## A Tail of Two Densities

This is Part 1 of a two-part post where I give an introduction to the mathematics of differential privacy.

Practically speaking, differential privacy is a technique of perturbing database queries so that query results do not leak too much information while still being relatively accurate.

This post however focuses on the mathematical aspects of differential privacy, which is a study of tail bounds of the divergence between two probability measures, with the end goal of applying it to stochastic gradient descent. This post should be suitable for anyone familiar with probability theory.

I start with the definition of $$\epsilon$$-differential privacy (corresponding to max divergence), followed by $$(\epsilon, \delta)$$-differential privacy (a.k.a. approximate differential privacy, corresponding to the $$\delta$$-approximate max divergence). I show a characterisation of the $$(\epsilon, \delta)$$-differential privacy as conditioned $$\epsilon$$-differential privacy. Also, as examples, I illustrate the $$\epsilon$$-dp with Laplace mechanism and, using some common tail bounds, the approximate dp with the Gaussian mechanism.

Then I continue to show the effect of combinatorial and sequential compositions of randomised queries (called mechanisms) on privacy by stating and proving the composition theorems for differential privacy, as well as the effect of mixing mechanisms, by presenting the subsampling theorem (a.k.a. amplification theorem).

In Part 2, I discuss the Rényi differential privacy, corresponding to the Rényi divergence, a study of the moment generating functions of the divergence between probability measures to derive the tail bounds.

Like in Part 1, I prove a composition theorem and a subsampling theorem.

I also attempt to reproduce a seemingly better moment bound for the Gaussian mechanism with subsampling, with one intermediate step which I am not able to prove.

After that I explain the Tensorflow implementation of differential privacy in its Privacy module, which focuses on the differentially private stochastic gradient descent algorithm (DP-SGD).

Finally I use the results from both Part 1 and Part 2 to obtain some privacy guarantees for composed subsampling queries in general, and for DP-SGD in particular. I also compare these privacy guarantees.

Acknowledgement. I would like to thank Stockholm AI for introducing me to the subject of differential privacy. Thanks to Amir Hossein Rahnama for hosting the discussions at Stockholm AI. Thanks to (in chronological order) Reynaldo Boulogne, Martin Abedi, Ilya Mironov, Kurt Johansson, Mark Bun, Salil Vadhan, Jonathan Ullman, Yuanyuan Xu and Yiting Li for communication and discussions. Also thanks to the r/MachineLearning community for comments and suggestions which result in improvement of readability of this post. The research was done while working at KTH Department of Mathematics.

If you are confused by any notations, ask me or try this. This post (including both Part 1 and Part2) is licensed under CC BY-SA and GNU FDL.

## The gist of differential privacy

If you only have one minute, here is what differential privacy is about:

Let $$p$$ and $$q$$ be two probability densities, we define the divergence variable1 of $$(p, q)$$ to be

$L(p || q) := \log {p(\xi) \over q(\xi)}$

where $$\xi$$ is a random variable distributed according to $$p$$.

Roughly speaking, differential privacy is the study of the tail bound of $$L(p || q)$$: for certain $$p$$s and $$q$$s, and for $$\epsilon > 0$$, find $$\delta(\epsilon)$$ such that

$\mathbb P(L(p || q) > \epsilon) < \delta(\epsilon),$

where $$p$$ and $$q$$ are the laws of the outputs of a randomised functions on two very similar inputs. Moreover, to make matters even simpler, only three situations need to be considered:

1. (General case) $$q$$ is in the form of $$q(y) = p(y + \Delta)$$ for some bounded constant $$\Delta$$.
2. (Compositions) $$p$$ and $$q$$ are combinatorial or sequential compositions of some simpler $$p_i$$'s and $$q_i$$'s respectively
3. (Subsampling) $$p$$ and $$q$$ are mixtures / averages of some simpler $$p_i$$'s and $$q_i$$'s respectively

In applications, the inputs are databases and the randomised functions are queries with an added noise, and the tail bounds give privacy guarantees. When it comes to gradient descent, the input is the training dataset, and the query updates the parameters, and privacy is achieved by adding noise to the gradients.

Now if you have an hour...

## $$\epsilon$$-dp

Definition (Mechanisms). Let $$X$$ be a space with a metric $$d: X \times X \to \mathbb N$$. A mechanism $$M$$ is a function that takes $$x \in X$$ as input and outputs a random variable on $$Y$$.

In this post, $$X = Z^m$$ is the space of datasets of $$m$$ rows for some integer $$m$$, where each item resides in some space $$Z$$. In this case the distance $$d(x, x') := \#\{i: x_i \neq x'_i\}$$ is the number of rows that differ between $$x$$ and $$x'$$.

Normally we have a query $$f: X \to Y$$, and construct the mechanism $$M$$ from $$f$$ by adding a noise:

$M(x) := f(x) + \text{noise}.$

Later, we will also consider mechanisms constructed from composition or mixture of other mechanisms.

In this post $$Y = \mathbb R^d$$ for some $$d$$.

Definition (Sensitivity). Let $$f: X \to \mathbb R^d$$ be a function. The sensitivity $$S_f$$ of $$f$$ is defined as

$S_f := \sup_{x, x' \in X: d(x, x') = 1} \|f(x) - f(x')\|_2,$

where $$\|y\|_2 = \sqrt{y_1^2 + ... + y_d^2}$$ is the $$\ell^2$$-norm.

Definition (Differential Privacy). A mechanism $$M$$ is called $$\epsilon$$-differential privacy ($$\epsilon$$-dp) if it satisfies the following condition: for all $$x, x' \in X$$ with $$d(x, x') = 1$$, and for all measureable set $$S \subset \mathbb R^n$$,

$\mathbb P(M(x) \in S) \le e^\epsilon P(M(x') \in S). \qquad (1)$

Practically speaking, this means given the results from perturbed query on two known databases that differs by one row, it is hard to determine which result is from which database.

An example of $$\epsilon$$-dp mechanism is the Laplace mechanism.

Definition. The Laplace distribution over $$\mathbb R$$ with parameter $$b > 0$$ has probability density function

$f_{\text{Lap}(b)}(x) = {1 \over 2 b} e^{- {|x| \over b}}.$

Definition. Let $$d = 1$$. The Laplace mechanism is defined by

$M(x) = f(x) + \text{Lap}(b).$

Claim. The Laplace mechanism with

$b \ge \epsilon^{-1} S_f \qquad (1.5)$

is $$\epsilon$$-dp.

Proof. Quite straightforward. Let $$p$$ and $$q$$ be the laws of $$M(x)$$ and $$M(x')$$ respectively.

${p (y) \over q (y)} = {f_{\text{Lap}(b)} (y - f(x)) \over f_{\text{Lap}(b)} (y - f(x'))} = \exp(b^{-1} (|y - f(x')| - |y - f(x)|))$

Using triangular inequality $$|A| - |B| \le |A - B|$$ on the right hand side, we have

${p (y) \over q (y)} \le \exp(b^{-1} (|f(x) - f(x')|)) \le \exp(\epsilon)$

where in the last step we use the condition (1.5). $$\square$$

## Approximate differential privacy

Unfortunately, $$\epsilon$$-dp does not apply to the most commonly used noise, the Gaussian noise. To fix this, we need to relax the definition a bit.

Definition. A mechanism $$M$$ is said to be $$(\epsilon, \delta)$$-differentially private if for all $$x, x' \in X$$ with $$d(x, x') = 1$$ and for all measureable $$S \subset \mathbb R^d$$

$\mathbb P(M(x) \in S) \le e^\epsilon P(M(x') \in S) + \delta. \qquad (2)$

Immediately we see that the $$(\epsilon, \delta)$$-dp is meaningful only if $$\delta < 1$$.

### Indistinguishability

To understand $$(\epsilon, \delta)$$-dp, it is helpful to study $$(\epsilon, \delta)$$-indistinguishability.

Definition. Two probability measures $$p$$ and $$q$$ on the same space are called $$(\epsilon, \delta)$$-ind(istinguishable) if for all measureable sets $$S$$:

\begin{aligned} p(S) \le e^\epsilon q(S) + \delta, \qquad (3) \\ q(S) \le e^\epsilon p(S) + \delta. \qquad (4) \end{aligned}

As before, we also call random variables $$\xi$$ and $$\eta$$ to be $$(\epsilon, \delta)$$-ind if their laws are $$(\epsilon, \delta)$$-ind. When $$\delta = 0$$, we call it $$\epsilon$$-ind.

Immediately we have

Claim 0. $$M$$ is $$(\epsilon, \delta)$$-dp (resp. $$\epsilon$$-dp) iff $$M(x)$$ and $$M(x')$$ are $$(\epsilon, \delta)$$-ind (resp. $$\epsilon$$-ind) for all $$x$$ and $$x'$$ with distance $$1$$.

Definition (Divergence Variable). Let $$p$$ and $$q$$ be two probability measures. Let $$\xi$$ be a random variable distributed according to $$p$$, we define a random variable $$L(p || q)$$ by

$L(p || q) := \log {p(\xi) \over q(\xi)},$

and call it the divergence variable of $$(p, q)$$.

One interesting and readily verifiable fact is

$\mathbb E L(p || q) = D(p || q)$

where $$D$$ is the KL-divergence.

Claim 1. If

\begin{aligned} \mathbb P(L(p || q) \le \epsilon) &\ge 1 - \delta, \qquad(5) \\ \mathbb P(L(q || p) \le \epsilon) &\ge 1 - \delta \end{aligned}

then $$p$$ and $$q$$ are $$(\epsilon, \delta)$$-ind.

Proof. We verify (3), and (4) can be shown in the same way. Let $$A := \{y \in Y: \log {p(y) \over q(y)} > \epsilon\}$$, then by (5) we have

$p(A) < \delta.$

So

$p(S) = p(S \cap A) + p(S \setminus A) \le \delta + e^\epsilon q(S \setminus A) \le \delta + e^\epsilon q(S).$

$$\square$$

This Claim translates differential privacy to the tail bound of divergence variables, and for the rest of this post all dp results are obtained by estimating this tail bound.

In the following we discuss the converse of Claim 1. The discussions are rather technical, and readers can skip to the next subsection on first reading.

The converse of Claim 1 is not true.

Claim 2. There exists $$\epsilon, \delta > 0$$, and $$p$$ and $$q$$ that are $$(\epsilon, \delta)$$-ind, such that

\begin{aligned} \mathbb P(L(p || q) \le \epsilon) &< 1 - \delta, \\ \mathbb P(L(q || p) \le \epsilon) &< 1 - \delta \end{aligned}

Proof. Here's a example. Let $$Y = \{0, 1\}$$, and $$p(0) = q(1) = 2 / 5$$ and $$p(1) = q(0) = 3 / 5$$. Then it is not hard to verify that $$p$$ and $$q$$ are $$(\log {4 \over 3}, {1 \over 3})$$-ind: just check (3) for all four possible $$S \subset Y$$ and (4) holds by symmetry. On the other hand,

$\mathbb P(L(p || q) \le \log {4 \over 3}) = \mathbb P(L(q || p) \le \log {4 \over 3}) = {2 \over 5} < {2 \over 3}.$

$$\square$$

A weaker version of the converse of Claim 1 is true (Kasiviswanathan-Smith 2015), though:

Claim 3. Let $$\alpha > 1$$. If $$p$$ and $$q$$ are $$(\epsilon, \delta)$$-ind, then

$\mathbb P(L(p || q) > \alpha \epsilon) < {1 \over 1 - \exp((1 - \alpha) \epsilon)} \delta.$

Proof. Define

$S = \{y: p(y) > e^{\alpha \epsilon} q(y)\}.$

Then we have

$e^{\alpha \epsilon} q(S) < p(S) \le e^\epsilon q(S) + \delta,$

where the first inequality is due to the definition of $$S$$, and the second due to the $$(\epsilon, \delta)$$-ind. Therefore

$q(S) \le {\delta \over e^{\alpha \epsilon} - e^\epsilon}.$

Using the $$(\epsilon, \delta)$$-ind again we have

$p(S) \le e^\epsilon q(S) + \delta = {1 \over 1 - e^{(1 - \alpha) \epsilon}} \delta.$

$$\square$$

This can be quite bad if $$\epsilon$$ is small.

To prove the composition theorems in the next section, we need a condition better than that in Claim 1 so that we can go back and forth between indistinguishability and such condition. In other words, we need a characterisation of indistinguishability.

Let us take a careful look at the condition in Claim 1 and call it C1:

C1. $$\mathbb P(L(p || q) \le \epsilon) \ge 1 - \delta$$ and $$\mathbb P(L(q || p) \le \epsilon) \ge 1 - \delta$$

It is equivalent to

C2. there exist events $$A, B \subset Y$$ with probabilities $$p(A)$$ and $$q(B)$$ at least $$1 - \delta$$ such that $$\log p(y) - \log q(y) \le \epsilon$$ for all $$y \in A$$ and $$\log q(y) - \log p(y) \le \epsilon$$ for all $$y \in B$$.

A similar-looking condition to C2 is the following:

C3. Let $$\Omega$$ be the underlying probability space. There exist two events $$E, F \subset \Omega$$ with $$\mathbb P(E), \mathbb P(F) \ge 1 - \delta$$, such that $$|\log p_{|E}(y) - \log q_{|F}(y)| \le \epsilon$$ for all $$y \in Y$$.

Here $$p_{|E}$$ (resp. $$q_{|F}$$) is $$p$$ (resp. $$q$$) conditioned on event $$E$$ (resp. $$F$$).

Remark. Note that the events in C2 and C3 are in different spaces, and therefore we can not write $$p_{|E}(S)$$ as $$p(S | E)$$ or $$q_{|F}(S)$$ as $$q(S | F)$$. In fact, if we let $$E$$ and $$F$$ in C3 be subsets of $$Y$$ with $$p(E), q(F) \ge 1 - \delta$$ and assume $$p$$ and $$q$$ have the same supports, then C3 degenerates to a stronger condition than C2. Indeed, in this case $$p_E(y) = p(y) 1_{y \in E}$$ and $$q_F(y) = q(y) 1_{y \in F}$$, and so $$p_E(y) \le e^\epsilon q_F(y)$$ forces $$E \subset F$$. We also obtain $$F \subset E$$ in the same way. This gives us $$E = F$$, and C3 becomes C2 with $$A = B = E = F$$.

As it turns out, C3 is the condition we need.

Claim 4. Two probability measures $$p$$ and $$q$$ are $$(\epsilon, \delta)$$-ind if and only if C3 holds.

Proof(Murtagh-Vadhan 2018). The "if" direction is proved in the same way as Claim 1. Without loss of generality we may assume $$\mathbb P(E) = \mathbb P(F) \ge 1 - \delta$$. To see this, suppose $$F$$ has higher probability than $$E$$, then we can substitute $$F$$ with a subset of $$F$$ that has the same probability as $$E$$ (with possible enlargement of the probability space).

Let $$\xi \sim p$$ and $$\eta \sim q$$ be two independent random variables, then

\begin{aligned} p(S) &= \mathbb P(\xi \in S | E) \mathbb P(E) + \mathbb P(\xi \in S; E^c) \\ &\le e^\epsilon \mathbb P(\eta \in S | F) \mathbb P(E) + \delta \\ &= e^\epsilon \mathbb P(\eta \in S | F) \mathbb P(F) + \delta\\ &\le e^\epsilon q(S) + \delta. \end{aligned}

The "only-if" direction is more involved.

We construct events $$E$$ and $$F$$ by constructing functions $$e, f: Y \to [0, \infty)$$ satisfying the following conditions:

1. $$0 \le e(y) \le p(y)$$ and $$0 \le f(y) \le q(y)$$ for all $$y \in Y$$.
2. $$|\log e(y) - \log f(y)| \le \epsilon$$ for all $$y \in Y$$.
3. $$e(Y), f(Y) \ge 1 - \delta$$.
4. $$e(Y) = f(Y)$$.

Here for a set $$S \subset Y$$, $$e(S) := \int_S e(y) dy$$, and the same goes for $$f(S)$$.

Let $$\xi \sim p$$ and $$\eta \sim q$$. Then we define $$E$$ and $$F$$ by

$\mathbb P(E | \xi = y) = e(y) / p(y) \\ \mathbb P(F | \eta = y) = f(y) / q(y).$

Remark inside proof. This can seem a bit confusing. Intuitively, we can think of it this way when $$Y$$ is finite: Recall a random variable on $$Y$$ is a function from the probability space $$\Omega$$ to $$Y$$. Let event $$G_y \subset \Omega$$ be defined as $$G_y = \xi^{-1} (y)$$. We cut $$G_y$$ into the disjoint union of $$E_y$$ and $$G_y \setminus E_y$$ such that $$\mathbb P(E_y) = e(y)$$. Then $$E = \bigcup_{y \in Y} E_y$$. So $$e(y)$$ can be seen as the "density" of $$E$$.

Indeed, given $$E$$ and $$F$$ defined this way, we have

$p_E(y) = {e(y) \over e(Y)} \le {\exp(\epsilon) f(y) \over e(Y)} = {\exp(\epsilon) f(y) \over f(Y)} = \exp(\epsilon) q_F(y).$

and

$\mathbb P(E) = \int \mathbb P(E | \xi = y) p(y) dy = e(Y) \ge 1 - \delta,$

and the same goes for $$\mathbb P(F)$$.

What remains is to construct $$e(y)$$ and $$f(y)$$ satisfying the four conditions.

Like in the proof of Claim 1, let $$S, T \subset Y$$ be defined as

\begin{aligned} S := \{y: p(y) > \exp(\epsilon) q(y)\},\\ T := \{y: q(y) > \exp(\epsilon) p(y)\}. \end{aligned}

Let

\begin{aligned} e(y) &:= \exp(\epsilon) q(y) 1_{y \in S} + p(y) 1_{y \notin S}\\ f(y) &:= \exp(\epsilon) p(y) 1_{y \in T} + q(y) 1_{y \notin T}. \qquad (6) \end{aligned}

By checking them on the three disjoint subsets $$S$$, $$T$$, $$(S \cup T)^c$$, it is not hard to verify that the $$e(y)$$ and $$f(y)$$ constructed this way satisfy the first two conditions. They also satisfy the third condition:

\begin{aligned} e(Y) &= 1 - (p(S) - \exp(\epsilon) q(S)) \ge 1 - \delta, \\ f(Y) &= 1 - (q(T) - \exp(\epsilon) p(T)) \ge 1 - \delta. \end{aligned}

If $$e(Y) = f(Y)$$ then we are done. Otherwise, without loss of generality, assume $$e(Y) < f(Y)$$, then all it remains to do is to reduce the value of $$f(y)$$ while preserving Condition 1, 2 and 3, until $$f(Y) = e(Y)$$.

As it turns out, this can be achieved by reducing $$f(y)$$ on the set $$\{y \in Y: q(y) > p(y)\}$$. To see this, let us rename the $$f(y)$$ defined in (6) $$f_+(y)$$, and construct $$f_-(y)$$ by

$f_-(y) := p(y) 1_{y \in T} + (q(y) \wedge p(y)) 1_{y \notin T}.$

It is not hard to show that not only $$e(y)$$ and $$f_-(y)$$ also satisfy conditions 1-3, but

$e(y) \ge f_-(y), \forall y \in Y,$

and thus $$e(Y) \ge f_-(Y)$$. Therefore there exists an $$f$$ that interpolates between $$f_-$$ and $$f_+$$ with $$f(Y) = e(Y)$$. $$\square$$

To prove the adaptive composition theorem for approximate differential privacy, we need a similar claim (We use index shorthand $$\xi_{< i} = \xi_{1 : i - 1}$$ and similarly for other notations):

Claim 5. Let $$\xi_{1 : i}$$ and $$\eta_{1 : i}$$ be random variables. Let

\begin{aligned} p_i(S | y_{1 : i - 1}) := \mathbb P(\xi_i \in S | \xi_{1 : i - 1} = y_{1 : i - 1})\\ q_i(S | y_{1 : i - 1}) := \mathbb P(\eta_i \in S | \eta_{1 : i - 1} = y_{1 : i - 1}) \end{aligned}

be the conditional laws of $$\xi_i | \xi_{< i}$$ and $$\eta_i | \eta_{< i}$$ respectively. Then the following are equivalent:

1. For any $$y_{< i} \in Y^{i - 1}$$, $$p_i(\cdot | y_{< i})$$ and $$q_i(\cdot | y_{< i})$$ are $$(\epsilon, \delta)$$-ind
2. There exists events $$E_i, F_i \subset \Omega$$ with $$\mathbb P(E_i | \xi_{<i} = y_{<i}) = \mathbb P(F_i | \eta_{<i} = y_{< i}) \ge 1 - \delta$$ for any $$y_{< i}$$, such that $$p_{i | E_i}(\cdot | y_{< i})$$ and $$q_{i | E_i} (\cdot | y_{< i})$$ are $$\epsilon$$-ind for any $$y_{< i}$$, where \begin{aligned} p_{i | E_i}(S | y_{1 : i - 1}) := \mathbb P(\xi_i \in S | E_i, \xi_{1 : i - 1} = y_{1 : i - 1})\\ q_{i | F_i}(S | y_{1 : i - 1}) := \mathbb P(\eta_i \in S | F_i, \eta_{1 : i - 1} = y_{1 : i - 1}) \end{aligned}

are $$p_i$$ and $$q_i$$ conditioned on $$E_i$$ and $$F_i$$ respectively.

Proof. Item 2 => Item 1: as in the Proof of Claim 4,

\begin{aligned} p_i(S | y_{< i}) &= p_{i | E_i} (S | y_{< i}) \mathbb P(E_i | \xi_{< i} = y_{< i}) + p_{i | E_i^c}(S | y_{< i}) \mathbb P(E_i^c | \xi_{< i} = y_{< i}) \\ &\le p_{i | E_i} (S | y_{< i}) \mathbb P(E_i | \xi_{< i} = y_{< i}) + \delta \\ &= p_{i | E_i} (S | y_{< i}) \mathbb P(F_i | \xi_{< i} = y_{< i}) + \delta \\ &\le e^\epsilon q_{i | F_i} (S | y_{< i}) \mathbb P(F_i | \xi_{< i} = y_{< i}) + \delta \\ &= e^\epsilon q_i (S | y_{< i}) + \delta. \end{aligned}

The direction from $$q_i(S | y_{< i}) \le e^\epsilon p_i(S | y_{< i}) + \delta$$ can be shown in the same way.

Item 1 => Item 2: as in the Proof of Claim 4 we construct $$e(y_{1 : i})$$ and $$f(y_{1 : i})$$ as "densities" of events $$E_i$$ and $$F_i$$.

Let

\begin{aligned} e(y_{1 : i}) &:= e^\epsilon q_i(y_i | y_{< i}) 1_{y_i \in S_i(y_{< i})} + p_i(y_i | y_{< i}) 1_{y_i \notin S_i(y_{< i})}\\ f(y_{1 : i}) &:= e^\epsilon p_i(y_i | y_{< i}) 1_{y_i \in T_i(y_{< i})} + q_i(y_i | y_{< i}) 1_{y_i \notin T_i(y_{< i})}\\ \end{aligned}

where

\begin{aligned} S_i(y_{< i}) = \{y_i \in Y: p_i(y_i | y_{< i}) > e^\epsilon q_i(y_i | y_{< i})\}\\ T_i(y_{< i}) = \{y_i \in Y: q_i(y_i | y_{< i}) > e^\epsilon p_i(y_i | y_{< i})\}. \end{aligned}

Then $$E_i$$ and $$F_i$$ are defined as

\begin{aligned} \mathbb P(E_i | \xi_{\le i} = y_{\le i}) &= {e(y_{\le i}) \over p_i(y_{\le i})},\\ \mathbb P(F_i | \xi_{\le i} = y_{\le i}) &= {f(y_{\le i}) \over q_i(y_{\le i})}. \end{aligned}

The rest of the proof is almost the same as the proof of Claim 4. $$\square$$

### Back to approximate differential privacy

By Claim 0 and 1 we have

Claim 6. If for all $$x, x' \in X$$ with distance $$1$$

$\mathbb P(L(M(x) || M(x')) \le \epsilon) \ge 1 - \delta,$

then $$M$$ is $$(\epsilon, \delta)$$-dp.

Note that in the literature the divergence variable $$L(M(x) || M(x'))$$ is also called the privacy loss.

By Claim 0 and Claim 4 we have

Claim 7. $$M$$ is $$(\epsilon, \delta)$$-dp if and only if for every $$x, x' \in X$$ with distance $$1$$, there exist events $$E, F \subset \Omega$$ with $$\mathbb P(E) = \mathbb P(F) \ge 1 - \delta$$, $$M(x) | E$$ and $$M(x') | F$$ are $$\epsilon$$-ind.

We can further simplify the privacy loss $$L(M(x) || M(x'))$$, by observing the translational and scaling invariance of $$L(\cdot||\cdot)$$:

\begin{aligned} L(\xi || \eta) &\overset{d}{=} L(\alpha \xi + \beta || \alpha \eta + \beta), \qquad \alpha \neq 0. \qquad (6.1) \end{aligned}

With this and the definition

$M(x) = f(x) + \zeta$

for some random variable $$\zeta$$, we have

$L(M(x) || M(x')) \overset{d}{=} L(\zeta || \zeta + f(x') - f(x)).$

Without loss of generality, we can consider $$f$$ with sensitivity $$1$$, for

$L(f(x) + S_f \zeta || f(x') + S_f \zeta) \overset{d}{=} L(S_f^{-1} f(x) + \zeta || S_f^{-1} f(x') + \zeta)$

so for any noise $$\zeta$$ that achieves $$(\epsilon, \delta)$$-dp for a function with sensitivity $$1$$, we have the same privacy guarantee by for an arbitrary function with sensitivity $$S_f$$ by adding a noise $$S_f \zeta$$.

With Claim 6 we can show that the Gaussian mechanism is approximately differentially private. But first we need to define it.

Definition (Gaussian mechanism). Given a query $$f: X \to Y$$, the Gaussian mechanism $$M$$ adds a Gaussian noise to the query:

$M(x) = f(x) + N(0, \sigma^2 I).$

Some tail bounds for the Gaussian distribution will be useful.

Claim 8 (Gaussian tail bounds). Let $$\xi \sim N(0, 1)$$ be a standard normal distribution. Then for $$t > 0$$

$\mathbb P(\xi > t) < {1 \over \sqrt{2 \pi} t} e^{- {t^2 \over 2}}, \qquad (6.3)$

and

$\mathbb P(\xi > t) < e^{- {t^2 \over 2}}. \qquad (6.5)$

Proof. Both bounds are well known. The first can be proved using

$\int_t^\infty e^{- {y^2 \over 2}} dy < \int_t^\infty {y \over t} e^{- {y^2 \over 2}} dy.$

The second is shown using Chernoff bound. For any random variable $$\xi$$,

$\mathbb P(\xi > t) < {\mathbb E \exp(\lambda \xi) \over \exp(\lambda t)} = \exp(\kappa_\xi(\lambda) - \lambda t), \qquad (6.7)$

where $$\kappa_\xi(\lambda) = \log \mathbb E \exp(\lambda \xi)$$ is the cumulant of $$\xi$$. Since (6.7) holds for any $$\lambda$$, we can get the best bound by minimising $$\kappa_\xi(\lambda) - \lambda t$$ (a.k.a. the Legendre transformation). When $$\xi$$ is standard normal, we get (6.5). $$\square$$

Remark. We will use the Chernoff bound extensively in the second part of this post when considering Rényi differential privacy.

Claim 9. The Gaussian mechanism on a query $$f$$ is $$(\epsilon, \delta)$$-dp, where

$\delta = \exp(- (\epsilon \sigma / S_f - (2 \sigma / S_f)^{-1})^2 / 2). \qquad (6.8)$

Conversely, to achieve give $$(\epsilon, \delta)$$-dp, we may set

$\sigma > \left(\epsilon^{-1} \sqrt{2 \log \delta^{-1}} + (2 \epsilon)^{- {1 \over 2}}\right) S_f \qquad (6.81)$

or

$\sigma > (\epsilon^{-1} (1 \vee \sqrt{(\log (2 \pi)^{-1} \delta^{-2})_+}) + (2 \epsilon)^{- {1 \over 2}}) S_f \qquad (6.82)$

or

$\sigma > \epsilon^{-1} \sqrt{\log e^\epsilon \delta^{-2}} S_f \qquad (6.83)$

or

$\sigma > \epsilon^{-1} (\sqrt{1 + \epsilon} \vee \sqrt{(\log e^\epsilon (2 \pi)^{-1} \delta^{-2})_+}) S_f. \qquad (6.84)$

Proof. As discussed before we only need to consider the case where $$S_f = 1$$. Fix arbitrary $$x, x' \in X$$ with $$d(x, x') = 1$$. Let $$\zeta = (\zeta_1, ..., \zeta_d) \sim N(0, I_d)$$.

By Claim 6 it suffices to bound

$\mathbb P(L(M(x) || M(x')) > \epsilon)$

We have by the linear invariance of $$L$$,

$L(M(x) || M(x')) = L(f(x) + \sigma \zeta || f(x') + \sigma \zeta) \overset{d}{=} L(\zeta|| \zeta + \Delta / \sigma),$

where $$\Delta := f(x') - f(x)$$.

Plugging in the Gaussian density, we have

$L(M(x) || M(x')) \overset{d}{=} \sum_i {\Delta_i \over \sigma} \zeta_i + \sum_i {\Delta_i^2 \over 2 \sigma^2} \overset{d}{=} {\|\Delta\|_2 \over \sigma} \xi + {\|\Delta\|_2^2 \over 2 \sigma^2}.$

where $$\xi \sim N(0, 1)$$.

Hence

$\mathbb P(L(M(x) || M(x')) > \epsilon) = \mathbb P(\zeta > {\sigma \over \|\Delta\|_2} \epsilon - {\|\Delta\|_2 \over 2 \sigma}).$

Since $$\|\Delta\|_2 \le S_f = 1$$, we have

$\mathbb P(L(M(x) || M(x')) > \epsilon) \le \mathbb P(\xi > \sigma \epsilon - (2 \sigma)^{-1}).$

Thus the problem is reduced to the tail bound of a standard normal distribution, so we can use Claim 8. Note that we implicitly require $$\sigma > (2 \epsilon)^{- 1 / 2}$$ here so that $$\sigma \epsilon - (2 \sigma)^{-1} > 0$$ and we can use the tail bounds.

Using (6.3) we have

$\mathbb P(L(M(x) || M(x')) > \epsilon) < \exp(- (\epsilon \sigma - (2 \sigma)^{-1})^2 / 2).$

This gives us (6.8).

To bound the right hand by $$\delta$$, we require

$\epsilon \sigma - {1 \over 2 \sigma} > \sqrt{2 \log \delta^{-1}}. \qquad (6.91)$

Solving this inequality we have

$\sigma > {\sqrt{2 \log \delta^{-1}} + \sqrt{2 \log \delta^{-1} + 2 \epsilon} \over 2 \epsilon}.$

Using $$\sqrt{2 \log \delta^{-1} + 2 \epsilon} \le \sqrt{2 \log \delta^{-1}} + \sqrt{2 \epsilon}$$, we can achieve the above inequality by having

$\sigma > \epsilon^{-1} \sqrt{2 \log \delta^{-1}} + (2 \epsilon)^{-{1 \over 2}}.$

This gives us (6.81).

Alternatively, we can use the concavity of $$\sqrt{\cdot}$$:

$(2 \epsilon)^{-1} (\sqrt{2 \log \delta^{-1}} + \sqrt{2 \log \delta^{-1} + 2 \epsilon}) \le \epsilon^{-1} \sqrt{\log e^\epsilon \delta^{-2}},$

which gives us (6.83)

Back to (6.9), if we use (6.5) instead, we need

$\log t + {t^2 \over 2} > \log {(2 \pi)^{- 1 / 2} \delta^{-1}}$

where $$t = \epsilon \sigma - (2 \sigma)^{-1}$$. This can be satisfied if

\begin{aligned} t &> 1 \qquad (6.93)\\ t &> \sqrt{\log (2 \pi)^{-1} \delta^{-2}}. \qquad (6.95) \end{aligned}

We can solve both inequalities as before and obtain

$\sigma > \epsilon^{-1} (1 \vee \sqrt{(\log (2 \pi)^{-1} \delta^{-2})_+}) + (2 \epsilon)^{- {1 \over 2}},$

or

$\sigma > \epsilon^{-1}(\sqrt{1 + \epsilon} \vee \sqrt{(\log e^\epsilon (2 \pi)^{-1} \delta^{-2})_+}).$

This gives us (6.82)(6.84). $$\square$$

When $$\epsilon \le \alpha$$ is bounded, by (6.83) (6.84) we can require either

$\sigma > \epsilon^{-1} (\sqrt{\log e^\alpha \delta^{-2}}) S_f$

or

$\sigma > \epsilon^{-1} (\sqrt{1 + \alpha} \vee \sqrt{(\log (2 \pi)^{-1} e^\alpha \delta^{-2})_+}) S_f.$

The second bound is similar to and slightly better than the one in Theorem A.1 of Dwork-Roth 2013, where $$\alpha = 1$$:

$\sigma > \epsilon^{-1} \left({3 \over 2} \vee \sqrt{(2 \log {5 \over 4} \delta^{-1})_+}\right) S_f.$

Note that the lower bound of $${3 \over 2}$$ is implicitly required in the proof of Theorem A.1.

## Composition theorems

So far we have seen how a mechanism made of a single query plus a noise can be proved to be differentially private. But we need to understand the privacy when composing several mechanisms, combinatorially or sequentially. Let us first define the combinatorial case:

Definition (Independent composition). Let $$M_1, ..., M_k$$ be $$k$$ mechanisms with independent noises. The mechanism $$M = (M_1, ..., M_k)$$ is called the independent composition of $$M_{1 : k}$$.

To define the adaptive composition, let us motivate it with an example of gradient descent. Consider the loss function $$\ell(x; \theta)$$ of a neural network, where $$\theta$$ is the parameter and $$x$$ the input, gradient descent updates its parameter $$\theta$$ at each time $$t$$:

$\theta_{t} = \theta_{t - 1} - \alpha m^{-1} \sum_{i = 1 : m} \nabla_\theta \ell(x_i; \theta) |_{\theta = \theta_{t - 1}}.$

We may add privacy by adding noise $$\zeta_t$$ at each step:

$\theta_{t} = \theta_{t - 1} - \alpha m^{-1} \sum_{i = 1 : m} \nabla_\theta \ell(x_i; \theta) |_{\theta = \theta_{t - 1}} + \zeta_t. \qquad (6.97)$

Viewed as a sequence of mechanism, we have that at each time $$t$$, the mechanism $$M_t$$ takes input $$x$$, and outputs $$\theta_t$$. But $$M_t$$ also depends on the output of the previous mechanism $$M_{t - 1}$$. To this end, we define the adaptive composition.

Definition (Adaptive composition). Let $$({M_i(y_{1 : i - 1})})_{i = 1 : k}$$ be $$k$$ mechanisms with independent noises, where $$M_1$$ has no parameter, $$M_2$$ has one parameter in $$Y$$, $$M_3$$ has two parameters in $$Y$$ and so on. For $$x \in X$$, define $$\xi_i$$ recursively by

\begin{aligned} \xi_1 &:= M_1(x)\\ \xi_i &:= M_i(\xi_1, \xi_2, ..., \xi_{i - 1}) (x). \end{aligned}

The adaptive composition of $$M_{1 : k}$$ is defined by $$M(x) := (\xi_1, \xi_2, ..., \xi_k)$$.

The definition of adaptive composition may look a bit complicated, but the point is to describe $$k$$ mechanisms such that for each $$i$$, the output of the first, second, ..., $$i - 1$$th mechanisms determine the $$i$$th mechanism, like in the case of gradient descent.

It is not hard to write down the differentially private gradient descent as a sequential composition:

$M_t(\theta_{1 : t - 1})(x) = \theta_{t - 1} - \alpha m^{-1} \sum_{i = 1 : m} \nabla_\theta \ell(x_i; \theta) |_{\theta = \theta_{t - 1}} + \zeta_t.$

In Dwork-Rothblum-Vadhan 2010 (see also Dwork-Roth 2013) the adaptive composition is defined in a more general way, but the definition is based on the same principle, and proofs in this post on adaptive compositions carry over.

It is not hard to see that the adaptive composition degenerates to independent composition when each $$M_i(y_{1 : i})$$ evaluates to the same mechanism regardless of $$y_{1 : i}$$, in which case the $$\xi_i$$s are independent.

In the following when discussing adaptive compositions we sometimes omit the parameters for convenience without risk of ambiguity, and write $$M_i(y_{1 : i})$$ as $$M_i$$, but keep in mind of the dependence on the parameters.

It is time to state and prove the composition theorems. In this section we consider $$2 \times 2 \times 2 = 8$$ cases, i.e. situations of three dimensions, where there are two choices in each dimension:

1. Composition of $$\epsilon$$-dp or more generally $$(\epsilon, \delta)$$-dp mechanisms
2. Composition of independent or more generally adaptive mechanisms

Note that in the first two dimensions the second choice is more general than the first.

The proofs of these composition theorems will be laid out as follows:

1. Claim 10 - Basic composition theorem for $$(\epsilon, \delta)$$-dp with adaptive mechanisms: by a direct proof with an induction argument
2. Claim 14 - Advanced composition theorem for $$\epsilon$$-dp with independent mechanisms: by factorising privacy loss and using Hoeffding's Inequality
3. Claim 16 - Advanced composition theorem for $$\epsilon$$-dp with adaptive mechanisms: by factorising privacy loss and using Azuma's Inequality
4. Claims 17 and 18 - Advanced composition theorem for $$(\epsilon, \delta)$$-dp with independent / adaptive mechanisms: by using characterisations of $$(\epsilon, \delta)$$-dp in Claims 4 and 5 as an approximation of $$\epsilon$$-dp and then using Proofs in Item 2 or 3.

Claim 10 (Basic composition theorem). Let $$M_{1 : k}$$ be $$k$$ mechanisms with independent noises such that for each $$i$$ and $$y_{1 : i - 1}$$, $$M_i(y_{1 : i - 1})$$ is $$(\epsilon_i, \delta_i)$$-dp. Then the adpative composition of $$M_{1 : k}$$ is $$(\sum_i \epsilon_i, \sum_i \delta_i)$$-dp.

Proof (Dwork-Lei 2009, see also Dwork-Roth 2013 Appendix B.1). Let $$x$$ and $$x'$$ be neighbouring points in $$X$$. Let $$M$$ be the adaptive composition of $$M_{1 : k}$$. Define

$\xi_{1 : k} := M(x), \qquad \eta_{1 : k} := M(x').$

Let $$p^i$$ and $$q^i$$ be the laws of $$(\xi_{1 : i})$$ and $$(\eta_{1 : i})$$ respectively.

Let $$S_1, ..., S_k \subset Y$$ and $$T_i := \prod_{j = 1 : i} S_j$$. We use two tricks.

1. Since $$\xi_i | \xi_{< i} = y_{< i}$$ and $$\eta_i | \eta_{< i} = y_{< i}$$ are $$(\epsilon_i, \delta_i)$$-ind, and a probability is no greater than $$1$$, \begin{aligned} \mathbb P(\xi_i \in S_i | \xi_{< i} = y_{< i}) &\le (e^{\epsilon_i} \mathbb P(\eta_i \in S_i | \eta_{< i} = y_{< i}) + \delta_i) \wedge 1 \\ &\le (e^{\epsilon_i} \mathbb P(\eta_i \in S_i | \eta_{< i} = y_{< i}) + \delta_i) \wedge (1 + \delta_i) \\ &= (e^{\epsilon_i} \mathbb P(\eta_i \in S_i | \eta_{< i} = y_{< i}) \wedge 1) + \delta_i \end{aligned}

2. Given $$p$$ and $$q$$ that are $$(\epsilon, \delta)$$-ind, define $\mu(x) = (p(x) - e^\epsilon q(x))_+.$

We have $\mu(S) \le \delta, \forall S$

In the following we define $$\mu^{i - 1} = (p^{i - 1} - e^\epsilon q^{i - 1})_+$$ for the same purpose.

We use an inductive argument to prove the theorem:

\begin{aligned} \mathbb P(\xi_{\le i} \in T_i) &= \int_{T_{i - 1}} \mathbb P(\xi_i \in S_i | \xi_{< i} = y_{< i}) p^{i - 1} (y_{< i}) dy_{< i} \\ &\le \int_{T_{i - 1}} (e^{\epsilon_i} \mathbb P(\eta_i \in S_i | \eta_{< i} = y_{< i}) \wedge 1) p^{i - 1}(y_{< i}) dy_{< i} + \delta_i\\ &\le \int_{T_{i - 1}} (e^{\epsilon_i} \mathbb P(\eta_i \in S_i | \eta_{< i} = y_{< i}) \wedge 1) (e^{\epsilon_1 + ... + \epsilon_{i - 1}} q^{i - 1}(y_{< i}) + \mu^{i - 1} (y_{< i})) dy_{< i} + \delta_i\\ &\le \int_{T_{i - 1}} e^{\epsilon_i} \mathbb P(\eta_i \in S_i | \eta_{< i} = y_{< i}) e^{\epsilon_1 + ... + \epsilon_{i - 1}} q^{i - 1}(y_{< i}) dy_{< i} + \mu_{i - 1}(T_{i - 1}) + \delta_i\\ &\le e^{\epsilon_1 + ... + \epsilon_i} \mathbb P(\eta_{\le i} \in T_i) + \delta_1 + ... + \delta_{i - 1} + \delta_i.\\ \end{aligned}

In the second line we use Trick 1; in the third line we use the induction assumption; in the fourth line we multiply the first term in the first braket with first term in the second braket, and the second term (i.e. $$1$$) in the first braket with the second term in the second braket (i.e. the $$\mu$$ term); in the last line we use Trick 2.

The base case $$i = 1$$ is true since $$M_1$$ is $$(\epsilon_1, \delta_1)$$-dp. $$\square$$

Claim 11. If $$p$$ and $$q$$ are $$\epsilon$$-ind, then

$D(p || q) + D(q || p) \le \epsilon(e^\epsilon - 1).$

Proof. Since $$p$$ and $$q$$ are $$\epsilon$$-ind, we have $$|\log p(x) - \log q(x)| \le \epsilon$$ for all $$x$$. Let $$S := \{x: p(x) > q(x)\}$$. Then we have on

\begin{aligned} D(p || q) + D(q || p) &= \int (p(x) - q(x)) (\log p(x) - \log q(x)) dx\\ &= \int_S (p(x) - q(x)) (\log p(x) - \log q(x)) dx + \int_{S^c} (q(x) - p(x)) (\log q(x) - \log p(x)) dx\\ &\le \epsilon(\int_S p(x) - q(x) dx + \int_{S^c} q(x) - p(x) dx) \end{aligned}

Since on $$S$$ we have $$q(x) \le p(x) \le e^\epsilon q(x)$$, and on $$S^c$$ we have $$p(x) \le q(x) \le e^\epsilon p(x)$$, we obtain

$D(p || q) + D(q || p) \le \epsilon \int_S (1 - e^{-\epsilon}) p(x) dx + \epsilon \int_{S^c} (e^{\epsilon} - 1) p(x) dx \le \epsilon (e^{\epsilon} - 1),$

where in the last step we use $$e^\epsilon - 1 \ge 1 - e^{- \epsilon}$$ and $$p(S) + p(S^c) = 1$$. $$\square$$

Claim 12. If $$p$$ and $$q$$ are $$\epsilon$$-ind, then

$D(p || q) \le a(\epsilon) \ge D(q || p),$

where

$a(\epsilon) = \epsilon (e^\epsilon - 1) 1_{\epsilon \le \log 2} + \epsilon 1_{\epsilon > \log 2} \le (\log 2)^{-1} \epsilon^2 1_{\epsilon \le \log 2} + \epsilon 1_{\epsilon > \log 2}. \qquad (6.98)$

Proof. Since $$p$$ and $$q$$ are $$\epsilon$$-ind, we have

$D(p || q) = \mathbb E_{\xi \sim p} \log {p(\xi) \over q(\xi)} \le \max_y {\log p(y) \over \log q(y)} \le \epsilon.$

Comparing the quantity in Claim 11 ($$\epsilon(e^\epsilon - 1)$$) with the quantity above ($$\epsilon$$), we arrive at the conclusion. $$\square$$

Claim 13 (Hoeffding's Inequality). Let $$L_i$$ be independent random variables with $$|L_i| \le b$$, and let $$L = L_1 + ... + L_k$$, then for $$t > 0$$,

$\mathbb P(L - \mathbb E L \ge t) \le \exp(- {t^2 \over 2 k b^2}).$

Claim 14 (Advanced Independent Composition Theorem) ($$\delta = 0$$). Fix $$0 < \beta < 1$$. Let $$M_1, ..., M_k$$ be $$\epsilon$$-dp, then the independent composition $$M$$ of $$M_{1 : k}$$ is $$(k a(\epsilon) + \sqrt{2 k \log \beta^{-1}} \epsilon, \beta)$$-dp.

Remark. By (6.98) we know that $$k a(\epsilon) + \sqrt{2 k \log \beta^{-1}} \epsilon = \sqrt{2 k \log \beta^{-1}} \epsilon + k O(\epsilon^2)$$ when $$\epsilon$$ is sufficiently small, in which case the leading term is of order $$O(\sqrt k \epsilon)$$ and we save a $$\sqrt k$$ in the $$\epsilon$$-part compared to the Basic Composition Theorem (Claim 10).

Remark. In practice one can try different choices of $$\beta$$ and settle with the one that gives the best privacy guarantee. See the discussions at the end of Part 2 of this post.

Proof. Let $$p_i$$, $$q_i$$, $$p$$ and $$q$$ be the laws of $$M_i(x)$$, $$M_i(x')$$, $$M(x)$$ and $$M(x')$$ respectively.

$\mathbb E L_i = D(p_i || q_i) \le a(\epsilon),$

where $$L_i := L(p_i || q_i)$$. Due to $$\epsilon$$-ind also have

$|L_i| \le \epsilon.$

Therefore, by Hoeffding's Inequality,

$\mathbb P(L - k a(\epsilon) \ge t) \le \mathbb P(L - \mathbb E L \ge t) \le \exp(- t^2 / 2 k \epsilon^2),$

where $$L := \sum_i L_i = L(p || q)$$.

Plugging in $$t = \sqrt{2 k \epsilon^2 \log \beta^{-1}}$$, we have

$\mathbb P(L(p || q) \le k a(\epsilon) + \sqrt{2 k \epsilon^2 \log \beta^{-1}}) \ge 1 - \beta.$

Similarly we also have

$\mathbb P(L(q || p) \le k a(\epsilon) + \sqrt{2 k \epsilon^2 \log \beta^{-1}}) \ge 1 - \beta.$

By Claim 1 we arrive at the conclusion. $$\square$$

Claim 15 (Azuma's Inequality). Let $$X_{0 : k}$$ be a supermartingale. If $$|X_i - X_{i - 1}| \le b$$, then

$\mathbb P(X_k - X_0 \ge t) \le \exp(- {t^2 \over 2 k b^2}).$

Azuma's Inequality implies a slightly weaker version of Hoeffding's Inequality. To see this, let $$L_{1 : k}$$ be independent variables with $$|L_i| \le b$$. Let $$X_i = \sum_{j = 1 : i} L_j - \mathbb E L_j$$. Then $$X_{0 : k}$$ is a martingale, and

$| X_i - X_{i - 1} | = | L_i - \mathbb E L_i | \le 2 b,$

since $$\|L_i\|_1 \le \|L_i\|_\infty$$. Hence by Azuma's Inequality,

$\mathbb P(L - \mathbb E L \ge t) \le \exp(- {t^2 \over 8 k b^2}).$

Of course here we have made no assumption on $$\mathbb E L_i$$. If instead we have some bound for the expectation, say $$|\mathbb E L_i| \le a$$, then by the same derivation we have

$\mathbb P(L - \mathbb E L \ge t) \le \exp(- {t^2 \over 2 k (a + b)^2}).$

It is not hard to see what Azuma is to Hoeffding is like adaptive composition to independent composition. Indeed, we can use Azuma's Inequality to prove the Advanced Adaptive Composition Theorem for $$\delta = 0$$.

Claim 16 (Advanced Adaptive Composition Theorem) ($$\delta = 0$$). Let $$\beta > 0$$. Let $$M_{1 : k}$$ be $$k$$ mechanisms with independent noises such that for each $$i$$ and $$y_{1 : i}$$, $$M_i(y_{1 : i})$$ is $$(\epsilon, 0)$$-dp. Then the adpative composition of $$M_{1 : k}$$ is $$(k a(\epsilon) + \sqrt{2 k \log \beta^{-1}} (\epsilon + a(\epsilon)), \beta)$$-dp.

Proof. As before, let $$\xi_{1 : k} \overset{d}{=} M(x)$$ and $$\eta_{1 : k} \overset{d}{=} M(x')$$, where $$M$$ is the adaptive composition of $$M_{1 : k}$$. Let $$p_i$$ (resp. $$q_i$$) be the law of $$\xi_i | \xi_{< i}$$ (resp. $$\eta_i | \eta_{< i}$$). Let $$p^i$$ (resp. $$q^i$$) be the law of $$\xi_{\le i}$$ (resp. $$\eta_{\le i}$$). We want to construct supermartingale $$X$$. To this end, let

$X_i = \log {p^i(\xi_{\le i}) \over q^i(\xi_{\le i})} - i a(\epsilon)$

We show that $$(X_i)$$ is a supermartingale:

\begin{aligned} \mathbb E(X_i - X_{i - 1} | X_{i - 1}) &= \mathbb E \left(\log {p_i (\xi_i | \xi_{< i}) \over q_i (\xi_i | \xi_{< i})} - a(\epsilon) | \log {p^{i - 1} (\xi_{< i}) \over q^{i - 1} (\xi_{< i})}\right) \\ &= \mathbb E \left( \mathbb E \left(\log {p_i (\xi_i | \xi_{< i}) \over q_i (\xi_i | \xi_{< i})} | \xi_{< i}\right) | \log {p^{i - 1} (\xi_{< i}) \over q^{i - 1} (\xi_{< i})}\right) - a(\epsilon) \\ &= \mathbb E \left( D(p_i (\cdot | \xi_{< i}) || q_i (\cdot | \xi_{< i})) | \log {p^{i - 1} (\xi_{< i}) \over q^{i - 1} (\xi_{< i})}\right) - a(\epsilon) \\ &\le 0, \end{aligned}

since by Claim 12 $$D(p_i(\cdot | y_{< i}) || q_i(\cdot | y_{< i})) \le a(\epsilon)$$ for all $$y_{< i}$$.

Since

$| X_i - X_{i - 1} | = | \log {p_i(\xi_i | \xi_{< i}) \over q_i(\xi_i | \xi_{< i})} - a(\epsilon) | \le \epsilon + a(\epsilon),$

by Azuma's Inequality,

$\mathbb P(\log {p^k(\xi_{1 : k}) \over q^k(\xi_{1 : k})} \ge k a(\epsilon) + t) \le \exp(- {t^2 \over 2 k (\epsilon + a(\epsilon))^2}). \qquad(6.99)$

Let $$t = \sqrt{2 k \log \beta^{-1}} (\epsilon + a(\epsilon))$$ we are done. $$\square$$

Claim 17 (Advanced Independent Composition Theorem). Fix $$0 < \beta < 1$$. Let $$M_1, ..., M_k$$ be $$(\epsilon, \delta)$$-dp, then the independent composition $$M$$ of $$M_{1 : k}$$ is $$(k a(\epsilon) + \sqrt{2 k \log \beta^{-1}} \epsilon, k \delta + \beta)$$-dp.

Proof. By Claim 4, there exist events $$E_{1 : k}$$ and $$F_{1 : k}$$ such that

1. The laws $$p_{i | E_i}$$ and $$q_{i | F_i}$$ are $$\epsilon$$-ind.
2. $$\mathbb P(E_i), \mathbb P(F_i) \ge 1 - \delta$$.

Let $$E := \bigcap E_i$$ and $$F := \bigcap F_i$$, then they both have probability at least $$1 - k \delta$$, and $$p_{i | E}$$ and $$q_{i | F}$$ are $$\epsilon$$-ind.

By Claim 14, $$p_{|E}$$ and $$q_{|F}$$ are $$(\epsilon' := k a(\epsilon) + \sqrt{2 k \epsilon^2 \log \beta^{-1}}, \beta)$$-ind. Let us shrink the bigger event between $$E$$ and $$F$$ so that they have equal probabilities. Then

\begin{aligned} p (S) &\le p_{|E}(S) \mathbb P(E) + \mathbb P(E^c) \\ &\le (e^{\epsilon'} q_{|F}(S) + \beta) \mathbb P(F) + k \delta\\ &\le e^{\epsilon'} q(S) + \beta + k \delta. \end{aligned}

$$\square$$

Claim 18 (Advanced Adaptive Composition Theorem). Fix $$0 < \beta < 1$$. Let $$M_{1 : k}$$ be $$k$$ mechanisms with independent noises such that for each $$i$$ and $$y_{1 : i}$$, $$M_i(y_{1 : i})$$ is $$(\epsilon, \delta)$$-dp. Then the adpative composition of $$M_{1 : k}$$ is $$(k a(\epsilon) + \sqrt{2 k \log \beta^{-1}} (\epsilon + a(\epsilon)), \beta + k \delta)$$-dp.

Remark. This theorem appeared in Dwork-Rothblum-Vadhan 2010, but I could not find a proof there. A proof can be found in Dwork-Roth 2013 (See Theorem 3.20 there). Here I prove it in a similar way, except that instead of the use of an intermediate random variable there, I use the conditional probability results from Claim 5, the approach mentioned in Vadhan 2017.

Proof. By Claim 5, there exist events $$E_{1 : k}$$ and $$F_{1 : k}$$ such that

1. The laws $$p_{i | E_i}(\cdot | y_{< i})$$ and $$q_{i | F_i}(\cdot | y_{< i})$$ are $$\epsilon$$-ind for all $$y_{< i}$$.
2. $$\mathbb P(E_i | y_{< i}), \mathbb P(F_i | y_{< i}) \ge 1 - \delta$$ for all $$y_{< i}$$.

Let $$E := \bigcap E_i$$ and $$F := \bigcap F_i$$, then they both have probability at least $$1 - k \delta$$, and $$p_{i | E}(\cdot | y_{< i}$$ and $$q_{i | F}(\cdot | y_{< i})$$ are $$\epsilon$$-ind.

By Advanced Adaptive Composition Theorem ($$\delta = 0$$), $$p_{|E}$$ and $$q_{|F}$$ are $$(\epsilon' := k a(\epsilon) + \sqrt{2 k \log \beta^{-1}} (\epsilon + a(\epsilon)), \beta)$$-ind.

The rest is the same as in the proof of Claim 17. $$\square$$

## Subsampling

Recall we have been considering databases in the space $$Z^m$$. Let $$n < m$$ be a positive integer, $$\mathcal I := \{I \subset [m]: |I| = n\}$$ be the set of subsets of $$[m]$$ of size $$n$$, and $$\gamma$$ a random subset sampled uniformly from $$\mathcal I$$. Let $$r = {n \over m}$$ which we call the subsampling rate. Then we may add a subsampling module to the noisy gradient descent algorithm (6.97) considered before

$\theta_{t} = \theta_{t - 1} - \alpha n^{-1} \sum_{i \in \gamma} \nabla_\theta h_\theta(x_i) |_{\theta = \theta_{t - 1}} + \zeta_t. \qquad (7)$

It turns out subsampling has an amplification effect on privacy.

Claim 19 (Ullman 2017). Fix $$r \in [0, 1]$$. Let $$n \le m$$ be two nonnegative integers with $$n = r m$$. Let $$N$$ be an $$(\epsilon, \delta)$$-dp machanism on $$Z^n$$. Define mechanism $$M$$ on $$Z^m$$ by

$M(x) = N(x_\gamma)$

Then $$M$$ is $$(\log (1 + r(e^\epsilon - 1)), r \delta)$$-dp.

Remark. Some seem to cite Kasiviswanathan-Lee-Nissim-Raskhodnikova-Smith 2005 for this result, but it is not clear to me how it appears there.

Proof. Let $$x, x' \in Z^n$$ such that they differ by one row $$x_i \neq x_i'$$. Naturally we would like to consider the cases where the index $$i$$ is picked and the ones where it is not separately. Let $$\mathcal I_\in$$ and $$\mathcal I_\notin$$ be these two cases:

\begin{aligned} \mathcal I_\in = \{J \subset \mathcal I: i \in J\}\\ \mathcal I_\notin = \{J \subset \mathcal I: i \notin J\}\\ \end{aligned}

We will use these notations later. Let $$A$$ be the event $$\{\gamma \ni i\}$$.

Let $$p$$ and $$q$$ be the laws of $$M(x)$$ and $$M(x')$$ respectively. We collect some useful facts about them. First due to $$N$$ being $$(\epsilon, \delta)$$-dp,

$p_{|A}(S) \le e^\epsilon q_{|A}(S) + \delta.$

Also,

$p_{|A}(S) \le e^\epsilon p_{|A^c}(S) + \delta.$

To see this, note that being conditional laws, $$p_A$$ and $$p_{A^c}$$ are averages of laws over $$\mathcal I_\in$$ and $$\mathcal I_\notin$$ respectively:

\begin{aligned} p_{|A}(S) = |\mathcal I_\in|^{-1} \sum_{I \in \mathcal I_\in} \mathbb P(N(x_I) \in S)\\ p_{|A^c}(S) = |\mathcal I_\notin|^{-1} \sum_{J \in \mathcal I_\notin} \mathbb P(N(x_J) \in S). \end{aligned}

Now we want to pair the $$I$$'s in $$\mathcal I_\in$$ and $$J$$'s in $$\mathcal I_\notin$$ so that they differ by one index only, which means $$d(x_I, x_J) = 1$$. Formally, this means we want to consider the set:

$\mathcal D := \{(I, J) \in \mathcal I_\in \times \mathcal I_\notin: |I \cap J| = n - 1\}.$

We may observe by trying out some simple cases that every $$I \in \mathcal I_\in$$ is paired with $$n$$ elements in $$\mathcal I_\notin$$, and every $$J \in \mathcal I_\notin$$ is paired with $$m - n$$ elements in $$\mathcal I_\in$$. Therefore

$p_{|A}(S) = |\mathcal D|^{-1} \sum_{(I, J) \in \mathcal D} \mathbb P(N(x_I \in S)) \le |\mathcal D|^{-1} \sum_{(I, J) \in \mathcal D} (e^\epsilon \mathbb P(N(x_J \in S)) + \delta) = e^\epsilon p_{|A^c} (S) + \delta.$

Since each of the $$m$$ indices is picked independently with probability $$r$$, we have

$\mathbb P(A) = r.$

Let $$t \in [0, 1]$$ to be determined. We may write

\begin{aligned} p(S) &= r p_{|A} (S) + (1 - r) p_{|A^c} (S)\\ &\le r(t e^\epsilon q_{|A}(S) + (1 - t) e^\epsilon q_{|A^c}(S) + \delta) + (1 - r) q_{|A^c} (S)\\ &= rte^\epsilon q_{|A}(S) + (r(1 - t) e^\epsilon + (1 - r)) q_{|A^c} (S) + r \delta\\ &= te^\epsilon r q_{|A}(S) + \left({r \over 1 - r}(1 - t) e^\epsilon + 1\right) (1 - r) q_{|A^c} (S) + r \delta \\ &\le \left(t e^\epsilon \wedge \left({r \over 1 - r} (1 - t) e^\epsilon + 1\right)\right) q(S) + r \delta. \qquad (7.5) \end{aligned}

We can see from the last line that the best bound we can get is when

$t e^\epsilon = {r \over 1 - r} (1 - t) e^\epsilon + 1.$

Solving this equation we obtain

$t = r + e^{- \epsilon} - r e^{- \epsilon}$

and plugging this in (7.5) we have

$p(S) \le (1 + r(e^\epsilon - 1)) q(S) + r \delta.$

$$\square$$

Since $$\log (1 + x) < x$$ for $$x > 0$$, we can rewrite the conclusion of the Claim to $$(r(e^\epsilon - 1), r \delta)$$-dp. Further more, if $$\epsilon < \alpha$$ for some $$\alpha$$, we can rewrite it as $$(r \alpha^{-1} (e^\alpha - 1) \epsilon, r \delta)$$-dp or $$(O(r \epsilon), r \delta)$$-dp.

Let $$\epsilon < 1$$. We see that if the mechanism $$N$$ is $$(\epsilon, \delta)$$-dp on $$Z^n$$, then $$M$$ is $$(2 r \epsilon, r \delta)$$-dp, and if we run it over $$k / r$$ minibatches, by Advanced Adaptive Composition theorem, we have $$(\sqrt{2 k r \log \beta^{-1}} \epsilon + 2 k r \epsilon^2, k \delta + \beta)$$-dp.

This is better than the privacy guarantee without subsampling, where we run over $$k$$ iterations and obtain $$(\sqrt{2 k \log \beta^{-1}} \epsilon + 2 k \epsilon^2, k \delta + \beta)$$-dp. So with subsampling we gain an extra $$\sqrt r$$ in the $$\epsilon$$-part of the privacy guarantee. But, smaller subsampling rate means smaller minibatch size, which would result in bigger variance, so there is a trade-off here.

Finally we define the differentially private stochastic gradient descent (DP-SGD) with the Gaussian mechanism (Abadi-Chu-Goodfellow-McMahan-Mironov-Talwar-Zhang 2016), which is (7) with the noise specialised to Gaussian and an added clipping operation to bound to sensitivity of the query to a chosen $$C$$:

$\theta_{t} = \theta_{t - 1} - \alpha \left(n^{-1} \sum_{i \in \gamma} \nabla_\theta \ell(x_i; \theta) |_{\theta = \theta_{t - 1}}\right)_{\text{Clipped at }C / 2} + N(0, \sigma^2 C^2 I),$

where

$y_{\text{Clipped at } \alpha} := y / (1 \vee {\|y\|_2 \over \alpha})$

is $$y$$ clipped to have norm at most $$\alpha$$.

Note that the clipping in DP-SGD is much stronger than making the query have sensitivity $$C$$. It makes the difference between the query results of two arbitrary inputs bounded by $$C$$, rather than neighbouring inputs.

In Part 2 of this post we will use the tools developed above to discuss the privacy guarantee for DP-SGD, among other things.

## References

• Abadi, Martín, Andy Chu, Ian Goodfellow, H. Brendan McMahan, Ilya Mironov, Kunal Talwar, and Li Zhang. "Deep Learning with Differential Privacy." Proceedings of the 2016 ACM SIGSAC Conference on Computer and Communications Security - CCS'16, 2016, 308--18. https://doi.org/10.1145/2976749.2978318.
• Dwork, Cynthia, and Aaron Roth. "The Algorithmic Foundations of Differential Privacy." Foundations and Trends® in Theoretical Computer Science 9, no. 3--4 (2013): 211--407. https://doi.org/10.1561/0400000042.
• Dwork, Cynthia, Guy N. Rothblum, and Salil Vadhan. "Boosting and Differential Privacy." In 2010 IEEE 51st Annual Symposium on Foundations of Computer Science, 51--60. Las Vegas, NV, USA: IEEE, 2010. https://doi.org/10.1109/FOCS.2010.12.
• Shiva Prasad Kasiviswanathan, Homin K. Lee, Kobbi Nissim, Sofya Raskhodnikova, and Adam Smith. "What Can We Learn Privately?" In 46th Annual IEEE Symposium on Foundations of Computer Science (FOCS'05). Pittsburgh, PA, USA: IEEE, 2005. https://doi.org/10.1109/SFCS.2005.1.
• Murtagh, Jack, and Salil Vadhan. "The Complexity of Computing the Optimal Composition of Differential Privacy." In Theory of Cryptography, edited by Eyal Kushilevitz and Tal Malkin, 9562:157--75. Berlin, Heidelberg: Springer Berlin Heidelberg, 2016. https://doi.org/10.1007/978-3-662-49096-9_7.
• Ullman, Jonathan. "Solution to CS7880 Homework 1.", 2017. http://www.ccs.neu.edu/home/jullman/cs7880s17/HW1sol.pdf
• Vadhan, Salil. "The Complexity of Differential Privacy." In Tutorials on the Foundations of Cryptography, edited by Yehuda Lindell, 347--450. Cham: Springer International Publishing, 2017. https://doi.org/10.1007/978-3-319-57048-8_7.

1. For those who have read about differential privacy and never heard of the term "divergence variable", it is closely related to the notion of "privacy loss", see the paragraph under Claim 6 in Back to approximate differential privacy. I defined the term this way so that we can focus on the more general stuff: compared to the privacy loss $$L(M(x) || M(x'))$$, the term $$L(p || q)$$ removes the "distracting information" that $$p$$ and $$q$$ are related to databases, queries, mechanisms etc., but merely probability laws. By removing the distraction, we simplify the analysis. And once we are done with the analysis of $$L(p || q)$$, we can apply the results obtained in the general setting to the special case where $$p$$ is the law of $$M(x)$$ and $$q$$ is the law of $$M(x')$$.