Sub-Gaussian distribution

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 $$X $$ is called subgaussian if there is a positive constant C such that for every $$t \geq 0$$,



\operatorname{P}(|X| \geq t) \leq 2 \exp{(-t^2/C^2)} $ . There are many equivalent definitions. For example, a random variable $$X$$ is sub-Gaussian iff its distribution function is upper bounded (up to a constant) by the distribution function of a Gaussian:


 * $$P(|X| \geq t) \leq cP(|Z| \geq t) \quad \forall t > 0$$

where $$c\ge 0$$ is constant and $$Z$$ is a mean zero Gaussian random variable.

Subgaussian norm
The subgaussian norm of $$X $$, denoted as $$\Vert X \Vert_{\psi_2} $$, is$$\Vert X \Vert_{\psi_2} = \inf\left\{ c>0 : \operatorname{E}\left[\exp{\left(\frac{X^2}{c^2}\right)}\right] \leq 2 \right\}.$$In other words, it is the Orlicz norm of $$X $$ generated by the Orlicz function $$\Phi(u)=e^{u^2}-1. $$ By condition $$(2)$$ below, subgaussian random variables can be characterized as those random variables with finite subgaussian norm.

Variance proxy
If there exists some $$s^2$$ such that $$\operatorname{E} [e^{(X-\operatorname{E}[X])t}] \leq e^{\frac{s^2t^2}{2}}$$ for all $$t$$, then $$s^2$$ is called a variance proxy, and the smallest such $$s^2$$ is called the optimal variance proxy and denoted by $$\Vert X\Vert_{\mathrm{vp}}^2$$.

Since $$\operatorname{E} [e^{(X-\operatorname{E}[X])t}] = e^{\frac{\sigma^2 t^2}{2}}$$ when $$X \sim \mathcal{N}(\mu, \sigma^2)$$ is Gaussian, we then have $$\|X\|^2_{vp} = \sigma^2$$, as it should.

Equivalent definitions
Let $$X $$ be a random variable. The following conditions are equivalent: (Proposition 2.5.2 )

\operatorname{P}(|X| \geq t) \leq 2 \exp{(-t^2/K_1^2)} $$ for all $$t \geq 0$$, where $$K_1$$ is a positive constant; Furthermore, the constant $$K$$ is the same in the definitions (1) to (5), up to an absolute constant. So for example, given a random variable satisfying (1) and (2), the minimal constants $$K_1, K_2$$ in the two definitions satisfy $$K_1 \leq cK_2, K_2 \leq c' K_1$$, where $$c, c'$$ are constants independent of the random variable.
 * 1) Tail probability bound: $$
 * 1) Finite subgaussian norm: $$\Vert X \Vert_{\psi_2} = K_2 < \infty$$.
 * 2) Moment: $$\operatorname{E} |X|^p \leq 2K_3^p \Gamma\left(\frac{p}{2}+1\right)$$ for all $$p \geq 1$$, where $$K_3$$ is a positive constant and $$\Gamma$$ is the Gamma function.
 * 3) Moment: $$\operatorname{E}|X|^p\leq K^p p^{p/2}$$ for all $$p \geq 1$$,
 * 4) Moment-generating function (of $$X $$), or variance proxy : $$\operatorname{E} [e^{(X-\operatorname{E}[X])t}] \leq e^{\frac{K^2t^2}{2}}$$ for all $$t$$, where $$K$$ is a positive constant.
 * 5) Moment-generating function (of $$X^2 $$): for some $$K > 0$$, $$\operatorname{E}[e^{X^2t^2}] \leq e^{K^2t^2}$$ for all $$t \in [-1/K, +1/K]$$.
 * 6)  Union bound: for some c > 0, $$ \ \operatorname{E}[\max\{|X_1 - \operatorname{E}[X]|,\ldots,|X_n - \operatorname{E}[X]|\}] \leq c \sqrt{\log n}$$ for all n > c, where $$X_1, \ldots, X_n$$ are i.i.d copies of X.
 * 7)  Subexponential: $$X^2$$ has a subexponential distribution.

Proof of equivalence
As an example, the first four definitions are equivalent by the proof below.

Proof. $$(1)\implies(3)$$ By the layer cake representation,$$\begin{align} \operatorname{E} |X|^p &= \int_0^\infty \operatorname{P}(|X|^p \geq t) dt\\ &= \int_0^\infty pt^{p-1}\operatorname{P}(|X| \geq t) dt\\ &\leq 2\int_0^\infty pt^{p-1}\exp\left(-\frac{t^2}{K_1^2}\right) dt\\ \end{align}$$

After a change of variables $$u=t^2/K_1^2$$, we find that$$\begin{align} \operatorname{E} |X|^p &\leq 2K_1^p \frac{p}{2}\int_0^\infty u^{\frac{p}{2}-1}e^{-u} du\\ &= 2K_1^p \frac{p}{2}\Gamma\left(\frac{p}{2}\right)\\ &= 2K_1^p \Gamma\left(\frac{p}{2}+1\right). \end{align}$$$$(3)\implies(2)$$ By the Taylor series $e^x = 1 + \sum_{p=1}^\infty \frac{x^p}{p!},$ $$\begin{align} \operatorname{E}[\exp{(\lambda X^2)}] &= 1 + \sum_{p=1}^\infty \frac{\lambda^p \operatorname{E}{[X^{2p}]}}{p!}\\ &\leq 1 + \sum_{p=1}^\infty \frac{2\lambda^p K_3^{2p} \Gamma\left(p+1\right)}{p!}\\ &= 1 + 2 \sum_{p=1}^\infty \lambda^p K_3^{2p}\\ &= 2 \sum_{p=0}^\infty \lambda^p K_3^{2p}-1\\ &= \frac{2}{1-\lambda K_3^2}-1 \quad\text{for }\lambda K_3^2 <1, \end{align}$$which is less than or equal to $$2$$ for $$\lambda \leq \frac{1}{3K_3^2}$$. Let $$K_2 \geq 3^{\frac{1}{2}}K_3$$, then $\operatorname{E}[\exp{(X^2/K_2^2)}] \leq 2.$

$$(2)\implies(1)$$ By Markov's inequality,$$\operatorname{P}(|X|\geq t) = \operatorname{P}\left( \exp\left(\frac{X^2}{K_2^2}\right) \geq \exp\left(\frac{t^2}{K_2^2}\right) \right) \leq \frac{\operatorname{E}[\exp{(X^2/K_2^2)}]}{\exp\left(\frac{t^2}{K_2^2}\right)} \leq 2 \exp\left(-\frac{t^2}{K_2^2}\right). $$$$(3) \iff (4)$$ by asymptotic formula for gamma function: $$\Gamma(p/2 + 1 ) \sim \sqrt{\pi p} \left(\frac{p}{2e} \right)^{p/2}$$.

From the proof, we can extract a cycle of three inequalities:

\operatorname{P}(|X| \geq t) \leq 2 \exp{(-t^2/K^2)} $$, then $$\operatorname{E} |X|^p \leq 2K^p \Gamma\left(\frac{p}{2}+1\right)$$ for all $$p \geq 1 $$. \operatorname{P}(|X| \geq t) \leq 2 \exp{(-t^2/K^2)} $$.
 * If $$
 * If $$\operatorname{E} |X|^p \leq 2K^p \Gamma\left(\frac{p}{2}+1\right)$$ for all $$p \geq 1 $$, then $$\|X \|_{\psi_2} \leq 3^{\frac{1}{2}}K $$.
 * If $$\|X \|_{\psi_2} \leq K $$, then $$

In particular, the constant $$K $$ 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 $$X$$.

Similarly, because up to a positive multiplicative constant, $$\Gamma(p/2 + 1) = p^{p/2} \times ((2e)^{-1/2}p^{1/2p})^p$$ for all $$p \geq 1$$, the definitions (3) and (4) are also equivalent up to a constant.

Basic properties
Proposition. Proposition. (Chernoff bound) If $$X$$ is subgaussian, then $$Pr(X \geq t) \leq e^{-\frac{t^2}{2\|X\|_{vp}^2}}$$ for all $$t \geq 0$$.
 * If $$X$$ is subgaussian, and $$k > 0$$, then $$\|kX\|_{\psi_2} = k \|X\|_{\psi_2}$$ and $$\|kX\|_{vp} = k \|X\|_{vp}$$.
 * If $$X, Y$$ are subgaussian, then $$\|X+Y\|_{vp}^2 \leq (\|X\|_{vp} + \|Y\|_{vp})^2$$.

Definition. $$X \lesssim X'$$ means that $$X \leq CX'$$, where the positive constant $$C$$ is independent of $$X$$ and $$X'$$.

Proposition. If $$X$$ is subgaussian, then $$\|X - E[X]\|_{\psi_2} \lesssim \|X\|_{\psi_2}$$.

Proof. By triangle inequality, $$\|X - E[X]\|_{\psi_2} \leq \|X\|_{\psi_2} + \|E[X]\|_{\psi_2}$$. Now we have $$\|E[X]\|_{\psi_2} = \sqrt{\ln 2} |E[X]| \leq \sqrt{\ln 2} E[|X|] \sim E[|X|]$$. By the equivalence of definitions (2) and (4) of subgaussianity, given above, we have $$E[|X|] \lesssim \|X\|_{\psi_2}$$.

Proposition. If $$X, Y$$ are subgaussian and independent, then $$\|X+Y\|_{vp}^2 \leq \|X\|_{vp}^2 + \|Y\|_{vp}^2$$.

Proof. If independent, then use that the cumulant of independent random variables is additive. That is, $$\ln \operatorname{E}[e^{t(X+Y)}] = \ln \operatorname{E}[e^{tX}] + \ln \operatorname{E}[e^{tY}]$$.

If not independent, then by Hölder's inequality, for any $$1/p + 1/q = 1$$ we have$$E[e^{t(X+Y)}] = \|e^{t(X+Y)}\|_1 \leq e^{\frac 12 t^2 (p \|X\|_{vp}^2 + q \|Y\|_{vp}^2)}$$Solving the optimization problem $$\begin{cases} \min p \|X\|_{vp}^2 + q \|Y\|_{vp}^2 \\ 1/p + 1/q = 1 \end{cases}$$, we obtain the result.

Corollary. Linear sums of subgaussian random variables are subgaussian.

Strictly subgaussian
Expanding the cumulant generating function:$$\frac 12 s^2 t^2 \geq \ln \operatorname{E}[e^{tX}] = \frac 12 \mathrm{Var}[X] t^2 + \kappa_3 t^3 + \cdots$$we find that $$\mathrm{Var}[X] \leq \|X\|_{\mathrm{vp}}^2$$. At the edge of possibility, we define that a random variable $$X$$ satisfying $$\mathrm{Var}[X]=\|X\|_{\mathrm{vp}}^2$$ is called strictly subgaussian.

Properties
Theorem. Let $$X$$ be a subgaussian random variable with mean zero. If all zeros of its characteristic function are real, then $$X$$ is strictly subgaussian.

Corollary. If $$X_1, \dots, X_n$$ 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
The optimal variance proxy $$\Vert X\Vert_{\mathrm{vp}}^2$$ is known for many standard probability distributions, including the beta, Bernoulli, Dirichlet, Kumaraswamy, triangular, truncated Gaussian, and truncated exponential.

Bernoulli distribution
Let $$p + q = 1$$ be two positive numbers. Let $$X $$ be a centered Bernoulli distribution $$p\delta_q + q \delta_{-p}$$, so that it has mean zero, then $$\Vert X\Vert_{\mathrm{vp}}^2 = \frac{p-q}{2(\log p-\log q)}$$. Its subgaussian norm is $$t$$ where $$t$$ is the unique positive solution to $$pe^{(q/t)^2} + qe^{(p/t)^2} = 2$$.

Let $$X $$ be a random variable with symmetric Bernoulli distribution (or Rademacher distribution). That is, $$X $$ takes values $$-1 $$ and $$1 $$ with probabilities $$1/2 $$ each. Since $$X^2=1 $$, it follows that$$\Vert X \Vert_{\psi_2} = \inf\left\{ c>0 : \operatorname{E}\left[\exp{\left(\frac{X^2}{c^2}\right)}\right] \leq 2 \right\} = \inf\left\{ c>0 : \exp{\left(\frac{1}{c^2}\right)} \leq 2 \right\}=\frac{1}{\sqrt{\ln 2}}, $$and hence $$X $$ is a subgaussian random variable.

Bounded distributions
Bounded distributions have no tail at all, so clearly they are subgaussian.

If $$X$$ is bounded within the interval $$[a, b]$$, Hoeffding's lemma states that $$\Vert X\Vert_{\mathrm{vp}}^2 \leq \left(\frac{b-a}{2}\right)^2 $$. Hoeffding's inequality is the Chernoff bound obtained using this fact.

Convolutions
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 $$X_1, X_2, \dots, X_n$$, we can construct an additive mixture $$X$$ as follows: first randomly pick a number $$i \in \{1, 2, \dots, n\}$$, then pick $$X_i$$.

Since $$\operatorname{E}\left[\exp{\left(\frac{X^2}{c^2}\right)}\right] = \sum_i p_i \operatorname{E}\left[\exp{\left(\frac{X_i^2}{c^2}\right)}\right] $$we have $$\|X\|_{\psi_2} \leq \max_i \|X_i\|_{\psi_2}$$, and so the mixture is subgaussian.

In particular, any gaussian mixture is subgaussian.

More generally, the mixture of infinitely many subgaussian distributions is also subgaussian, if the subgaussian norm has a finite supremum: $$\|X\|_{\psi_2} \leq \sup_i \|X_i\|_{\psi_2}$$.

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 $$X $$ be a random vector taking values in $$\R^n$$.

Define.


 * $$\|X\|_{\psi_2} := \sup_{v \in S^{n-1}}\|v^T X\|_{\psi_2}$$, where $$S^{n-1}$$ is the unit sphere in $$\R^n$$.
 * $$X $$ is subgaussian iff $$\|X\|_{\psi_2} < \infty$$.

Theorem. (Theorem 3.4.6 ) For any positive integer $$n$$, the uniformly distributed random vector $$X \sim U(\sqrt{n} S^{n-1}) $$ is subgaussian, with $$\|X\|_{\psi_2} \lesssim{} 1$$.

This is not so surprising, because as $$n \to \infty$$, the projection of $$ U(\sqrt{n} S^{n-1}) $$ to the first coordinate converges in distribution to the standard normal distribution.

Maximum inequalities
Proposition. If $$X_1, \dots, X_n$$ are mean-zero subgaussians, with $$\|X_i \|_{vp}^2 \leq \sigma^2$$, then for any $$\delta > 0$$, we have $$\max(X_1, \dots, X_n) \leq \sigma\sqrt{2 \ln \frac{n}{\delta}}$$ with probability $$\geq 1-\delta$$.

Proof. By the Chernoff bound, $$Pr(X_i \geq \sigma \sqrt{2 \ln(n/\delta)}) \leq \delta/n$$. Now apply the union bound.

Proposition. (Exercise 2.5.10 ) If $$X_1, X_2, \dots$$ are subgaussians, with $$\|X_i \|_{\psi_2} \leq K$$, then $$E\left[\sup_n \frac{|X_n|}{\sqrt{1+\ln n}}\right] \lesssim K, \quad E\left[\max_{1 \leq n \leq N} |X_n|\right] \lesssim K \sqrt{\ln N}$$Further, the bound is sharp, since when $$X_1, X_2, \dots$$ are IID samples of $$\mathcal N(0, 1)$$ we have $$E\left[\max_{1 \leq n \leq N} |X_n|\right] \gtrsim \sqrt{\ln N}$$.

Theorem. (over a finite set) If $$X_1, \dots, X_n$$ are subgaussian, with $$\|X_i \|_{vp}^2 \leq \sigma^2$$, then$$\begin{aligned} E[\max_i (X_i - E[X_i])] \leq \sigma\sqrt{ 2\ln n}, &\quad P(\max_i X_i > t) \leq n e^{-\frac{t^2}{2\sigma^2}}, \\ E[\max_i |X_i - E[X_i]|] \leq \sigma\sqrt{ 2\ln (2n)}, &\quad P(\max_i |X_i| > t) \leq 2 n e^{-\frac{t^2}{2\sigma^2}} \end{aligned}$$Theorem. (over a convex polytope) Fix a finite set of vectors $$v_1, \dots, v_n$$. If $$X$$ is a random vector, such that each $$\| v_i^T X \|_{vp}^2 \leq \sigma^2$$, then the above 4 inequalities hold, with $$\max_{v \in \mathrm{conv}(v_1, \dots, v_n)}v^T X$$ replacing $$\max_i X_i$$.

Here, $$\mathrm{conv}(v_1, \dots, v_n)$$ is the convex polytope spanned by the vectors $$v_1, \dots, v_n$$.

Theorem. (over a ball) If $$X$$ is a random vector in $$\R^d$$, such that $$\|v^T X\|_{vp}^2 \leq \sigma^2$$ for all $$v$$ on the unit sphere $$S$$, then $$E[\max_{v \in S} v^T X] = E[\max_{v \in S} |v^T X|] \leq 4\sigma \sqrt{d} $$For any $$\delta > 0$$, with probability at least $$1-\delta$$,$$\max_{v \in S} v^T X = \max_{v \in S} | v^T X | \leq 4 \sigma \sqrt{d}+2 \sigma \sqrt{2 \log (1 / \delta)}$$

Inequalities
Theorem. (Theorem 2.6.1 ) There exists a positive constant $$C$$ such that given any number of independent mean-zero subgaussian random variables $$X_1, \dots,X_n$$, $$\left\|\sum_{i=1}^n X_i\right\|_{\psi_2}^2 \leq C \sum_{i=1}^n\left\|X_i\right\|_{\psi_2}^2$$Theorem. (Hoeffding's inequality) (Theorem 2.6.3 ) There exists a positive constant $$c$$ such that given any number of independent mean-zero subgaussian random variables $$X_1, \dots,X_N$$,$$\mathbb{P}\left(\left|\sum_{i=1}^N X_i\right| \geq t\right) \leq 2 \exp \left(-\frac{c t^2}{\sum_{i=1}^N\left\|X_i\right\|_{\psi_2}^2}\right) \quad \forall t > 0$$Theorem. (Bernstein's inequality) (Theorem 2.8.1 ) There exists a positive constant $$c$$ such that given any number of independent mean-zero subexponential random variables $$X_1, \dots,X_N$$,$$\mathbb{P}\left(\left|\sum_{i=1}^N X_i\right| \geq t\right) \leq 2 \exp \left(-c \min \left(\frac{t^2}{\sum_{i=1}^N\left\|X_i\right\|_{\psi_1}^2}, \frac{t}{\max _i\left\|X_i\right\|_{\psi_1}}\right)\right)$$ Theorem. (Khinchine inequality) (Exercise 2.6.5 ) There exists a positive constant $$C$$ such that given any number of independent mean-zero variance-one subgaussian random variables $$X_1, \dots,X_N$$, any $$p \geq 2$$, and any $$a_1, \dots, a_N \in \R$$,$$\left(\sum_{i=1}^N a_i^2\right)^{1 / 2} \leq\left\|\sum_{i=1}^N a_i X_i\right\|_{L^p} \leq C K \sqrt{p}\left(\sum_{i=1}^N a_i^2\right)^{1 / 2}$$

Hanson-Wright inequality
The Hanson-Wright inequality states that if a random vector $$X $$ is subgaussian in a certain sense, then any quadratic form $$A$$ of this vector, $$X^TAX$$, is also subgaussian/subexponential. Further, the upper bound on the tail of $$X^TAX$$, is uniform.

A weak version of the following theorem was proved in (Hanson, Wright, 1971). 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. ''' There exists a constant $$c$$, such that:

Let $$n $$ be a positive integer. Let $$X_1, ..., X_n$$ be independent random variables, such that each satisfies $$E[X_i] = 0$$. Combine them into a random vector $$X = (X_1, \dots, X_n)$$. For any $$n\times n$$ matrix $$A$$, we have$$P(|X^T AX - E[X^TAX]| > t ) \leq \max\left( 2 e^{-\frac{ct^2}{K^4\|A\|_F^2}}, 2 e^{-\frac{ct}{K^2\|A\|}} \right) = 2 \exp \left[-c \min \left(\frac{t^2}{K^4\|A\|_F^2}, \frac{t}{K^2\|A\|}\right)\right] $$where $$K = \max_i \|X_i\|_{\psi_2}$$, and $$\|A\|_F = \sqrt{\sum_{ij} A_{ij}^2}$$ is the Frobenius norm of the matrix, and $$\|A\| = \max_{\|x\|_2=1} \|Ax\|_2$$ is the operator norm of the matrix.

In words, the quadratic form $$X^TAX$$ has its tail uniformly bounded by an exponential, or a gaussian, whichever is larger.

In the statement of the theorem, the constant $$c$$ is an "absolute constant", meaning that it has no dependence on $$n, X_1, \dots, X_n, A$$. It is a mathematical constant much like pi and e.

Consequences
Theorem (subgaussian concentration).  There exists a constant $$c$$, such that:

Let $$n, m $$ be positive integers. Let $$X_1, ..., X_n$$ be independent random variables, such that each satisfies $$E[X_i] = 0, E[X_i^2] = 1$$. Combine them into a random vector $$X = (X_1, \dots, X_n)$$. For any $$m\times n$$ matrix $$A$$, we have$$P( | \| AX\|_2 - \|A\|_F | > t ) \leq 2 e^{-\frac{ct^2}{K^4\|A\|^2}}$$In words, the random vector $$A X$$ is concentrated on a spherical shell of radius $$\|A \|_F$$, such that $$\| AX\|_2 - \|A \|_F$$ is subgaussian, with subgaussian norm $$\leq \sqrt{3/c} \|A\| K^2$$.