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

${\textstyle \operatorname {P} (|X|\geq t)\leq 2\exp {(-t^{2}/C^{2})}}$.

There are many equivalent definitions. For example, a random variable ${\displaystyle X}$ is sub-Gaussian iff its distribution function is upper bounded (up to a constant) by the distribution function of a Gaussian:

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

where ${\displaystyle c\geq 0}$ is constant and ${\displaystyle Z}$ is a mean zero Gaussian random variable.^{[1]: Theorem 2.6}

Definitions

Subgaussian norm

The subgaussian norm of ${\displaystyle X}$, denoted as ${\displaystyle \Vert X\Vert _{\psi _{2}}}$, is

In other words, it is the Orlicz norm of ${\displaystyle X}$ generated by the Orlicz function ${\displaystyle \Phi (u)=e^{u^{2}}-1.}$ By condition ${\displaystyle (2)}$ below, subgaussian random variables can be characterized as those random variables with finite subgaussian norm.

Variance proxy

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

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

Equivalent definitions

Let ${\displaystyle X}$ be a random variable. The following conditions are equivalent: (Proposition 2.5.2 ^[2])

1. Tail probability bound: ${\displaystyle \operatorname {P} (|X|\geq t)\leq 2\exp {(-t^{2}/K_{1}^{2})}}$ for all ${\displaystyle t\geq 0}$, where ${\displaystyle K_{1}}$ is a positive constant;
2. Finite subgaussian norm: ${\displaystyle \Vert X\Vert _{\psi _{2}}=K_{2}<\infty }$.
3. Moment: ${\displaystyle \operatorname {E} |X|^{p}\leq 2K_{3}^{p}\Gamma \left({\frac {p}{2}}+1\right)}$ for all ${\displaystyle p\geq 1}$, where ${\displaystyle K_{3}}$ is a positive constant and ${\displaystyle \Gamma }$ is the Gamma function.
4. Moment: ${\displaystyle \operatorname {E} |X|^{p}\leq K^{p}p^{p/2}}$ for all ${\displaystyle p\geq 1}$,
5. Moment-generating function (of ${\displaystyle X}$), or variance proxy^[3][4] : ${\displaystyle \operatorname {E} [e^{(X-\operatorname {E} [X])t}]\leq e^{\frac {K^{2}t^{2}}{2}}}$ for all ${\displaystyle t}$, where ${\displaystyle K}$ is a positive constant.
6. Moment-generating function (of ${\displaystyle X^{2}}$): for some ${\displaystyle K>0}$, ${\displaystyle \operatorname {E} [e^{X^{2}t^{2}}]\leq e^{K^{2}t^{2}}}$ for all ${\displaystyle t\in [-1/K,+1/K]}$.
7. Union bound: for some c > 0, ${\displaystyle \ \operatorname {E} [\max\{|X_{1}-\operatorname {E} [X]|,\ldots ,|X_{n}-\operatorname {E} [X]|\}]\leq c{\sqrt {\log n}}}$ for all n > c, where ${\displaystyle X_{1},\ldots ,X_{n}}$ are i.i.d copies of X.
8. Subexponential: ${\displaystyle X^{2}}$ has a subexponential distribution.

Furthermore, the constant ${\displaystyle K}$ 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 ${\displaystyle K_{1},K_{2}}$ in the two definitions satisfy ${\displaystyle K_{1}\leq cK_{2},K_{2}\leq c'K_{1}}$, where ${\displaystyle c,c'}$ are constants independent of the random variable.

Proof of equivalence

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

Proof. ${\displaystyle (1)\implies (3)}$ By the layer cake representation,

After a change of variables ${\displaystyle u=t^{2}/K_{1}^{2}}$, we find that

${\displaystyle (3)\implies (2)}$ By the Taylor series ${\textstyle e^{x}=1+\sum _{p=1}^{\infty }{\frac {x^{p}}{p!}},}$which is less than or equal to ${\displaystyle 2}$ for ${\displaystyle \lambda \leq {\frac {1}{3K_{3}^{2}}}}$. Let ${\displaystyle K_{2}\geq 3^{\frac {1}{2}}K_{3}}$, then ${\textstyle \operatorname {E} [\exp {(X^{2}/K_{2}^{2})}]\leq 2.}$

${\displaystyle (2)\implies (1)}$ By Markov's inequality,

${\displaystyle (3)\iff (4)}$ by asymptotic formula for gamma function: ${\displaystyle \Gamma (p/2+1)\sim {\sqrt {\pi p}}\left({\frac {p}{2e}}\right)^{p/2}}$.

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

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

In particular, the constant ${\displaystyle K}$ 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 ${\displaystyle X}$.

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

Basic properties

Proposition.

• If ${\displaystyle X}$ is subgaussian, and ${\displaystyle k>0}$, then ${\displaystyle \|kX\|_{\psi _{2}}=k\|X\|_{\psi _{2}}}$ and ${\displaystyle \|kX\|_{vp}=k\|X\|_{vp}}$.
• If ${\displaystyle X,Y}$ are subgaussian, then ${\displaystyle \|X+Y\|_{vp}^{2}\leq (\|X\|_{vp}+\|Y\|_{vp})^{2}}$.

Proposition. (Chernoff bound) If ${\displaystyle X}$ is subgaussian, then ${\displaystyle Pr(X\geq t)\leq e^{-{\frac {t^{2}}{2\|X\|_{vp}^{2}}}}}$ for all ${\displaystyle t\geq 0}$.

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

Proposition. If ${\displaystyle X}$ is subgaussian, then ${\displaystyle \|X-E[X]\|_{\psi _{2}}\lesssim \|X\|_{\psi _{2}}}$.

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

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

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

If not independent, then by Hölder's inequality, for any ${\displaystyle 1/p+1/q=1}$ we have

Solving the optimization problem ${\displaystyle {\begin{cases}\min p\|X\|_{vp}^{2}+q\|Y\|_{vp}^{2}\\1/p+1/q=1\end{cases}}}$, we obtain the result.

Corollary. Linear sums of subgaussian random variables are subgaussian.

Strictly subgaussian

Expanding the cumulant generating function:

we find that ${\displaystyle \mathrm {Var} [X]\leq \|X\|_{\mathrm {vp} }^{2}}$. At the edge of possibility, we define that a random variable ${\displaystyle X}$ satisfying ${\displaystyle \mathrm {Var} [X]=\|X\|_{\mathrm {vp} }^{2}}$ is called strictly subgaussian.

Properties

Theorem.^[5] Let ${\displaystyle X}$ be a subgaussian random variable with mean zero. If all zeros of its characteristic function are real, then ${\displaystyle X}$ is strictly subgaussian.

Corollary. If ${\displaystyle X_{1},\dots ,X_{n}}$ 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

	${\displaystyle \|X\|_{\psi _{2}}}$	${\displaystyle \|X\|_{vp}^{2}}$	strictly subgaussian?
gaussian distribution ${\displaystyle {\mathcal {N}}(0,1)}$	${\displaystyle {\sqrt {8/3}}}$	${\displaystyle 1}$	Yes
mean-zero Bernoulli distribution ${\displaystyle p\delta _{q}+q\delta _{-p}}$	solution to ${\displaystyle pe^{(q/t)^{2}}+qe^{(p/t)^{2}}=2}$	${\displaystyle {\frac {p-q}{2(\log p-\log q)}}}$	Iff ${\displaystyle p=0,1/2,1}$
symmetric Bernoulli distribution ${\displaystyle {\frac {1}{2}}\delta _{1/2}+{\frac {1}{2}}\delta _{-1/2}}$	${\displaystyle {\frac {1}{2{\sqrt {\ln 2}}}}}$	${\displaystyle 1/4}$	Yes
uniform distribution ${\displaystyle U(-1,1)}$	solution to ${\displaystyle \int _{0}^{1}e^{x^{2}/t^{2}}dx=2}$, approximately 0.7727	${\displaystyle 1/3}$	Yes
arbitrary distribution on interval ${\displaystyle [a,b]}$		${\displaystyle \leq \left({\frac {b-a}{2}}\right)^{2}}$

The optimal variance proxy ${\displaystyle \Vert X\Vert _{\mathrm {vp} }^{2}}$ 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 ${\displaystyle p+q=1}$ be two positive numbers. Let ${\displaystyle X}$ be a centered Bernoulli distribution ${\displaystyle p\delta _{q}+q\delta _{-p}}$, so that it has mean zero, then ${\displaystyle \Vert X\Vert _{\mathrm {vp} }^{2}={\frac {p-q}{2(\log p-\log q)}}}$.^[9] Its subgaussian norm is ${\displaystyle t}$ where ${\displaystyle t}$ is the unique positive solution to ${\displaystyle pe^{(q/t)^{2}}+qe^{(p/t)^{2}}=2}$.

Let ${\displaystyle X}$ be a random variable with symmetric Bernoulli distribution (or Rademacher distribution). That is, ${\displaystyle X}$ takes values ${\displaystyle -1}$ and ${\displaystyle 1}$ with probabilities ${\displaystyle 1/2}$ each. Since ${\displaystyle X^{2}=1}$, it follows that

and hence ${\displaystyle X}$ is a subgaussian random variable.

Bounded distributions

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

If ${\displaystyle X}$ is bounded within the interval ${\displaystyle [a,b]}$, then since ${\displaystyle \mathrm {Var} [X]\leq \left({\frac {b-a}{2}}\right)^{2}}$, we have ${\displaystyle \Vert X\Vert _{\mathrm {vp} }^{2}\leq \left({\frac {b-a}{2}}\right)^{2}}$. Now, applying a Chernoff bound, we have Hoeffding's inequality.

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

Since ${\displaystyle \operatorname {E} \left[\exp {\left({\frac {X^{2}}{c^{2}}}\right)}\right]=\sum _{i}p_{i}\operatorname {E} \left[\exp {\left({\frac {X_{i}^{2}}{c^{2}}}\right)}\right]}$we have ${\displaystyle \|X\|_{\psi _{2}}\leq \max _{i}\|X_{i}\|_{\psi _{2}}}$, 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: ${\displaystyle \|X\|_{\psi _{2}}\leq \sup _{i}\|X_{i}\|_{\psi _{2}}}$.

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 ${\displaystyle X}$ be a random vector taking values in ${\displaystyle \mathbb {R} ^{n}}$.

Define.

• ${\displaystyle \|X\|_{\psi _{2}}:=\sup _{v\in S^{n-1}}\|v^{T}X\|_{\psi _{2}}}$, where ${\displaystyle S^{n-1}}$ is the unit sphere in ${\displaystyle \mathbb {R} ^{n}}$.
• ${\displaystyle X}$ is subgaussian iff ${\displaystyle \|X\|_{\psi _{2}}<\infty }$.

Theorem. (Theorem 3.4.6 ^[2]) For any positive integer ${\displaystyle n}$, the uniformly distributed random vector ${\displaystyle X\sim U({\sqrt {n}}S^{n-1})}$ is subgaussian, with ${\displaystyle \|X\|_{\psi _{2}}\lesssim {}1}$.

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

Maximum inequalities

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

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

Proposition. (Exercise 2.5.10 ^[2]) If ${\displaystyle X_{1},X_{2},\dots }$ are subgaussians, with ${\displaystyle \|X_{i}\|_{\psi _{2}}\leq K}$, then

Further, the bound is sharp, since when ${\displaystyle X_{1},X_{2},\dots }$ are IID samples of ${\displaystyle {\mathcal {N}}(0,1)}$ we have ${\displaystyle E\left[\max _{1\leq n\leq N}|X_{n}|\right]\gtrsim {\sqrt {\ln N}}}$.^[10]

^[11]

Theorem. (over a finite set) If ${\displaystyle X_{1},\dots ,X_{n}}$ are subgaussian, with ${\displaystyle \|X_{i}\|_{vp}^{2}\leq \sigma ^{2}}$, then

Theorem. (over a convex polytope) Fix a finite set of vectors ${\displaystyle v_{1},\dots ,v_{n}}$. If ${\displaystyle X}$ is a random vector, such that each ${\displaystyle \|v_{i}^{T}X\|_{vp}^{2}\leq \sigma ^{2}}$, then the above 4 inequalities hold, with ${\displaystyle \max _{v\in \mathrm {conv} (v_{1},\dots ,v_{n})}v^{T}X}$ replacing ${\displaystyle \max _{i}X_{i}}$.

Here, ${\displaystyle \mathrm {conv} (v_{1},\dots ,v_{n})}$ is the convex polytope spanned by the vectors ${\displaystyle v_{1},\dots ,v_{n}}$.

Theorem. (over a ball) If ${\displaystyle X}$ is a random vector in ${\displaystyle \mathbb {R} ^{d}}$, such that ${\displaystyle \|v^{T}X\|_{vp}^{2}\leq \sigma ^{2}}$ for all ${\displaystyle v}$ on the unit sphere ${\displaystyle S}$, then

For any ${\displaystyle \delta >0}$, with probability at least ${\displaystyle 1-\delta }$,

Inequalities

Theorem. (Theorem 2.6.1 ^[2]) There exists a positive constant ${\displaystyle C}$ such that given any number of independent mean-zero subgaussian random variables ${\displaystyle X_{1},\dots ,X_{n}}$,

Theorem. (Hoeffding's inequality) (Theorem 2.6.3 ^[2]) There exists a positive constant ${\displaystyle c}$ such that given any number of independent mean-zero subgaussian random variables ${\displaystyle X_{1},\dots ,X_{N}}$,Theorem. (Bernstein's inequality) (Theorem 2.8.1 ^[2]) There exists a positive constant ${\displaystyle c}$ such that given any number of independent mean-zero subexponential random variables ${\displaystyle X_{1},\dots ,X_{N}}$, Theorem. (Khinchine inequality) (Exercise 2.6.5 ^[2]) There exists a positive constant ${\displaystyle C}$ such that given any number of independent mean-zero variance-one subgaussian random variables ${\displaystyle X_{1},\dots ,X_{N}}$, any ${\displaystyle p\geq 2}$, and any ${\displaystyle a_{1},\dots ,a_{N}\in \mathbb {R} }$,

Hanson-Wright inequality

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

A weak version of the following theorem was proved in (Hanson, Wright, 1971).^[12] 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.^[13][14] There exists a constant ${\displaystyle c}$, such that:

Let ${\displaystyle n}$ be a positive integer. Let ${\displaystyle X_{1},...,X_{n}}$ be independent random variables, such that each satisfies ${\displaystyle E[X_{i}]=0}$. Combine them into a random vector ${\displaystyle X=(X_{1},\dots ,X_{n})}$. For any ${\displaystyle n\times n}$ matrix ${\displaystyle A}$, we have

where ${\displaystyle K=\max _{i}\|X_{i}\|_{\psi _{2}}}$, and ${\displaystyle \|A\|_{F}={\sqrt {\sum _{ij}A_{ij}^{2}}}}$ is the Frobenius norm of the matrix, and ${\displaystyle \|A\|=\max _{\|x\|_{2}=1}\|Ax\|_{2}}$ is the operator norm of the matrix.

In words, the quadratic form ${\displaystyle X^{T}AX}$ has its tail uniformly bounded by an exponential, or a gaussian, whichever is larger.

In the statement of the theorem, the constant ${\displaystyle c}$ is an "absolute constant", meaning that it has no dependence on ${\displaystyle n,X_{1},\dots ,X_{n},A}$. It is a mathematical constant much like pi and e.

Consequences

Theorem (subgaussian concentration).^[13] There exists a constant ${\displaystyle c}$, such that:

Let ${\displaystyle n,m}$ be positive integers. Let ${\displaystyle X_{1},...,X_{n}}$ be independent random variables, such that each satisfies ${\displaystyle E[X_{i}]=0,E[X_{i}^{2}]=1}$. Combine them into a random vector ${\displaystyle X=(X_{1},\dots ,X_{n})}$. For any ${\displaystyle m\times n}$ matrix ${\displaystyle A}$, we have

In words, the random vector ${\displaystyle AX}$ is concentrated on a spherical shell of radius ${\displaystyle \|A\|_{F}}$, such that ${\displaystyle \|AX\|_{2}-\|A\|_{F}}$ is subgaussian, with subgaussian norm ${\displaystyle \leq {\sqrt {3/c}}\|A\|K^{2}}$.

See also

• Platykurtic distribution

Notes

References

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