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author | Yuchen Pei <me@ypei.me> | 2019-03-21 20:18:28 +0100 |
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committer | Yuchen Pei <me@ypei.me> | 2019-03-21 20:18:28 +0100 |
commit | c4f90a8507d6525d3cd6aafb5fa767d893b61d3e (patch) | |
tree | ea729ea488e39f07df05968795f52ceb186ae0af | |
parent | bcd3b5a6f51d751b7d7e44c652ad70df83bacfd0 (diff) |
edits
-rw-r--r-- | posts/2019-03-14-great-but-manageable-expectations.md | 5 |
1 files changed, 2 insertions, 3 deletions
diff --git a/posts/2019-03-14-great-but-manageable-expectations.md b/posts/2019-03-14-great-but-manageable-expectations.md index 6a7e973..9315844 100644 --- a/posts/2019-03-14-great-but-manageable-expectations.md +++ b/posts/2019-03-14-great-but-manageable-expectations.md @@ -381,11 +381,10 @@ $$G_\lambda(r p + (1 - r) q || q) = \sum_{k = 1 : \lambda} {\lambda \choose k} r $(r p + (1 - r) q)^\lambda$ using binomial expansion. $\square$ **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 $d(x, x') = 1$. By Conjecture 1, it suffices to prove the following: -If $r \le c_1 \sigma^{-1}$ and $\lambda \le c_2 \sigma^2$, then +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 $O(r^2 \lambda^2 / \sigma^2) = O(1)$). |