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| -rw-r--r-- | posts/2019-03-14-great-but-manageable-expectations.md | 81 |
1 files changed, 36 insertions, 45 deletions
diff --git a/posts/2019-03-14-great-but-manageable-expectations.md b/posts/2019-03-14-great-but-manageable-expectations.md index 8b4187c..d812043 100644 --- a/posts/2019-03-14-great-but-manageable-expectations.md +++ b/posts/2019-03-14-great-but-manageable-expectations.md @@ -327,25 +327,43 @@ 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. -**Conjecture 0**. For a subsampled Gaussian mechanism +**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 $p_i$, $q_i$, $\mu_i$, $\nu_i$ 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. + +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 26**. Let $\lambda$ be a positive integer, then +**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 Conjecture 0 -(Incomplete)**. I will break the proof into two parts and discuss each -one. +**Proof of Claim 26**. 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 $d(x, x') = 1$. +I will break the proof into two parts: -1. The MGF of the privacy loss is bounded by that of +1. The MGF of the privacy loss $L(p || q)$ is bounded by that of $L(r \mu_1 + (1 - r) \mu_0 || \mu_0)$ where $\mu_i = N(i, \sigma^2)$. 2. If $r \le c_1 \sigma^{-1}$ and $\lambda \le c_2 \sigma^2$, then @@ -357,21 +375,9 @@ one. $c_1$, $c_2$ and the function $C(c_1, c_2)$ are important to the practicality and usefulness of Conjecture 0. -Item 1 can be deduced from the following conjecture, which, as the only -gap in the Proof of Conjecture 0, is heuristically reasonable, hence why -I am inclined to believe Conjecture 0 is true. - -**Conjecture 1**. 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 is their average over -$i$\". - -To see how Conjecture 1 implies Item 1, note that +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 $$D_\lambda(q_I || p_I) = D_\lambda(p_I || q_I) \begin{cases} @@ -383,28 +389,14 @@ and that $p = |\mathcal I|^{-1} \sum_{I \in \mathcal I} p_I$ and $q = |\mathcal I|^{-1} \sum_{I \in \mathcal I} q_I$ and $|\mathcal I_\in| = r |\mathcal I|$. -Alternatively, one may try to prove a weaker version of Conjecture 1, by -specialising on mixture of Gaussians. - -One way one may try to prove Conjecture 1 is to prove an equivalent -statement: - -**Conjecture 2**. Let $p_1$, $q_1$, $p_2$, $q_2$, -$\mu_1$, $\mu_2$, $\nu_1$, $\nu_2$ be probability densities, and suppose - -$$\int {p_i^\lambda \over q_i^{\lambda - 1}} \le \int {\mu_i^\lambda \over \nu_i^{\lambda - 1}}, \qquad i = 1, 2$$ - -then - -$$\int {(p_1 + p_2)^\lambda \over (q_1 + q_2)^{\lambda - 1}} \le \int {(\mu_1 + \mu_1)^\lambda \over (\nu_1 + \nu_2)^{\lambda - 1}}.$$ - -Indeed, on one hand, Conjecture 2 is a special case of Conjecture 1 and -on the other hand, given Conjecture 2, Conjecture 1 can be shown using -induction by replacing one pair of densities a time. +**Remark in the proof**. As we can see here, instead of trying to prove Conjecture 1, +it suffices to prove a weaker version of it, by specialising on mixture of Gaussians, +to have a Claim 26 without any conjectural assumptions. +I have in fact posted the Conjecture on [Stackexchange](https://math.stackexchange.com/questions/3147963/an-inequality-related-to-the-renyi-divergence). Now let us prove Item 2. -Using Claim 26 and Example 1, we have +Using Claim 27 and Example 1, we have $$\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)\\ @@ -487,7 +479,7 @@ 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 Conjecture 0 we have: -**Conjecture 3**. Let $\epsilon, c_1, c_2 > 0$, +**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 @@ -500,7 +492,7 @@ $$\sigma \ge \sqrt{2 c_2^{-1}} \epsilon^{- {1 \over 2}} \sqrt{\log \delta^{-1}}, we can achieve $(\epsilon, \delta)$-dp. -**Proof**. By Conjecture 0 and the Moment Composition Theorem +**Proof**. By Claim 26 and the Moment Composition Theorem (Claim 22), for $\lambda = c_2 \sigma^2$, substituting $T = {c_2 \over 2 C(c_1, c_2)} \epsilon \sigma^2$, we have @@ -508,7 +500,7 @@ $$\mathbb P(L(p || q) \ge \epsilon) \le \exp(k C(c_1, c_2) - \lambda \epsilon) = $\square$ -**Remark**. Conjecture 3 is my understanding / version of +**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: @@ -626,8 +618,7 @@ bounds. Let us first compare Route 1 and Route 2 without specialising to the Gaussian mechanism. -**Disclaimer**. What follows is my original research and -has not been reviewed by anyone. +**Disclaimer**. What follows is a bit messy and has not been reviewed by anyone. Suppose each mechanism $N_i$ satisfies $(\epsilon', \delta(\epsilon'))$-dp. Let |