I am studying differential privacy and I got stuck again in proof of a lemma. Which is:

"$D_{\infty}^\delta(Y||Z) \leq \epsilon$ if and only if there exists a random variable $Y'$ such that $\Delta(Y,Y') \leq \delta$ and $D_\infty(Y||Z) \leq \epsilon $."

I have a problem understanding the reverse proof.


Be $Y, Z$ two random variables.

  1. $\Delta (Y,Z) \overset{def}{=} \underset{S}{max} \ \ \ | Pr[Y\in S]-Pr[Z\in S]|$
  2. $D_{\infty}(Y||Z)=\underset{S\subseteq Supp(Y)}{max}\Big[ln\frac{Pr[Y\in S]}{Pr[Z \in S]}\Big]$, which is the KL-Divergence between two distributions $Y,Z$
  3. $D_{\infty}^\delta(Y||Z)=\underset{S\subseteq Supp(Y):Pr[Y\in S]\geq \delta}{max}\Big[ln\frac{Pr[Y\in S]-\delta}{Pr[Z \in S]}\Big]$


Suppose that $D_{\infty}^\delta(Y||Z) \leq \epsilon$. Sea $S=\{y:Pr[Y=y] > e^\epsilon \cdot Pr[Z=y]\}$. Then

\begin{equation*} \sum_{y \in S}(Pr[Y=y]-e^\epsilon \cdot Pr[Z=y]) = Pr[Y \in S]-e^\epsilon \cdot Pr[Z \in S] \leq \delta \end{equation*}

(I understand until here)

Moreover, if we let $T=\{y:Pr[Y=y] \leq Pr[Z=y]\}$, then :

\begin{equation*} \begin{split} \sum_{y\in T}(Pr[Z=y]-Pr[Y=y]) &= \sum _{y \notin T}(Pr[Y=y]-Pr[Z=Y]) \ \ \ (I-got-stuck-here) \\ & \geq \sum _{y \in S}(Pr[Y=y]-Pr[Z=Y])\\ & \geq \sum _{y \in S}(Pr[Y=y] > e^\epsilon \cdot Pr[Z=y]) \end{split} \end{equation*}

(I don't' understand why: $\sum_{y\in T}(Pr[Z=y]-Pr[Y=y]) = \sum _{y \notin T}(Pr[Y=y]-Pr[Z=Y]$)

Thus we can obtain $Y'$ from $Y$ by lowering the probabilities on $S$ and raising the probabilities on $T$ To satisfy:

  1. For all $y\in S$, $Pr[Y'=y]=e^\epsilon \cdot Pr[Z=y] < Pr[Y=y]]$
  2. For all $y \in T$, $Pr[Y=y]\leq Pr[Y'=y]\leq Pr[Z=y]$
  3. For all $y\notin S \cup T$, $Pr[Y'=y]=Pr[Y=y] \leq e^{\epsilon} \cdot Pr[Z=y]$

Then $D_{\infty}^\delta(Y'||Z) \leq \epsilon$ by inspection

Reference: Dwork, C. & Roth, A. (2014). The Algorithmic Foundations of Differential Privacy. Foundations and Trends in Theoretical Computer Science, page 45.


1 Answer 1


For any set $T$, the reason $$\sum_{y\in T}(Pr[Z=y]-Pr[Y=y]) = \sum _{y \notin T}(Pr[Y=y]-Pr[Z=y])$$ is that both $\Pr[Z=y]$ and $\Pr(Y=y)$ add to 1 over the entire sample, so $Pr[Y=y]-Pr[Z=y]$ adds to zero over the entire sample.

So $$\sum_{y\in T}(Pr[Z=y]-Pr[Y=y]) + \sum _{y \notin T}(Pr[Z=y]-Pr[Y=y])=0$$ giving $$\sum_{y\in T}(Pr[Z=y]-Pr[Y=y]) - \sum _{y \notin T}(Pr[Y=y]-Pr[Z=y])=0$$ and $$\sum_{y\in T}(Pr[Z=y]-Pr[Y=y]) = \sum _{y \notin T}(Pr[Y=y]-Pr[Z=y])$$

Incidentally, $D_\infty(Y||Z)$ as you have defined it is not the KL-divergence. The KL-divergence is the expected value of the log likelihood ratio, not the supremum.

  • $\begingroup$ Thanks for the quick answer, Thomas. It makes sense. Yeah, that is not the KL divergence, but is similar (The Expression). $\endgroup$ Commented Jul 2, 2020 at 5:07

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