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# Great but Manageable Expectations Published on 2019-03-14

This is Part 2 of a two-part blog post on differential privacy. Continuing from Part 1, I discuss the Rényi differential privacy, corresponding to the Rényi divergence, a study of the moment generating functions of the divergence between probability measures in order 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.

If you are confused by any notations, ask me or try this.

## Rényi divergence and differential privacy

Recall in the proof of Gaussian mechanism privacy guarantee (Claim 8) we used the Chernoff bound for the Gaussian noise. Why not use the Chernoff bound for the divergence variable / privacy loss directly, since the latter is closer to the core subject than the noise? This leads us to the study of Rényi divergence.

So far we have seen several notions of divergence used in differential privacy: the max divergence which is $$\epsilon$$-ind in disguise:

$D_\infty(p || q) := \max_y \log {p(y) \over q(y)},$

the $$\delta$$-approximate max divergence that defines the $$(\epsilon, \delta)$$-ind:

$D_\infty^\delta(p || q) := \max_y \log{p(y) - \delta \over q(y)},$

and the KL-divergence which is the expectation of the divergence variable:

$D(p || q) = \mathbb E L(p || q) = \int \log {p(y) \over q(y)} p(y) dy.$

The Rényi divergence is an interpolation between the max divergence and the KL-divergence, defined as the log moment generating function / cumulants of the divergence variable:

$D_\lambda(p || q) = (\lambda - 1)^{-1} \log \mathbb E \exp((\lambda - 1) L(p || q)) = (\lambda - 1)^{-1} \log \int {p(y)^\lambda \over q(y)^{\lambda - 1}} dy.$

Indeed, when $$\lambda \to \infty$$ we recover the max divergence, and when $$\lambda \to 1$$, by recognising $$D_\lambda$$ as a derivative in $$\lambda$$ at $$\lambda = 1$$, we recover the KL divergence. In this post we only consider $$\lambda > 1$$.

Using the Rényi divergence we may define:

Definition (Rényi differential privacy) (Mironov 2017). An mechanism $$M$$ is $$(\lambda, \rho)$$/-Rényi differentially private/ ($$(\lambda, \rho)$$-rdp) if for all $$x$$ and $$x'$$ with distance $$1$$,

$D_\lambda(M(x) || M(x')) \le \rho.$

For convenience we also define two related notions, $$G_\lambda (f || g)$$ and $$\kappa_{f, g} (t)$$ for $$\lambda > 1$$, $$t > 0$$ and positive functions $$f$$ and $$g$$:

$G_\lambda(f || g) = \int f(y)^{\lambda} g(y)^{1 - \lambda} dy; \qquad \kappa_{f, g} (t) = \log G_{t + 1}(f || g).$

For probability densities $$p$$ and $$q$$, $$G_{t + 1}(p || q)$$ and $$\kappa_{p, q}(t)$$ are the $$t$$th moment generating function and cumulant of the divergence variable $$L(p || q)$$, and

$D_\lambda(p || q) = (\lambda - 1)^{-1} \kappa_{p, q}(\lambda - 1).$

In the following, whenever you see $$t$$, think of it as $$\lambda - 1$$.

Example 1 (RDP for the Gaussian mechanism). Using the scaling and translation invariance of $$L$$ (6.1), we have that the divergence variable for two Gaussians with the same variance is

$L(N(\mu_1, \sigma^2 I) || N(\mu_2, \sigma^2 I)) \overset{d}{=} L(N(0, I) || N((\mu_2 - \mu_1) / \sigma, I)).$

With this we get

$D_\lambda(N(\mu_1, \sigma^2 I) || N(\mu_2, \sigma^2 I)) = {\lambda \|\mu_2 - \mu_1\|_2^2 \over 2 \sigma^2} = D_\lambda(N(\mu_2, \sigma^2 I) || N(\mu_1, \sigma^2 I)).$

Again due to the scaling invariance of $$L$$, we only need to consider $$f$$ with sensitivity $$1$$, see the discussion under (6.1). The Gaussian mechanism on query $$f$$ is thus $$(\lambda, \lambda / 2 \sigma^2)$$-rdp for any $$\lambda > 1$$.

From the example of Gaussian mechanism, we see that the relation between $$\lambda$$ and $$\rho$$ is like that between $$\epsilon$$ and $$\delta$$. Given $$\lambda$$ (resp. $$\rho$$) and parameters like variance of the noise and the sensitivity of the query, we can write $$\rho = \rho(\lambda)$$ (resp. $$\lambda = \lambda(\rho)$$).

Using the Chernoff bound (6.7), we can bound the divergence variable:

$\mathbb P(L(p || q) \ge \epsilon) \le {\mathbb E \exp(t L(p || q)) \over \exp(t \epsilon))} = \exp (\kappa_{p, q}(t) - \epsilon t). \qquad (7.7)$

For a function $$f: I \to \mathbb R$$, denote its Legendre transform by

$f^*(\epsilon) := \sup_{t \in I} (\epsilon t - f(t)).$

By taking infimum on the RHS of (7.7), we obtain

Claim 20. Two probability densities $$p$$ and $$q$$ are $$(\epsilon, \exp(-\kappa_{p, q}^*(\epsilon)))$$-ind.

Given a mechanism $$M$$, let $$\kappa_M(t)$$ denote an upper bound for the cumulant of its privacy loss:

$\log \mathbb E \exp(t L(M(x) || M(x'))) \le \kappa_M(t), \qquad \forall x, x'\text{ with } d(x, x') = 1.$

For example, we can set $$\kappa_M(t) = t \rho(t + 1)$$. Using the same argument we have the following:

Claim 21. If $$M$$ is $$(\lambda, \rho)$$-rdp, then

1. it is also $$(\epsilon, \exp((\lambda - 1) (\rho - \epsilon)))$$-dp for any $$\epsilon \ge \rho$$.
2. Alternatively, $$M$$ is $$(\epsilon, - \exp(\kappa_M^*(\epsilon)))$$-dp for any $$\epsilon > 0$$.
3. Alternatively, for any $$0 < \delta \le 1$$, $$M$$ is $$(\rho + (\lambda - 1)^{-1} \log \delta^{-1}, \delta)$$-dp.

Example 2 (Gaussian mechanism). We can apply the above argument to the Gaussian mechanism on query $$f$$ and get:

$\delta \le \inf_{\lambda > 1} \exp((\lambda - 1) ({\lambda \over 2 \sigma^2} - \epsilon))$

By assuming $$\sigma^2 > (2 \epsilon)^{-1}$$ we have that the infimum is achieved when $$\lambda = (1 + 2 \epsilon / \sigma^2) / 2$$ and

$\delta \le \exp(- ((2 \sigma)^{-1} - \epsilon \sigma)^2 / 2)$

which is the same result as (6.8), obtained using the Chernoff bound of the noise.

However, as we will see later, compositions will yield different results from those obtained from methods in Part 1 when considering Rényi dp.

### Moment Composition

Claim 22 (Moment Composition Theorem). Let $$M$$ be the adaptive composition of $$M_{1 : k}$$. Suppose for any $$y_{< i}$$, $$M_i(y_{< i})$$ is $$(\lambda, \rho)$$-rdp. Then $$M$$ is $$(\lambda, k\rho)$$-rdp.

Proof. Rather straightforward. As before let $$p_i$$ and $$q_i$$ be the conditional laws of adpative composition of $$M_{1 : i}$$ at $$x$$ and $$x'$$ respectively, and $$p^i$$ and $$q^i$$ be the joint laws of $$M_{1 : i}$$ at $$x$$ and $$x'$$ respectively. Denote

$D_i = \mathbb E \exp((\lambda - 1)\log {p^i(\xi_{1 : i}) \over q^i(\xi_{1 : i})})$

Then

\begin{aligned} D_i &= \mathbb E\mathbb E \left(\exp((\lambda - 1)\log {p_i(\xi_i | \xi_{< i}) \over q_i(\xi_i | \xi_{< i})}) \exp((\lambda - 1)\log {p^{i - 1}(\xi_{< i}) \over q^{i - 1}(\xi_{< i})}) \big| \xi_{< i}\right) \\ &= \mathbb E \mathbb E \left(\exp((\lambda - 1)\log {p_i(\xi_i | \xi_{< i}) \over q_i(\xi_i | \xi_{< i})}) | \xi_{< i}\right) \exp\left((\lambda - 1)\log {p^{i - 1}(\xi_{< i}) \over q^{i - 1}(\xi_{< i})}\right)\\ &\le \mathbb E \exp((\lambda - 1) \rho) \exp\left((\lambda - 1)\log {p^{i - 1}(\xi_{< i}) \over q^{i - 1}(\xi_{< i})}\right)\\ &= \exp((\lambda - 1) \rho) D_{i - 1}. \end{aligned}

Applying this recursively we have

$D_k \le \exp(k(\lambda - 1) \rho),$

and so

$(\lambda - 1)^{-1} \log \mathbb E \exp((\lambda - 1)\log {p^k(\xi_{1 : i}) \over q^k(\xi_{1 : i})}) = (\lambda - 1)^{-1} \log D_k \le k \rho.$

Since this holds for all $$x$$ and $$x'$$, we are done. $$\square$$

This, together with the scaling property of the legendre transformation:

$(k f)^*(x) = k f^*(x / k)$

yields

Claim 23. The $$k$$-fold adaptive composition of $$(\lambda, \rho(\lambda))$$-rdp mechanisms is $$(\epsilon, \exp(- k \kappa^*(\epsilon / k)))$$-dp, where $$\kappa(t) := t \rho(t + 1)$$.

Example 3 (Gaussian mechanism). We can apply the above claim to Gaussian mechanism. Again, without loss of generality we assume $$S_f = 1$$. But let us do it manually to get the same results. If we apply the Moment Composition Theorem to the an adaptive composition of Gaussian mechanisms on the same query, then since each $$M_i$$ is $$(\lambda, (2 \sigma^2)^{-1} \lambda)$$-rdp, the composition $$M$$ is $$(\lambda, (2 \sigma^2)^{-1} k \lambda)$$-rdp. Processing this using the Chernoff bound as in the previous example, we have

$\delta = \exp(- ((2 \sigma / \sqrt k)^{-1} - \epsilon \sigma / \sqrt k)^2 / 2),$

Substituting $$\sigma$$ with $$\sigma / \sqrt k$$ in (6.81), we conclude that if

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

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

As we will see in the discussions at the end of this post, this result is different from (and probably better than) the one obtained by using the Advanced Composition Theorem (Claim 18).

### Subsampling

We also have a subsampling theorem for the Rényi dp.

Claim 24. Fix $$r \in [0, 1]$$. Let $$m \le n$$ be two nonnegative integers with $$m = r n$$. Let $$N$$ be a $$(\lambda, \rho)$$-rdp machanism on $$X^m$$. Let $$\mathcal I := \{J \subset [n]: |J| = m\}$$ be the set of subsets of $$[n]$$ of size $$m$$. Define mechanism $$M$$ on $$X^n$$ by

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

where $$\gamma$$ is sampled uniformly from $$\mathcal I$$. Then $$M$$ is $$(\lambda, {1 \over \lambda - 1} \log (1 + r(e^{(\lambda - 1) \rho} - 1)))$$-rdp.

To prove Claim 24, we need a useful lemma:

Claim 25. Let $$p_{1 : n}$$ and $$q_{1 : n}$$ be nonnegative integers, and $$\lambda > 1$$. Then

${(\sum p_i)^\lambda \over (\sum q_i)^{\lambda - 1}} \le \sum_i {p_i^\lambda \over q_i^{\lambda - 1}}. \qquad (8)$

Proof. Let

$r(i) := p_i / P, \qquad u(i) := q_i / Q$

where

$P := \sum p_i, \qquad Q := \sum q_i$

then $$r$$ and $$u$$ are probability mass functions. Plugging in $$p_i = r(i) P$$ and $$q_i = u(i) Q$$ into the objective (8), it suffices to show

$1 \le \sum_i {r(i)^\lambda \over u(i)^{\lambda - 1}} = \mathbb E_{\xi \sim u} \left({r(\xi) \over u(\xi)}\right)^\lambda$

This is true due to Jensen's Inequality:

$\mathbb E_{\xi \sim u} \left({r(\xi) \over u(\xi)}\right)^\lambda \ge \left(\mathbb E_{\xi \sim u} {r(\xi) \over u(\xi)} \right)^\lambda = 1.$

$$\square$$

Proof of Claim 24. Define $$\mathcal I$$ as before.

Let $$p$$ and $$q$$ be the laws of $$M(x)$$ and $$M(x')$$ respectively. For any $$I \in \mathcal I$$, let $$p_I$$ and $$q_I$$ be the laws of $$N(x_I)$$ and $$N(x_I')$$ respectively. Then we have

\begin{aligned} p(y) &= n^{-1} \sum_{I \in \mathcal I} p_I(y) \\ q(y) &= n^{-1} \sum_{I \in \mathcal I} q_I(y), \end{aligned}

where $$n = |\mathcal I|$$.

The MGF of $$L(p || q)$$ is thus

$\mathbb E((\lambda - 1) L(p || q)) = n^{-1} \int {(\sum_I p_I(y))^\lambda \over (\sum_I q_I(y))^{\lambda - 1}} dy \le n^{-1} \sum_I \int {p_I(y)^\lambda \over q_I(y)^{\lambda - 1}} dy \qquad (9)$

where in the last step we used Claim 25. As in the proof of Claim 19, we divide $$\mathcal I$$ into disjoint sets $$\mathcal I_\in$$ and $$\mathcal I_\notin$$. Furthermore we denote by $$n_\in$$ and $$n_\notin$$ their cardinalities. Then the right hand side of (9) becomes

$n^{-1} \sum_{I \in \mathcal I_\in} \int {p_I(y)^\lambda \over q_I(y)^{\lambda - 1}} dy + n^{-1} \sum_{I \in \mathcal I_\notin} \int {p_I(y)^\lambda \over q_I(y)^{\lambda - 1}} dy$

The summands in the first are the MGF of $$L(p_I || q_I)$$, and the summands in the second term are $$1$$, so

\begin{aligned} \mathbb E((\lambda - 1) L(p || q)) &\le n^{-1} \sum_{I \in \mathcal I_\in} \mathbb E \exp((\lambda - 1) L(p_I || q_I)) + (1 - r) \\ &\le n^{-1} \sum_{I \in \mathcal I_\in} \exp((\lambda - 1) D_\lambda(p_I || q_I)) + (1 - r) \\ &\le r \exp((\lambda - 1) \rho) + (1 - r). \end{aligned}

Taking log and dividing by $$(\lambda - 1)$$ on both sides we have

$D_\lambda(p || q) \le (\lambda - 1)^{-1} \log (1 + r(\exp((\lambda - 1) \rho) - 1)).$

$$\square$$

As before, we can rewrite the conclusion of Lemma 6 using $$1 + z \le e^z$$ and obtain $$(\lambda, (\lambda - 1)^{-1} r (e^{(\lambda - 1) \rho} - 1))$$-rdp, which further gives $$(\lambda, \alpha^{-1} (e^\alpha - 1) r \rho)$$-rdp (or $$(\lambda, O(r \rho))$$-rdp) if $$(\lambda - 1) \rho < \alpha$$ for some $$\alpha$$.

It is not hard to see that the subsampling theorem in moment method, even though similar to the results of that in the usual method, does not help due to lack of an analogue of advanced composition theorem of the moments.

Example 4 (Gaussian mechanism). Applying the moment subsampling theorem to the Gaussian mechanism, we obtain $$(\lambda, O(r \lambda / \sigma^2))$$-rdp for a subsampled Gaussian mechanism with rate $$r$$. Abadi-Chu-Goodfellow-McMahan-Mironov-Talwar-Zhang 2016 (ACGMMTZ16 in the following), however, gains an extra $$r$$ in the bound given certain assumptions.

## ACGMMTZ16

What follows is my understanding of this result. I call it a conjecture because there is a gap which I am not able to reproduce their proof or prove it myself. This does not mean the result is false. On the contrary, I am inclined to believe it is true.

Claim 26. Assuming Conjecture 1 (see below) is true. For a subsampled Gaussian mechanism with ratio $$r$$, if $$r = O(\sigma^{-1})$$ and $$\lambda = O(\sigma^2)$$, then we have $$(\lambda, O(r^2 \lambda / \sigma^2))$$-rdp.

Wait, why is there a conjecture? Well, I have tried but not been able to prove the following, which is a hidden assumption in the original proof:

Conjecture 1. Let $$M$$ be the Gaussian mechanism with subsampling rate $$r$$, and $$p$$ and $$q$$ be the laws of $$M(x)$$ and $$M(x')$$ respectively, where $$d(x, x') = 1$$. Then

$D_\lambda (p || q) \le D_\lambda (r \mu_1 + (1 - r) \mu_0 || \mu_0)$

where $$\mu_i = N(i, \sigma^2)$$.

Remark. Conjecture 1 is heuristically reasonable. To see this, let us use the notations $$p_I$$ and $$q_I$$ to be $$q$$ and $$p$$ conditioned on the subsampling index $$I$$, just like in the proof of the subsampling theorems (Claim 19 and 24). Then for $$I \in \mathcal I_\in$$,

$D_\lambda(p_I || q_I) \le D_\lambda(\mu_0 || \mu_1),$

and for $$I \in \mathcal I_\notin$$,

$D_\lambda(p_I || q_I) = 0 = D_\lambda(\mu_0 || \mu_0).$

Since we are taking an average over $$\mathcal I$$, of which $$r |\mathcal I|$$ are in $$\mathcal I_\in$$ and $$(1 - r) |\mathcal I|$$ are in $$\mathcal I_\notin$$, (9.3) says "the inequalities carry over averaging".

A more general version of Conjecture 1 has been proven false. The counter example for the general version does not apply here, so it is still possible Conjecture 1 is true.

Let $$p_\in$$ (resp. $$q_\in$$) be the average of $$p_I$$ (resp. $$q_I$$) over $$I \in \mathcal I_\in$$, and $$p_\notin$$ (resp. $$q_\notin$$) be the average of $$p_I$$ (resp. $$q_I$$) over $$I \in \mathcal I_\notin$$.

Immediately we have $$p_\notin = q_\notin$$, hence

$D_\lambda(p_\notin || q_\notin) = 0 = D_\lambda(\mu_0 || \mu_0). \qquad(9.7)$

By Claim 25, we have

$D_\lambda(p_\in || q_\in) \le D_\lambda (\mu_1 || \mu_0). \qquad(9.9)$

So one way to prove Conjecture 1 is perhaps prove a more specialised comparison theorem than the false conjecture:

Given (9.7) and (9.9), show that

$D_\lambda(r p_\in + (1 - r) p_\notin || r q_\in + (1 - r) q_\notin) \le D_\lambda(r \mu_1 + (1 - r) \mu_0 || \mu_0).$

[End of Remark]

<!— Conjecture 1 $Probably [FALSE](https://math.stackexchange.com/a/3152296/149540), to be removed$. Let $$p_i$$, $$q_i$$, $$\mu_i$$, $$\nu_i$$ be probability densities on the same space for $$i = 1 : n$$. If $$D_\lambda(p_i || q_i) \le D_\lambda(\mu_i || \nu_i)$$ for all $$i$$, then

$D_\lambda(n^{-1} \sum_i p_i || n^{-1} \sum_i q_i) \le D_\lambda(n^{-1} \sum_i \mu_i || n^{-1} \sum_i \nu_i).$

Basically, it is saying \"if for each $$i$$, $$p_i$$ and $$q_i$$ are closer to each other than $$\mu_i$$ and $$\nu_i$$, then so are their averages over $$i$$\". So it is heuristically reasonable. But it is probably [**FALSE**](https://math.stackexchange.com/a/3152296/149540). This does not mean Claim 26 is false, as it might still be possible that Conjecture 2 (see below) is true.

This conjecture is equivalent to its special case when $$n = 2$$ by an induction argument (replacing one pair of densities at a time). –>

Recall the definition of $$G_\lambda$$ under the definition of Rényi differential privacy. The following Claim will be useful.

Claim 27. Let $$\lambda$$ be a positive integer, then

$G_\lambda(r p + (1 - r) q || q) = \sum_{k = 1 : \lambda} {\lambda \choose k} r^k (1 - r)^{\lambda - k} G_k(p || q).$

Proof. Quite straightforward, by expanding the numerator $$(r p + (1 - r) q)^\lambda$$ using binomial expansion. $$\square$$

Proof of Claim 26. By Conjecture 1, it suffices to prove the following:

If $$r \le c_1 \sigma^{-1}$$ and $$\lambda \le c_2 \sigma^2$$ for some positive constant $$c_1$$ and $$c_2$$, then there exists $$C = C(c_1, c_2)$$ such that $$G_\lambda (r \mu_1 + (1 - r) \mu_0 || \mu_0) \le C$$ (since $$O(r^2 \lambda^2 / \sigma^2) = O(1)$$).

Remark in the proof. Note that the choice of $$c_1$$, $$c_2$$ and the function $$C(c_1, c_2)$$ are important to the practicality and usefulness of Claim 26.

<!— Part 1 can be derived using Conjecture 1, but since Conjecture 1 is probably false, let us rename Part 1 itself Conjecture 2, which needs to be verified by other means. We use the notations $$p_I$$ and $$q_I$$ to be $$q$$ and $$p$$ conditioned on the subsampling index $$I$$, just like in the proof of the subsampling theorems (Claim 19 and 24). Then

$D_\lambda(q_I || p_I) = D_\lambda(p_I || q_I) \begin{cases} \le D_\lambda(\mu_0 || \mu_1) = D_\lambda(\mu_1 || \mu_0), & I \in \mathcal I_\in\\ = D_\lambda(\mu_0 || \mu_0) = D_\lambda(\mu_1 || \mu_1) = 0 & I \in \mathcal I_\notin \end{cases}$

Since $$p = |\mathcal I|^{-1} \sum_{I \in \mathcal I} p_I$$ and $$q = |\mathcal I|^{-1} \sum_{I \in \mathcal I} q_I$$ and $$|\mathcal I_\in| = r |\mathcal I|$$, by Conjecture 1, we have Part 1.

Remark in the proof. As we can see here, instead of trying to prove Conjecture 1, it suffices to prove a weaker version of it, by specialising on mixture of Gaussians, in order to have a Claim 26 without any conjectural assumptions. I have in fact posted the Conjecture on [Stackexchange](https://math.stackexchange.com/questions/3147963/an-inequality-related-to-the-renyi-divergence).

Now let us verify Part 2. –>

Using Claim 27 and Example 1, we have

\begin{aligned} G_\lambda(r \mu_1 + (1 - r) \mu_0 || \mu_0)) &= \sum_{j = 0 : \lambda} {\lambda \choose j} r^j (1 - r)^{\lambda - j} G_j(\mu_1 || \mu_0)\\ &=\sum_{j = 0 : \lambda} {\lambda \choose j} r^j (1 - r)^{\lambda - j} \exp(j (j - 1) / 2 \sigma^2). \qquad (9.5) \end{aligned}

Denote by $$n = \lceil \sigma^2 \rceil$$. It suffices to show

$\sum_{j = 0 : n} {n \choose j} (c_1 n^{- 1 / 2})^j (1 - c_1 n^{- 1 / 2})^{n - j} \exp(c_2 j (j - 1) / 2 n) \le C$

Note that we can discard the linear term $$- c_2 j / \sigma^2$$ in the exponential term since we want to bound the sum from above.

We examine the asymptotics of this sum when $$n$$ is large, and treat the sum as an approximation to an integration of a function $$\phi: [0, 1] \to \mathbb R$$. For $$j = x n$$, where $$x \in (0, 1)$$, $$\phi$$ is thus defined as (note we multiply the summand with $$n$$ to compensate the uniform measure on $$1, ..., n$$:

\begin{aligned} \phi_n(x) &:= n {n \choose j} (c_1 n^{- 1 / 2})^j (1 - c_1 n^{- 1 / 2})^{n - j} \exp(c_2 j^2 / 2 n) \\ &= n {n \choose x n} (c_1 n^{- 1 / 2})^{x n} (1 - c_1 n^{- 1 / 2})^{(1 - x) n} \exp(c_2 x^2 n / 2) \end{aligned}

Using Stirling's approximation

$n! \approx \sqrt{2 \pi n} n^n e^{- n},$

we can approach the binomial coefficient:

${n \choose x n} \approx (\sqrt{2 \pi x (1 - x)} x^{x n} (1 - x)^{(1 - x) n})^{-1}.$

We also approximate

$(1 - c_1 n^{- 1 / 2})^{(1 - x) n} \approx \exp(- c_1 \sqrt{n} (1 - x)).$

With these we have

$\phi_n(x) \approx {1 \over \sqrt{2 \pi x (1 - x)}} \exp\left(- {1 \over 2} x n \log n + (x \log c_1 - x \log x - (1 - x) \log (1 - x) + {1 \over 2} c_2 x^2) n + {1 \over 2} \log n\right).$

This vanishes as $$n \to \infty$$, and since $$\phi_n(x)$$ is bounded above by the integrable function $${1 \over \sqrt{2 \pi x (1 - x)}}$$ (c.f. the arcsine law), and below by $$0$$, we may invoke the dominant convergence theorem and exchange the integral with the limit and get

\begin{aligned} \lim_{n \to \infty} &G_n (r \mu_1 + (1 - r) \mu_0 || \mu_0)) \\ &\le \lim_{n \to \infty} \int \phi_n(x) dx = \int \lim_{n \to \infty} \phi_n(x) dx = 0. \end{aligned}

Thus we have that the generating function of the divergence variable $$L(r \mu_1 + (1 - r) \mu_0 || \mu_0)$$ is bounded.

Can this be true for better orders

$r \le c_1 \sigma^{- d_r},\qquad \lambda \le c_2 \sigma^{d_\lambda}$

for some $$d_r \in (0, 1]$$ and $$d_\lambda \in [2, \infty)$$? If we follow the same approximation using these exponents, then letting $$n = c_2 \sigma^{d_\lambda}$$,

\begin{aligned} {n \choose j} &r^j (1 - r)^{n - j} G_j(\mu_0 || \mu_1) \le \phi_n(x) \\ &\approx {1 \over \sqrt{2 \pi x (1 - x)}} \exp\left({1 \over 2} c_2^{2 \over d_\lambda} x^2 n^{2 - {2 \over d_\lambda}} - {d_r \over 2} x n \log n + (x \log c_1 - x \log x - (1 - x) \log (1 - x)) n + {1 \over 2} \log n\right). \end{aligned}

So we see that to keep the divergence moments bounded it is possible to have any $$r = O(\sigma^{- d_r})$$ for $$d_r \in (0, 1)$$, but relaxing $$\lambda$$ may not be safe.

If we relax $$r$$, then we get

$G_\lambda(r \mu_1 + (1 - r) \mu_0 || \mu_0) = O(r^{2 / d_r} \lambda^2 \sigma^{-2}) = O(1).$

Note that now the constant $$C$$ depends on $$d_r$$ as well. Numerical experiments seem to suggest that $$C$$ can increase quite rapidly as $$d_r$$ decreases from $$1$$. $$\square$$

In the following for consistency we retain $$k$$ as the number of epochs, and use $$T := k / r$$ to denote the number of compositions / steps / minibatches. With Claim 26 we have:

Claim 28. Assuming Conjecture 1 is true. Let $$\epsilon, c_1, c_2 > 0$$, $$r \le c_1 \sigma^{-1}$$, $$T = {c_2 \over 2 C(c_1, c_2)} \epsilon \sigma^2$$. then the DP-SGD with subsampling rate $$r$$, and $$T$$ steps is $$(\epsilon, \delta)$$-dp for

$\delta = \exp(- {1 \over 2} c_2 \sigma^2 \epsilon).$

In other words, for

$\sigma \ge \sqrt{2 c_2^{-1}} \epsilon^{- {1 \over 2}} \sqrt{\log \delta^{-1}},$

we can achieve $$(\epsilon, \delta)$$-dp.

Proof. By Claim 26 and the Moment Composition Theorem (Claim 22), for $$\lambda = c_2 \sigma^2$$, substituting $$T = {c_2 \over 2 C(c_1, c_2)} \epsilon \sigma^2$$, we have

$\mathbb P(L(p || q) \ge \epsilon) \le \exp(k C(c_1, c_2) - \lambda \epsilon) = \exp\left(- {1 \over 2} c_2 \sigma^2 \epsilon\right).$

$$\square$$

Remark. Claim 28 is my understanding / version of Theorem 1 in [ACGMMTZ16], by using the same proof technique. Here I quote the original version of theorem with notions and notations altered for consistency with this post:

There exists constants $$c_1', c_2' > 0$$ so that for any $$\epsilon < c_1' r^2 T$$, DP-SGD is $$(\epsilon, \delta)$$-differentially private for any $$\delta > 0$$ if we choose

$\sigma \ge c_2' {r \sqrt{T \log (1 / \delta)} \over \epsilon}. \qquad (10)$

I am however unable to reproduce this version, assuming Conjecture 1 is true, for the following reasons:

1. In the proof in the paper, we have $$\epsilon = c_1' r^2 T$$ instead of "less than" in the statement of the Theorem. If we change it to $$\epsilon < c_1' r^2 T$$ then the direction of the inequality becomes opposite to the direction we want to prove: $\exp(k C(c_1, c_2) - \lambda \epsilon) \ge ...$
2. The condition $$r = O(\sigma^{-1})$$ of Claim 26 whose result is used in the proof of this theorem is not mentioned in the statement of the proof. The implication is that (10) becomes an ill-formed condition as the right hand side also depends on $$\sigma$$.

## Tensorflow implementation

The DP-SGD is implemented in TensorFlow Privacy. In the following I discuss the package in the current state (2019-03-11). It is divided into two parts: optimizers which implements the actual differentially private algorithms, and analysis which computes the privacy guarantee.

The analysis part implements a privacy ledger that "keeps a record of all queries executed over a given dataset for the purpose of computing privacy guarantees". On the other hand, all the computation is done in rdp_accountant.py. At this moment, rdp_accountant.py only implements the computation of the privacy guarantees for DP-SGD with Gaussian mechanism. In the following I will briefly explain the code in this file.

Some notational correspondences: their alpha is our $$\lambda$$, their q is our $$r$$, their A_alpha (in the comments) is our $$\kappa_{r N(1, \sigma^2) + (1 - r) N(0, \sigma^2)} (\lambda - 1)$$, at least when $$\lambda$$ is an integer.

• The function _compute_log_a presumably computes the cumulants $$\kappa_{r N(1, \sigma^2) + (1 - r) N(0, \sigma^2), N(0, \sigma^2)}(\lambda - 1)$$. It calls _compute_log_a_int or _compute_log_a_frac depending on whether $$\lambda$$ is an integer.
• The function _compute_log_a_int computes the cumulant using (9.5).
• When $$\lambda$$ is not an integer, we can't use (9.5). I have yet to decode how _compute_log_a_frac computes the cumulant (or an upper bound of it) in this case
• The function _compute_delta computes $$\delta$$s for a list of $$\lambda$$s and $$\kappa$$s using Item 1 of Claim 25 and return the smallest one, and the function _compute_epsilon computes epsilon uses Item 3 in Claim 25 in the same way.

In optimizers, among other things, the DP-SGD with Gaussian mechanism is implemented in dp_optimizer.py and gaussian_query.py. See the definition of DPGradientDescentGaussianOptimizer in dp_optimizer.py and trace the calls therein.

At this moment, the privacy guarantee computation part and the optimizer part are separated, with rdp_accountant.py called in compute_dp_sgd_privacy.py with user-supplied parameters. I think this is due to the lack of implementation in rdp_accountant.py of any non-DPSGD-with-Gaussian privacy guarantee computation. There is already an issue on this, so hopefully it won't be long before the privacy guarantees can be automatically computed given a DP-SGD instance.

## Comparison among different methods

So far we have seen three routes to compute the privacy guarantees for DP-SGD with the Gaussian mechanism:

1. Claim 9 (single Gaussian mechanism privacy guarantee) -> Claim 19 (Subsampling theorem) -> Claim 18 (Advanced Adaptive Composition Theorem)
2. Example 1 (RDP for the Gaussian mechanism) -> Claim 22 (Moment Composition Theorem) -> Example 3 (Moment composition applied to the Gaussian mechanism)
3. Claim 26 (RDP for Gaussian mechanism with specific magnitudes for subsampling rate) -> Claim 28 (Moment Composition Theorem and translation to conventional DP)

Which one is the best?

To make fair comparison, we may use one parameter as the metric and set all others to be the same. For example, we can

1. Given the same $$\epsilon$$, $$r$$ (in Route 1 and 3), $$k$$, $$\sigma$$, compare the $$\delta$$s
2. Given the same $$\epsilon$$, $$r$$ (in Route 1 and 3), $$k$$, $$\delta$$, compare the $$\sigma$$s
3. Given the same $$\delta$$, $$r$$ (in Route 1 and 3), $$k$$, $$\sigma$$, compare the $$\epsilon$$s.

I find that the first one, where $$\delta$$ is used as a metric, the best. This is because we have the tightest bounds and the cleanest formula when comparing the $$\delta$$. For example, the Azuma and Chernoff bounds are both expressed as a bound for $$\delta$$. On the other hand, the inversion of these bounds either requires a cost in the tightness (Claim 9, bounds on $$\sigma$$) or in the complexity of the formula (Claim 16 Advanced Adaptive Composition Theorem, bounds on $$\epsilon$$).

So if we use $$\sigma$$ or $$\epsilon$$ as a metric, either we get a less fair comparison, or have to use a much more complicated formula as the bounds.

Let us first compare Route 1 and Route 2 without specialising to the Gaussian mechanism.

Warning. What follows is a bit messy.

Suppose each mechanism $$N_i$$ satisfies $$(\epsilon', \delta(\epsilon'))$$-dp. Let $$\tilde \epsilon := \log (1 + r (e^{\epsilon'} - 1))$$, then we have the subsampled mechanism $$M_i(x) = N_i(x_\gamma)$$ is $$(\tilde \epsilon, r \tilde \delta(\tilde \epsilon))$$-dp, where

$\tilde \delta(\tilde \epsilon) = \delta(\log (r^{-1} (\exp(\tilde \epsilon) - 1) + 1))$

Using the Azuma bound in the proof of Advanced Adaptive Composition Theorem (6.99):

$\mathbb P(L(p^k || q^k) \ge \epsilon) \le \exp(- {(\epsilon - r^{-1} k a(\tilde\epsilon))^2 \over 2 r^{-1} k (\tilde\epsilon + a(\tilde\epsilon))^2}).$

So we have the final bound for Route 1:

$\delta_1(\epsilon) = \min_{\tilde \epsilon: \epsilon > r^{-1} k a(\tilde \epsilon)} \exp(- {(\epsilon - r^{-1} k a(\tilde\epsilon))^2 \over 2 r^{-1} k (\tilde\epsilon + a(\tilde\epsilon))^2}) + k \tilde \delta(\tilde \epsilon).$

As for Route 2, since we do not gain anything from subsampling in RDP, we do not subsample at all.

By Claim 23, we have the bound for Route 2:

$\delta_2(\epsilon) = \exp(- k \kappa^* (\epsilon / k)).$

On one hand, one can compare $$\delta_1$$ and $$\delta_2$$ with numerical experiments. On the other hand, if we further specify $$\delta(\epsilon')$$ in Route 1 as the Chernoff bound for the cumulants of divergence variable, i.e.

$\delta(\epsilon') = \exp(- \kappa^* (\epsilon')),$

we have

$\delta_1 (\epsilon) = \min_{\tilde \epsilon: a(\tilde \epsilon) < r k^{-1} \epsilon} \exp(- {(\epsilon - r^{-1} k a(\tilde\epsilon))^2 \over 2 r^{-1} k (\tilde\epsilon + a(\tilde\epsilon))^2}) + k \exp(- \kappa^* (b(\tilde\epsilon))),$

where

$b(\tilde \epsilon) := \log (r^{-1} (\exp(\tilde \epsilon) - 1) + 1) \le r^{-1} \tilde\epsilon.$

We note that since $$a(\tilde \epsilon) = \tilde\epsilon(e^{\tilde \epsilon} - 1) 1_{\tilde\epsilon < \log 2} + \tilde\epsilon 1_{\tilde\epsilon \ge \log 2}$$, we may compare the two cases separately.

Note that we have $$\kappa^*$$ is a monotonously increasing function, therefore

$\kappa^* (b(\tilde\epsilon)) \le \kappa^*(r^{-1} \tilde\epsilon).$

So for $$\tilde \epsilon \ge \log 2$$, we have

$k \exp(- \kappa^*(b(\tilde\epsilon))) \ge k \exp(- \kappa^*(r^{-1} \tilde \epsilon)) \ge k \exp(- \kappa^*(k^{-1} \epsilon)) \ge \delta_2(\epsilon).$

For $$\tilde\epsilon < \log 2$$, it is harder to compare, as now

$k \exp(- \kappa^*(b(\tilde\epsilon))) \ge k \exp(- \kappa^*(\epsilon / \sqrt{r k})).$

It is tempting to believe that this should also be greater than $$\delta_2(\epsilon)$$. But I can not say for sure. At least in the special case of Gaussian, we have

$k \exp(- \kappa^*(\epsilon / \sqrt{r k})) = k \exp(- (\sigma \sqrt{\epsilon / k r} - (2 \sigma)^{-1})^2) \ge \exp(- k ({\sigma \epsilon \over k} - (2 \sigma)^{-1})^2) = \delta_2(\epsilon)$

when $$\epsilon$$ is sufficiently small. However we still need to consider the case where $$\epsilon$$ is not too small. But overall it seems most likely Route 2 is superior than Route 1.

So let us compare Route 2 with Route 3:

Given the condition to obtain the Chernoff bound

${\sigma \epsilon \over k} > (2 \sigma)^{-1}$

we have

$\delta_2(\epsilon) > \exp(- k (\sigma \epsilon / k)^2) = \exp(- \sigma^2 \epsilon^2 / k).$

For this to achieve the same bound

$\delta_3(\epsilon) = \exp\left(- {1 \over 2} c_2 \sigma^2 \epsilon\right)$

we need $$k < {2 \epsilon \over c_2}$$. This is only possible if $$c_2$$ is small or $$\epsilon$$ is large, since $$k$$ is a positive integer.

So taking at face value, Route 3 seems to achieve the best results. However, it also has some similar implicit conditions that need to be satisfied: First $$T$$ needs to be at least $$1$$, meaning

${c_2 \over C(c_1, c_2)} \epsilon \sigma^2 \ge 1.$

Second, $$k$$ needs to be at least $$1$$ as well, i.e.

$k = r T \ge {c_1 c_2 \over C(c_1, c_2)} \epsilon \sigma \ge 1.$

Both conditions rely on the magnitudes of $$\epsilon$$, $$\sigma$$, $$c_1$$, $$c_2$$, and the rate of growth of $$C(c_1, c_2)$$. The biggest problem in this list is the last, because if we know how fast $$C$$ grows then we'll have a better idea what are the constraints for the parameters to achieve the result in Route 3.

## Further questions

Here is a list of what I think may be interesting topics or potential problems to look at, with no guarantee that they are all awesome untouched research problems:

1. Prove Conjecture 1
2. Find a theoretically definitive answer whether the methods in Part 1 or Part 2 yield better privacy guarantees.
3. Study the non-Gaussian cases, general or specific. Let $$p$$ be some probability density, what is the tail bound of $$L(p(y) || p(y + \alpha))$$ for $$|\alpha| \le 1$$? Can you find anything better than Gaussian? For a start, perhaps the nice tables of Rényi divergence in Gil-Alajaji-Linder 2013 may be useful?
4. Find out how useful Claim 26 is. Perhaps start with computing the constant $$C$$ nemerically.
5. Help with the aforementioned issue in the Tensorflow privacy package.
• 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.
• Erven, Tim van, and Peter Harremoës. "R\'enyi Divergence and Kullback-Leibler Divergence." IEEE Transactions on Information Theory 60, no. 7 (July 2014): 3797–3820. https://doi.org/10.1109/TIT.2014.2320500.
• Gil, M., F. Alajaji, and T. Linder. "Rényi Divergence Measures for Commonly Used Univariate Continuous Distributions." Information Sciences 249 (November 2013): 124–31. https://doi.org/10.1016/j.ins.2013.06.018.
• Mironov, Ilya. "Renyi Differential Privacy." 2017 IEEE 30th Computer Security Foundations Symposium (CSF), August 2017, 263–75. https://doi.org/10.1109/CSF.2017.11.