# Appendix B. Rényi Differential Privacy

Here, we present a series of theorems demonstrating the relationship between RDP and $(\u03f5,\delta )$ differential privacy.

# Theorem: RDP Is Immune to Postprocessing

For a mechanism $f$ that is $(\alpha ,\u03f5)$-RDP, and a randomized mapping $g$, the composition $g\left(f\right(x\left)\right)$ is also $(\alpha ,\u03f5)$-RDP.

## Proof

See Mironov^{1}
for a proof of this claim, building on van Ervan and Harremoës.^{2}

Now, let’s demonstrate the equivalence between RDP and $(\u03f5,\delta )$-DP. First, we will need some mathematical preliminaries.

# Theorem: Young’s Inequality

If $a,b\ge 0$ and $\frac{1}{p}+\frac{1}{q}=1$ with $p,q>1$, then $ab\le \frac{{a}^{p}}{p}+\frac{{b}^{q}}{q}$

## Proof via Calculus

Define $f\left(x\right)=\frac{{x}^{p}}{p}+\frac{1}{q}-x$ and take the derivative to determine the minimum of the function.

The derivative is ${f}^{\text{'}}\left(x\right)={x}^{p-1}-1$, and setting this equal to 0:

We know this is a minimum by the second derivative test: ${f}^{\text{'}\text{'}}\left(x\right)=(p-1){x}^{p-2}\ge 0$.

So we know that the minimum of $f\left(x\right)$ occurs at $x=1$ and that this minimum is $f\left(1\right)=0$.

This means that:

Without loss of generality, assume $a>b$. Then ${a}^{p}{b}^{-q}\ge 1$. This further implies that:

and we know that $\frac{1}{p}+\frac{1}{q}=1$, so:

Multiplying both sides ...

Get *Hands-On Differential Privacy* now with the O’Reilly learning platform.

O’Reilly members experience books, live events, courses curated by job role, and more from O’Reilly and nearly 200 top publishers.