Mixing
bounding min-entropy gain in differential privacy
$begingroup$
In privacy-related computer science literature, we say that a randomized algorithm $mathcal{K}$ that produces a model $theta$ from a sample $X=(x_1,...,x_n)$ is $epsilon$-differentially private iff
$$
forall theta, quad mathbb{P}[mathcal{K}(X)=theta | X=X_0] leq e^epsilon mathbb{P}[mathcal{K} (X)=theta|X=X_1] tag{1}
$$
where $X_0$ and $X_1$ are any Hamming-1 neighbors (differ by a single entry)
Using Bayes rule, we can rewrite (1) as a bounded Bayes ratio
$$
frac{mathbb{P}[X=X_0|theta]}{mathbb{P}[X=X_1|theta]} leq e^epsilon frac{mathbb{P}[X=X_0]}{mathbb{P}[X=X_1]} tag{2}
$$
In order to relate this to the Bayes (irreducible) error of an adversary trying to estimate $X$ from $theta$, one can leverage the min-entropy
$$
H_infty(X) = -log max_x mathbb{P}[X=x] \ text{and} \
H_infty(X|theta) = -log mathbb{E}_theta max_x mathbb{P}[X=x|theta]
$$
First part of the question : Can we translate (2) into a bound involving $H_infty(X|theta)$ and $H_infty(X)$ ?
Second part of the question : Since $H_infty(X|theta)$ takes the expectation over $theta$, it looks like a weaker guarantee than (1), (bounded leakage only holding in expectation, with some $theta$ leaking more than the bound). Is there a well-accepted entropy definition that keeps the same expressivity as (1), e.g.,
$$
H_text{bla}(X|theta) = -log max_{theta,x} mathbb{P}[X=x|theta]
$$
or something like that ?
thanks !
entropy ratio log-likelihood
asked Jan 16 at 11:19
Jerome FJerome F
254
$endgroup$
add a comment |
$begingroup$
In privacy-related computer science literature, we say that a randomized algorithm $mathcal{K}$ that produces a model $theta$ from a sample $X=(x_1,...,x_n)$ is $epsilon$-differentially private iff
$$
forall theta, quad mathbb{P}[mathcal{K}(X)=theta | X=X_0] leq e^epsilon mathbb{P}[mathcal{K} (X)=theta|X=X_1] tag{1}
$$
where $X_0$ and $X_1$ are any Hamming-1 neighbors (differ by a single entry)
Using Bayes rule, we can rewrite (1) as a bounded Bayes ratio
$$
frac{mathbb{P}[X=X_0|theta]}{mathbb{P}[X=X_1|theta]} leq e^epsilon frac{mathbb{P}[X=X_0]}{mathbb{P}[X=X_1]} tag{2}
$$
In order to relate this to the Bayes (irreducible) error of an adversary trying to estimate $X$ from $theta$, one can leverage the min-entropy
$$
H_infty(X) = -log max_x mathbb{P}[X=x] \ text{and} \
H_infty(X|theta) = -log mathbb{E}_theta max_x mathbb{P}[X=x|theta]
$$
First part of the question : Can we translate (2) into a bound involving $H_infty(X|theta)$ and $H_infty(X)$ ?
Second part of the question : Since $H_infty(X|theta)$ takes the expectation over $theta$, it looks like a weaker guarantee than (1), (bounded leakage only holding in expectation, with some $theta$ leaking more than the bound). Is there a well-accepted entropy definition that keeps the same expressivity as (1), e.g.,
$$
H_text{bla}(X|theta) = -log max_{theta,x} mathbb{P}[X=x|theta]
$$
or something like that ?
thanks !
entropy ratio log-likelihood
asked Jan 16 at 11:19
Jerome FJerome F
254
$endgroup$
add a comment |
$begingroup$
In privacy-related computer science literature, we say that a randomized algorithm $mathcal{K}$ that produces a model $theta$ from a sample $X=(x_1,...,x_n)$ is $epsilon$-differentially private iff
$$
forall theta, quad mathbb{P}[mathcal{K}(X)=theta | X=X_0] leq e^epsilon mathbb{P}[mathcal{K} (X)=theta|X=X_1] tag{1}
$$
where $X_0$ and $X_1$ are any Hamming-1 neighbors (differ by a single entry)
Using Bayes rule, we can rewrite (1) as a bounded Bayes ratio
$$
frac{mathbb{P}[X=X_0|theta]}{mathbb{P}[X=X_1|theta]} leq e^epsilon frac{mathbb{P}[X=X_0]}{mathbb{P}[X=X_1]} tag{2}
$$
In order to relate this to the Bayes (irreducible) error of an adversary trying to estimate $X$ from $theta$, one can leverage the min-entropy
$$
H_infty(X) = -log max_x mathbb{P}[X=x] \ text{and} \
H_infty(X|theta) = -log mathbb{E}_theta max_x mathbb{P}[X=x|theta]
$$
First part of the question : Can we translate (2) into a bound involving $H_infty(X|theta)$ and $H_infty(X)$ ?
Second part of the question : Since $H_infty(X|theta)$ takes the expectation over $theta$, it looks like a weaker guarantee than (1), (bounded leakage only holding in expectation, with some $theta$ leaking more than the bound). Is there a well-accepted entropy definition that keeps the same expressivity as (1), e.g.,
$$
H_text{bla}(X|theta) = -log max_{theta,x} mathbb{P}[X=x|theta]
$$
or something like that ?
thanks !
entropy ratio log-likelihood
asked Jan 16 at 11:19
Jerome FJerome F
254
$endgroup$
In privacy-related computer science literature, we say that a randomized algorithm $mathcal{K}$ that produces a model $theta$ from a sample $X=(x_1,...,x_n)$ is $epsilon$-differentially private iff
$$
forall theta, quad mathbb{P}[mathcal{K}(X)=theta | X=X_0] leq e^epsilon mathbb{P}[mathcal{K} (X)=theta|X=X_1] tag{1}
$$
where $X_0$ and $X_1$ are any Hamming-1 neighbors (differ by a single entry)
Using Bayes rule, we can rewrite (1) as a bounded Bayes ratio
$$
frac{mathbb{P}[X=X_0|theta]}{mathbb{P}[X=X_1|theta]} leq e^epsilon frac{mathbb{P}[X=X_0]}{mathbb{P}[X=X_1]} tag{2}
$$
In order to relate this to the Bayes (irreducible) error of an adversary trying to estimate $X$ from $theta$, one can leverage the min-entropy
$$
H_infty(X) = -log max_x mathbb{P}[X=x] \ text{and} \
H_infty(X|theta) = -log mathbb{E}_theta max_x mathbb{P}[X=x|theta]
$$
First part of the question : Can we translate (2) into a bound involving $H_infty(X|theta)$ and $H_infty(X)$ ?
Second part of the question : Since $H_infty(X|theta)$ takes the expectation over $theta$, it looks like a weaker guarantee than (1), (bounded leakage only holding in expectation, with some $theta$ leaking more than the bound). Is there a well-accepted entropy definition that keeps the same expressivity as (1), e.g.,
$$
H_text{bla}(X|theta) = -log max_{theta,x} mathbb{P}[X=x|theta]
$$
or something like that ?
thanks !
entropy ratio log-likelihood
entropy ratio log-likelihood
asked Jan 16 at 11:19
Jerome FJerome F
254
asked Jan 16 at 11:19
Jerome FJerome F
254
asked Jan 16 at 11:19
Jerome FJerome F
254
asked Jan 16 at 11:19
Jerome FJerome F
254
asked Jan 16 at 11:19
Jerome FJerome F
254
254
add a comment |
add a comment |
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