From ad5cc0a1a11ca22935a6894f483a77005c419719 Mon Sep 17 00:00:00 2001 From: Yuchen Pei Date: Thu, 14 Mar 2019 15:00:59 +0100 Subject: changed hypothesis to conjecture --- ...2019-03-14-great-but-manageable-expectations.md | 48 +++++++++++----------- 1 file changed, 24 insertions(+), 24 deletions(-) diff --git a/posts/2019-03-14-great-but-manageable-expectations.md b/posts/2019-03-14-great-but-manageable-expectations.md index a10e793..578bb09 100644 --- a/posts/2019-03-14-great-but-manageable-expectations.md +++ b/posts/2019-03-14-great-but-manageable-expectations.md @@ -303,12 +303,12 @@ assumptions. ACGMMTZ16 --------- -What follows is my understanding of this result. I call it a hypothesis +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. -**Hypothesis 0**. For a subsampled Gaussian mechanism +**Conjecture 0**. 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. @@ -322,7 +322,7 @@ $$G_\lambda(r p + (1 - r) q || q) = \sum_{k = 1 : \lambda} {\lambda \choose k} r **Proof**. Quite straightforward, by expanding the numerator $(r p + (1 - r) q)^\lambda$ using binomial expansion. $\square$ -**Proof of Hypothesis 0 +**Proof of Conjecture 0 (Incomplete)**. I will break the proof into two parts and discuss each one. @@ -336,13 +336,13 @@ one. **Remark in the proof**. Note that the choice of $c_1$, $c_2$ and the function $C(c_1, c_2)$ are important to the -practicality and usefulness of Hypothesis 0. +practicality and usefulness of Conjecture 0. -Item 1 can be deduced from the following hypothesis, which, as the only -gap in the Proof of Hypothesis 0, is heuristically reasonable, hence why -I am inclined to believe Hypothesis 0 is true. +Item 1 can be deduced from the following conjecture, which, as the only +gap in the Proof of Conjecture 0, is heuristically reasonable, hence why +I am inclined to believe Conjecture 0 is true. -**Hypothesis 1**. Let $p_i$, $q_i$, $\mu_i$, $\nu_i$ be +**Conjecture 1**. Let $p_i$, $q_i$, $\mu_i$, $\nu_i$ be probability densities on the same space for $i = 1 : n$. If $D_\lambda(p_i || q_i) \le D_\lambda(\mu_i || \nu_i)$ for all $i$, then @@ -352,7 +352,7 @@ Basically, it is saying \"if for each $i$, $p_i$ and $q_i$ are closer to each other than $\mu_i$ and $\nu_i$, then so is their average over $i$\". -To see how Hypothesis 1 implies Item 1, note that +To see how Conjecture 1 implies Item 1, note that $$D_\lambda(q_I || p_I) = D_\lambda(p_I || q_I) \begin{cases} @@ -364,13 +364,13 @@ 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 Hypothesis 1, by +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 Hypothesis 1 is to prove an equivalent +One way one may try to prove Conjecture 1 is to prove an equivalent statement: -**Hypothesis 2**. Let $p_1$, $q_1$, $p_2$, $q_2$, +**Conjecture 2**. Let $p_1$, $q_1$, $p_2$, $q_2$, $\mu_1$, $\mu_2$, $\nu_1$, $\nu_2$ be probability densities, and suppose $$\int {p_i^\lambda \over q_i^{\lambda - 1}} \le \int {\mu_i^\lambda \over \nu_i^{\lambda - 1}}, \qquad i = 1, 2$$ @@ -379,8 +379,8 @@ 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, Hypothesis 2 is a special case of Hypothesis 1 and -on the other hand, given Hypothesis 2, Hypothesis 1 can be shown using +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. Now let us prove Item 2. @@ -466,9 +466,9 @@ 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 Hypothesis 0 we have: +minibatches. With Conjecture 0 we have: -**Hypothesis 3**. Let $\epsilon, c_1, c_2 > 0$, +**Conjecture 3**. 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 @@ -481,7 +481,7 @@ $$\sigma \ge \sqrt{2 c_2^{-1}} \epsilon^{- {1 \over 2}} \sqrt{\log \delta^{-1}}, we can achieve $(\epsilon, \delta)$-dp. -**Proof**. By Hypothesis 0 and the Moment Composition Theorem +**Proof**. By Conjecture 0 and the Moment Composition Theorem (Claim 22), for $\lambda = c_2 \sigma^2$, substituting $T = {c_2 \over 2 C(c_1, c_2)} \epsilon \sigma^2$, we have @@ -489,7 +489,7 @@ $$\mathbb P(L(p || q) \ge \epsilon) \le \exp(k C(c_1, c_2) - \lambda \epsilon) = $\square$ -**Remark**. Hypothesis 3 is my understanding / version of +**Remark**. Conjecture 3 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: @@ -500,7 +500,7 @@ for consistency with this post: $$\sigma \ge c_2' {r \sqrt{T \log (1 / \delta)} \over \epsilon}. \qquad (10)$$ -I am however unable to reproduce this version, assuming Hypothesis 0 is +I am however unable to reproduce this version, assuming Conjecture 0 is true, for the following reasons: 1. In the proof in the paper, we have $\epsilon = c_1' r^2 T$ instead @@ -509,7 +509,7 @@ true, for the following reasons: opposite to the direction we want to prove: $$\exp(k C(c_1, c_2) - \lambda \epsilon) \ge ...$$ -2. The implicit condition $r = O(\sigma^{-1})$ of Hypothesis 0 whose +2. The implicit condition $r = O(\sigma^{-1})$ of Conjecture 0 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 @@ -576,8 +576,8 @@ DP-SGD with the Gaussian mechanism: 2. Example 1 (RDP for the Gaussian mechanism) -\> Claim 22 (Moment Composition Theorem) -\> Example 3 (Moment composition applied to the Gaussian mechanism) -3. Hypothesis 0 (RDP for Gaussian mechanism with specific magnitudes - for subsampling rate) -\> Hypothesis 3 (Moment Composition Theorem +3. Conjecture 0 (RDP for Gaussian mechanism with specific magnitudes + for subsampling rate) -\> Conjecture 3 (Moment Composition Theorem and translation to conventional DP) Which one is the best? @@ -716,7 +716,7 @@ 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 Hypothesis 2 +1. Prove Conjecture 2 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 @@ -724,7 +724,7 @@ untouched research problems: $L(p(y) || p(y + \alpha))$ 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 Hypothesis 0 is. Perhaps start with computing +4. Find out how useful Conjecture 0 is. Perhaps start with computing the constant $C$ nemerically. 5. Help with [the aforementioned issue](https://github.com/tensorflow/privacy/issues/23) in the -- cgit v1.2.3