In probability theory, a subgaussian distribution, the distribution of a subgaussian random variable, is a probability distribution with strong tail decay. More specifically, the tails of a subgaussian distribution are dominated by (i.e. decay at least as fast as) the tails of a Gaussian. This property gives subgaussian distributions their name.
Often in analysis, we divide an object (such as a random variable) into two parts, a central bulk and a distant tail, then analyze each separately. In probability, this division usually goes like "Everything interesting happens near the center. The tail event is so rare, we may safely ignore that." Subgaussian distributions are worthy of study, because the gaussian distribution is well-understood, and so we can give sharp bounds on the rarity of the tail event. Similarly, the subexponential distributions are also worthy of study.
Formally, the probability distribution of a random variable is called subgaussian if there is a positive constantC such that for every ,
.
There are many equivalent definitions. For example, a random variable is sub-Gaussian iff its distribution function is bounded from above (up to a constant) by the distribution function of a Gaussian:
where is constant and is a mean zero Gaussian random variable.[1]: Theorem 2.6
Definitions
Subgaussian norm
The subgaussian norm of , denoted as , isIn other words, it is the Orlicz norm of generated by the Orlicz function By condition below, subgaussian random variables can be characterized as those random variables with finite subgaussian norm.
Variance proxy
If there exists some such that for all , then is called a variance proxy, and the smallest such is called the optimal variance proxy and denoted by .
Since when is Gaussian, we then have , as it should.
Equivalent definitions
Let be a random variable. The following conditions are equivalent: (Proposition 2.5.2 [2])
Tail probability bound: for all , where is a positive constant;
Furthermore, the constant is the same in the definitions (1) to (5), up to an absolute constant. So for example, given a random variable satisfying (1) and (2), the minimal constants in the two definitions satisfy , where are constants independent of the random variable.
Proof of equivalence
As an example, the first four definitions are equivalent by the proof below.
From the proof, we can extract a cycle of three inequalities:
If , then for all .
If for all , then .
If , then .
In particular, the constant provided by the definitions are the same up to a constant factor, so we can say that the definitions are equivalent up to a constant independent of .
Similarly, because up to a positive multiplicative constant, for all , the definitions (3) and (4) are also equivalent up to a constant.
Basic properties
Proposition.
If is subgaussian, and , then and .
If are subgaussian, then .
Proposition. (Chernoff bound) If is subgaussian, then for all .
Definition. means that , where the positive constant is independent of and .
Proposition. If is subgaussian, then .
Proof. By triangle inequality, . Now we have . By the equivalence of definitions (2) and (4) of subgaussianity, given above, we have .
Proposition. If are subgaussian and independent, then .
Proof. If independent, then use that the cumulant of independent random variables is additive. That is, .
If not independent, then by Hölder's inequality, for any we haveSolving the optimization problem , we obtain the result.
Corollary. Linear sums of subgaussian random variables are subgaussian.
Strictly subgaussian
Expanding the cumulant generating function:we find that . At the edge of possibility, we define that a random variable satisfying is called strictly subgaussian.
Properties
Theorem.[5] Let be a subgaussian random variable with mean zero. If all zeros of its characteristic function are real, then is strictly subgaussian.
Corollary. If are independent and strictly subgaussian, then any linear sum of them is strictly subgaussian.
Examples
By calculating the characteristic functions, we can show that some distributions are strictly subgaussian: symmetric uniform distribution, symmetric Bernoulli distribution.
Since a symmetric uniform distribution is strictly subgaussian, its convolution with itself is strictly subgaussian. That is, the symmetric triangular distribution is strictly subgaussian.
Since the symmetric Bernoulli distribution is strictly subgaussian, any symmetric Binomial distribution is strictly subgaussian.
Examples
strictly subgaussian?
gaussian distribution
Yes
mean-zero Bernoulli distribution
solution to
Iff
symmetric Bernoulli distribution
Yes
uniform distribution
solution to , approximately 0.7727
Yes
arbitrary distribution on interval
The optimal variance proxy is known for many standard probability distributions, including the beta, Bernoulli, Dirichlet[6], Kumaraswamy, triangular[7], truncated Gaussian, and truncated exponential.[8]
Bernoulli distribution
Let be two positive numbers. Let be a centered Bernoulli distribution , so that it has mean zero, then .[5] Its subgaussian norm is where is the unique positive solution to .
Let be a random variable with symmetric Bernoulli distribution (or Rademacher distribution). That is, takes values and with probabilities each. Since , it follows thatand hence is a subgaussian random variable.
Bounded distributions
Bounded distributions have no tail at all, so clearly they are subgaussian.
Since the sum of subgaussian random variables is still subgaussian, the convolution of subgaussian distributions is still subgaussian. In particular, any convolution of the normal distribution with any bounded distribution is subgaussian.
Mixtures
Given subgaussian distributions , we can construct an additive mixture as follows: first randomly pick a number , then pick .
Since we have , and so the mixture is subgaussian.
More generally, the mixture of infinitely many subgaussian distributions is also subgaussian, if the subgaussian norm has a finite supremum: .
Subgaussian random vectors
So far, we have discussed subgaussianity for real-valued random variables. We can also define subgaussianity for random vectors. The purpose of subgaussianity is to make the tails decay fast, so we generalize accordingly: a subgaussian random vector is a random vector where the tail decays fast.
Let be a random vector taking values in .
Define.
, where is the unit sphere in .
is subgaussian iff .
Theorem. (Theorem 3.4.6 [2]) For any positive integer , the uniformly distributed random vector is subgaussian, with .
This is not so surprising, because as , the projection of to the first coordinate converges in distribution to the standard normal distribution.
Maximum inequalities
Proposition. If are mean-zero subgaussians, with , then for any , we have with probability .
Proof. By the Chernoff bound, . Now apply the union bound.
Proposition. (Exercise 2.5.10 [2]) If are subgaussians, with , then Further, the bound is sharp, since when are IID samples of we have .[9]
Theorem. (over a finite set) If are subgaussian, with , thenTheorem. (over a convex polytope) Fix a finite set of vectors . If is a random vector, such that each , then the above 4 inequalities hold, with replacing .
Here, is the convex polytope spanned by the vectors .
Theorem. (over a ball) If is a random vector in , such that for all on the unit sphere , then For any , with probability at least ,
Inequalities
Theorem. (Theorem 2.6.1 [2]) There exists a positive constant such that given any number of independent mean-zero subgaussian random variables , Theorem. (Hoeffding's inequality) (Theorem 2.6.3 [2]) There exists a positive constant such that given any number of independent mean-zero subgaussian random variables ,Theorem. (Bernstein's inequality) (Theorem 2.8.1 [2]) There exists a positive constant such that given any number of independent mean-zero subexponential random variables ,Theorem. (Khinchine inequality) (Exercise 2.6.5 [2]) There exists a positive constant such that given any number of independent mean-zero variance-one subgaussian random variables , any , and any ,
Hanson-Wright inequality
The Hanson-Wright inequality states that if a random vector is subgaussian in a certain sense, then any quadratic form of this vector, , is also subgaussian/subexponential. Further, the upper bound on the tail of , is uniform.
A weak version of the following theorem was proved in (Hanson, Wright, 1971).[11] There are many extensions and variants. Much like the central limit theorem, the Hanson-Wright inequality is more a cluster of theorems with the same purpose, than a single theorem. The purpose is to take a subgaussian vector and uniformly bound its quadratic forms.
Theorem.[12][13] There exists a constant , such that:
Let be a positive integer. Let be independent random variables, such that each satisfies . Combine them into a random vector . For any matrix , we havewhere , and is the Frobenius norm of the matrix, and is the operator norm of the matrix.
In words, the quadratic form has its tail uniformly bounded by an exponential, or a gaussian, whichever is larger.
In the statement of the theorem, the constant is an "absolute constant", meaning that it has no dependence on . It is a mathematical constant much like pi and e.
Consequences
Theorem (subgaussian concentration).[12] There exists a constant , such that:
Let be positive integers. Let be independent random variables, such that each satisfies . Combine them into a random vector . For any matrix , we haveIn words, the random vector is concentrated on a spherical shell of radius , such that is subgaussian, with subgaussian norm .
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