Perhaps I am missing something here, but I'm not seeing the value of having an upper tail bound for a Gaussian random variable. In my statistics class, we motivate tail inequalities for situations in which we have very limited information about a distribution yet we wish to understand the limitations of the uncertainty given our limited understanding of the distribution. E.g. Markov's inequality allows us to make statements about the probability of being a certain distance away from the (finite) mean of a non-negative random variable when all we know is the mean (and that the Random Variable is non-negative). Similarly for Chebychev's inequality: when all we know is the (finite) mean and variance of the distribution, we can still make statements about the probability of being a certain distance from the mean.

This is significantly different from when we talk about tail bounds for Gaussian random variables (or sums/averages of i.i.d. Gaussians): if we assume them to be Gaussian, then we know the entire PDF. Our original motivation for deriving a bound is now redundant -- why bound something when we can actually compute the probability directly using the PDF?

Perhaps there is value in the fact that computing such integrals can be difficult, and having a simpler (but less precise) bound allows for fast/simple approximations?


1 Answer 1


Yes, we don't have a clean expression of the CDF for a gaussian, so the tail bounds of CDF are quite useful. Here is one example use:

Let's say that you are trying to compute $\lim_{n \rightarrow \infty}\frac{1}{n} \log P(\bar{S}_n > t)$, where $\bar{S}_n$ is a sample mean of n i.i.d. Gaussians. If you know that the tail bound for a gaussian is approximately $e^{-\frac{t^2}{2 \sigma}}$ up to logarithmic factors, you can compute this value. Whereas, computing this from the integral can only give you numerical approximations.

(P.S. The limit I described above is a canonical one; it is part of large deviations theory which quantifies exactly how rare rare events happen in the limit)

  • $\begingroup$ Perhaps my understanding of numerical methods is lacking here, but when you say "Whereas, computing this from the integral can only give you numerical approximations.", is the accuracy of the approximation from computing the integral not just a function of the time that we give the numerical solver? Is the benefit only for situations when you cannot afford to spend time evaluating the integral, or is it not as simple as that? $\endgroup$
    – Conor Igoe
    Sep 10, 2018 at 21:41
  • $\begingroup$ @ConorIgoe sometimes the symbolic bound is needed as input to the next step of a larger computation, where it will be further manipulated. $\endgroup$
    – japreiss
    Jan 3, 2019 at 0:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .