Normal approximation to Laplace distribution?

I heard the phrase "normal approximation to the Laplace distribution" recently and did a double take. The normal distribution does not approximate the Laplace!

Normal and Laplace distributions

A normal distribution has the familiar bell curve shape. A Laplace distribution, also known as a double exponential distribution, it pointed in the middle, like a pole holding up a circus tent.

A normal distribution has very thin tails, i.e. probability density drops very rapidly as you move further from the middle, like exp(-x^2). The Laplace distribution has moderate tails [1], going to zero like exp(-|x|).

So normal and Laplace distributions are qualitatively very different, both in the center and in the tails. So why would you want to replace one by the other?

Statistics meets differential privacy

The normal distribution is convenient to use in mathematical statistics. Whether it is realistic in application depends on context, but it's convenient and conventional. The Laplace distribution is convenient and conventional in differential privacy. There's no need to ask whether it is realistic because Laplace noise is added deliberately; the distribution assumption is exactly correct by construction. (See this post for details.)

When mathematical statistics and differential privacy combine, it could be convenient to "approximate" a Laplace distribution by a normal distribution [2].

Solving for parameters

So if you wanted to replace a Laplace distribution with a normal distribution, which one would you choose? Both distributions are symmetric about their means, so it's natural to pick the means to be the same. So without loss of generality, we'll assume both distribution have mean 0. The question then becomes how to choose the scale parameters.

You could just set the two scale parameters to be the same, but that's similar to the Greek letter fallacy, assuming two parameters have the same meaning just because they have the same symbol. Because the two distributions have different tail weights, their scale parameters serve different functions.

One way to replace a Laplace distribution with a normal would be to pick the scale parameter of the normal so that both two quantiles match. For example, you might want both distributions to have have 95% of their probability mass in the same interval.

I've written before about how to solve for scale parameters given two quantiles. We find two quantiles of the Laplace distribution, then use the method in that post to find the corresponding normal distribution scale (standard deviation).

The Laplace distribution with scale s has density

f(x) = exp(-|x|/s)/2s.

If we want to solve for the quantile x such that Prob(X > x) = p, we have

x = -s log(2 - 2p).

Using the formula derived in the previously mentioned post,

If = 2x / I^-1(x)

where I is the cumulative distribution function of the standard normal.

Related posts

• Adding Gaussian or Laplacian noise for privacy
• Generating Laplace random samples
• Data privacy consulting

[1] The normal distribution is the canonical example of a thin-tailed distribution, while exponential tails are conventionally the boundary between thick and thin. "Thick tailed" and "thin tailed" are often taken to mean thicker than exponential and thinner that exponential respectively.

[2] You could use a Gaussian mechanism rather than a Laplace mechanism for similar reasons, but this makes the differential privacy theory more complicated. Rather than working with I-differential privacy you have to work with (I, I)-differential privacy. The latter is messier and harder to interpret.