x)} (Z (M ^ {(T (x) , p)} (x)) - f \left(x ^ {\prime}\right))}\right) = \mathcal {L} ^ {(T (x))}. \tag {11}
$$
We then instantiate the Brownian noise reduction, in which case
$$
p _ {t} (z - f (x)) / p _ {t} (z) = \exp \left(- \frac {1}{2 t} (z - f (x)) ^ {2} + \frac {1}{2 t} z ^ {2}\right) = \exp \left(\frac {f (x)}{t} z - \frac {f (x) ^ {2}}{2 t}\right).
$$
Without loss of generality, consider the function  $g(y) = f(y) - f(x)$  so that we apply
$$
\log \left(\frac {p _ {t} (z - g (x))}{p _ {t} (z - g (x ^ {\prime}))}\right) = - \frac {g (x ^ {\prime})}{| g (x ^ {\prime}) | t} z | g (x ^ {\prime}) | + \frac {g (x ^ {\prime}) ^ {2}}{2 t}.
$$
We then use the Brownian motion  $(B(t))_{t\geq 0}$  to get the privacy loss  $\mathcal{L}_{\mathtt{BM}}^{(1:T(x))}$
$$
\begin{array}{l} \mathcal {L} _ {\mathrm {B M}} ^ {(1: T (x))} = \frac {(f (x) - f \left(x ^ {\prime}\right) ^ {2}}{2 t} - \frac {f (x) - f \left(x ^ {\prime}\right)}{\left| f (x) - f \left(x ^ {\prime}\right) \right| t} | f (x) - f \left(x ^ {\prime}\right) | \cdot B \left(M ^ {(T (x), p)} (x)\right) \\ = \frac {(f (x) - f (x ^ {\prime}) ^ {2}}{2 t} - \frac {1}{t} | f (x) - f (x ^ {\prime}) | \cdot W \left(M ^ {T (x), p} (x)\right), \\ \end{array}
$$
where  $(W(t) = z\cdot B(t))$  is a standard Brownian motion for  $|z| = 1$
There is another noise reduction mechanism based on Laplace noise, originally from Koufogiannis et al. [2017] and shown to be ex-post private in Ligett et al. [2017]. We first show that it is indeed a noise reduction mechanism with a stopping function and consider the resulting privacy loss random variable. We focus on the univariate case, yet the analysis extends to the multivariate case.
We construct a Markov process  $(X(t))_{t\geq 0}$  with  $X(0) = 0$  such that, for each  $t > 0$ ,  $X(t)$  has the Laplace distribution with parameter  $t$ , which has density  $p(z) = \frac{1}{2t}\exp (-|z| / t)$  for  $z\in \mathbb{R}$ . The process we construct has independent increments, i.e. for any  $0\leq t^{(1)} < t^{(2)} < \dots < t^{(k)}$  the following differences are independent,
$$
X (t ^ {(0)}), X (t ^ {(1)}) - X (t ^ {(0)}), \dots , X (t ^ {(k)}) - X (t ^ {(k - 1)}).
$$
Hence, the process is Markovian. The idea of constructing such a process is that a Laplace random variable with parameter  $t > 0$  is a symmetric, infinitely divisible random variable without a Gaussian component whose Lévy measure has density
$$
g (z; t) = \frac {e ^ {- | z | / t}}{| z |}, z \neq 0.
$$
Let  $U$  be an infinitely divisible random measure on  $[0, \infty)$  with Lebesgue control measure and local Lévy density
$$
\psi (z; t) = t ^ {- 2} e ^ {- | z | / t}, z \neq 0, t > 0. \tag {12}
$$
We define  $X(t) = U([0, t])$ , where  $t \geq 0$ . Then  $X(0) = 0$  and for  $t > 0$ ,  $X(t)$  is a symmetric, infinitely divisible random variable without a Gaussian component, whose Lévy measure has the density equal to
$$
\int_ {0} ^ {t} \psi (z; u) d u = \int_ {0} ^ {t} u ^ {- 2} e ^ {- | z | / u} d u = \int_ {1 / t} ^ {\infty} e ^ {- | z | s} d s = g (z; t).
$$
That is  $X(t)$  is a Laplace random variable with parameter  $t$  and the resulting process has independent increments by construction. Note that the process  $(X(t))_{t\geq 0}$  has infinitely many jumps near 0. On an interval  $[t^{(1)},t^{(2)}]$  with  $0 < t^{(1)} < t^{(1)}$ , it is flat with probability  $(t^{(1)} / t^{(2)})^2$ .
We can then describe the Laplace noise reduction algorithm in a way similar to the Brownian Noise Reduction.
Definition C.1 (Laplace Noise Reduction). Let  $f: \mathcal{X} \to \mathbb{R}^d$  be a function and  $(t^{(k)})_{k \geq 1}$  a sequence of time values. Let  $(X(t))_{t \geq 0}$  be the Markov process described above independently in each coordinate and  $T: (\mathbb{R}^d)^* \to \mathbb{N}$  be a stopping function. The Laplace noise reduction associated with  $f$ , time values  $(t^{(k)})_{k \geq 1}$ , and stopping function  $T$  is the algorithm  $LNR: \mathcal{X} \to ((0, t^{(1)}] \times \mathbb{R}^d)^*$  given by
$$
L N R (x) = L N R ^ {(1: T (x))} (x) = \left(t ^ {(k)}, f (x) + X \left(t ^ {(k)}\right)\right) _ {k \leq T (x)}.
$$
We then aim to prove the following
Lemma C.1. The privacy loss  $\mathcal{L}_{LNR}$  for the Laplace noise reduction associated with  $f$ , time values  $(t^{(k)})_{k\geq 1}$ , and stopping function  $T$  can be written as
$$
\mathcal {L} _ {L N R} = \mathcal {L} _ {L N R} ^ {(1: T (x))} = \mathcal {L} _ {L N R} ^ {(T (x))}.
$$
Furthermore we have
$$
\mathcal {L} _ {L N R} ^ {(1: T (x))} = - \frac {1}{M ^ {(T (x) , p)} (x)} \left(| X (M ^ {(T (x), p)} (x)) | - | X (M ^ {(T (x), p)} (x)) - \Delta_ {1} (f) |\right).
$$
Proof. Because the Laplace process has independent increments, we get the same expression for the privacy loss as in (11) where we substitute the Laplace process for  $Z(t)$ , which has the following density
$$
p _ {t} (z - f (x)) / p _ {t} (z) = \exp \left(- \frac {1}{t} \left(| z - f (x) | - | z |\right)\right).
$$
We will again use  $g(y) = f(y) - f(x)$  for neighboring datasets  $x, x'$  with  $f(x') > f(x)$  without loss of generality to get
$$
\log \left(p _ {t} (z - g (x)) / p _ {t} (z - g (x ^ {\prime}))\right) = - \frac {1}{t} \left(| z | - | z - g (x ^ {\prime}) |\right).
$$
Therefore, we have from (11) that
$$
\begin{array}{l} \mathcal {L} _ {\mathrm {L N R}} ^ {(1: T (x))} = \mathcal {L} _ {\mathrm {L N R}} ^ {(T (x))} \\ = - \frac {1}{M ^ {(T (x) , p)} (x)} \left(| X (M ^ {(T (x), p)} (x)) | - | X (M ^ {(T (x), p)} (x)) - | f (x) - f (x ^ {\prime}) | |\right), \\ \end{array}
$$
We now want to show a similar result that we had for the Brownian noise reduction mechanism in Lemma 6.1, but for the Laplace process. Instead of allowing general time step functions  $\phi^{(j)}$ , we will simply look at time values  $0 < t^{(k)} < \dots < t^{(1)} = b$  so that
$$
\mathcal {L} _ {\mathrm {L N R}} ^ {(1: T (x))} = - \frac {1}{t ^ {(T (x))}} \left(| X (t ^ {(T (x))}) | - | X (t ^ {(T (x))}) - | f (x) - f (x ^ {\prime}) | |\right).
$$
This form of the privacy loss for the Laplace noise reduction might be helpful in determining whether we can get a similar backward martingale as in the Brownian noise reduction case in Lemma 6.1. We leave the problem open to try to get a composition bound for Laplace noise reduction mechanisms that improves over simply adding up the ex-post privacy bounds as in Theorem B.1.