# Re: Random delay as a countermeasure to timing attacks

*From*: daw@xxxxxxxxxxxxxxxxxxxxxxxx (David Wagner)*Date*: Mon, 6 Nov 2006 18:55:28 +0000 (UTC)

David Wagner wrote:

One possible hypothesis:

- the signal has a Gaussian distribution,

- the noise (the delay you add, plus any other random noise)

has a Gaussian distribution, and

- all of these contributions are independent.

Then it's easy to see that the S/N ratio goes up linearly

with the standard deviation of the noise, and goes down

proportional to the square root of the number of measurements,

leading to your desired result.

Oops, I think I meant that the signal is 0 or 1

(has a Bernoulli distribution). That's the simplest case,

because then you are just distinguishing between two

distributions: X and 1+X, where X ~ N(\mu,\sigma^2) for

some values of \mu,\sigma. You should be able to compute

the variation distance between these two distributions (as

a function of \sigma) using calculus, and I believe you'll

find that you need \sigma ~ 1 to have some non-negligible

chance of distinguishing. (If \sigma << 1, you're out of luck.)

.

**Follow-Ups**:**Re: Random delay as a countermeasure to timing attacks***From:*Unruh

**Re: Random delay as a countermeasure to timing attacks***From:*François Grieu

**References**:**Random delay as a countermeasure to timing attacks***From:*Francois Grieu

**Re: Random delay as a countermeasure to timing attacks***From:*David Wagner

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