# Differential privacy of “randomized responses”

We define randomized responses as follows:

In a question that can be responded with a "Yes" or "No", a respondent is asked to flip a fair coin, in secret, and answer the truth if it comes up tails. Otherwise he/she flips another coin in secret, and answers "Yes" if comes up tails, "No" otherwise.

According to the introduction this paper [1], the $\epsilon$-differential privacy of surveys relying on randomized responses is: $$\ln(0.75/(1-0.75)) = \ln(3)$$ I am wondering how was this calculated. Note that the proofs contained in paper are related to the RAPPOR protocol.

[1]Úlfar Erlingsson, Vasyl Pihur, Aleksandra Korolova,"RAPPOR: Randomized Aggregatable Privacy-Preserving Ordinal Response" https://research.google.com/pubs/archive/42852.pdf

• There is a full proof in the paper, so you need to be more specific. – kodlu Feb 22 '18 at 3:50
• The proofs contained in paper are related to the RAPPOR protocol. What I am asking is how the differential privacy of the "randomized responses" was calculated. I have edited my question accordingly. – Nikos Fotiou Feb 22 '18 at 11:09

Fix a respondent. A case analysis shows that $$Pr[Response = Yes|Truth = Yes] = 3/4$$ Specifically, when the truth is “Yes” the outcome will be “Yes” if the first coin comes up tails (probability 1/2) or the first and second come up heads (probability 1/4), while $$Pr[Response = Yes|Truth = No] = 1/4$$ (first comes up heads and second comes up tails; probability 1/4). Applying similar reasoning to the case of a “No” answer, we obtain: $$\frac{Pr[Response = Yes|Truth = Yes]}{Pr[Response = Yes|Truth = No]}= \frac{3/4}{1/4}= \frac{Pr[Response = No|Truth = No]}{ Pr[Response = No|Truth = Yes]} = 3$$