Sub-Gaussian distribution

Short description: Probability distribution whose tail probability is less than some gaussian, in some sense.

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 $t\ge 0$,

$\mathrm{P}(|X|\ge t)\le 2\exp (-t^2/C^2)$.

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

$P(|X|\ge t)\le cP(|Z|\ge t)\mathrm{\forall }t>0$

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

Definitions

Subgaussian norm

The subgaussian norm of $X$, denoted as $\Vert X{\Vert }_{{\psi }_2}$, is$$\Vert X{\Vert }_{{\psi }_2}=\inf \{c>0:\mathrm{E}[\exp \left(\frac{X^2}{c^2}\right)]\le 2\}.$$In other words, it is the Orlicz norm of $X$ generated by the Orlicz function $\mathrm{\Phi }(u)=e^{u^2}-1.$ 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 $\mathrm{E}[e^{(X-\mathrm{E}[X])t}]\le e^{\frac{s^2{t}^2}{2}}$ 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 $\Vert X{\Vert }_{\mathrm{v}\mathrm{p}}^2$.

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

Equivalent definitions

Let $X$ be a random variable with zero mean. Let $K_1,K_2,K_3,\dots$ be positive constants. The following conditions are equivalent: (Proposition 2.5.2 )

1. Tail probability bound: $\mathrm{P}(|X|\ge t)\le 2\exp (-t^2/K_1^2)$ for all $t\ge 0$;
2. Finite subgaussian norm: $\Vert X{\Vert }_{{\psi }_2}=K_2<\mathrm{\infty }$;
3. Moment: $\mathrm{E}|X{|}^p\le 2K_3^p\mathrm{\Gamma }(\frac{p}{2}+1)$ for all $p\ge 1$, where $\mathrm{\Gamma }$ is the Gamma function;
4. Moment: $\mathrm{E}|X{|}^p\le K^p{p}^{p/2}$ for all $p\ge 1$;
5. Moment-generating function (of $X$), or variance proxy : $\mathrm{E}[e^{(X-\mathrm{E}[X])t}]\le e^{\frac{K^2{t}^2}{2}}$ for all $t$;
6. Moment-generating function (of $X^2$): $\mathrm{E}[e^{X^2{t}^2}]\le e^{K^2{t}^2}$ for all $t\in [-1/K,+1/K]$;
7. Union bound: for some c > 0, $\mathrm{E}[\max \{|X_1-\mathrm{E}[X]|,\dots ,|X_n-\mathrm{E}[X]|\}]\le c\sqrt{\log n}$ for all n > c, where $X_1,\dots ,X_n$ are i.i.d copies of X;
8. Subexponential: $X^2$ has a subexponential distribution.

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 $K_1\le c{K}_2,K_2\le c^{\prime }{K}_1$, where $c,c^{\prime }$ are constants independent of the random variable.

Proof of equivalence

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

Proof. $(1)⟹(3)$ By the layer cake representation,$$\begin{array}{rl}\mathrm{E}|X{|}^p& =\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathrm{P}(|X{|}^p\ge t)dt\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}p{t}^{p-1}\mathrm{P}(|X|\ge t)dt\\ & \le 2\underset{0}{\overset{\mathrm{\infty }}{\int }}p{t}^{p-1}\exp (-\frac{t^2}{K_1^2})dt\\ \end{array}$$

After a change of variables $u=t^2/K_1^2$, we find that$$\begin{array}{rl}\mathrm{E}|X{|}^p& \le 2K_1^p\frac{p}{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}{u}^{\frac{p}{2}-1}{e}^{-u}du\\ & =2K_1^p\frac{p}{2}\mathrm{\Gamma }\left(\frac{p}{2}\right)\\ & =2K_1^p\mathrm{\Gamma }(\frac{p}{2}+1).\end{array}$$$(3)⟹(2)$ By the Taylor series $e^x=1+{\sum }_{p=1}^{\mathrm{\infty }}\frac{x^p}{p!},$$$\begin{array}{rl}\mathrm{E}[\exp (\lambda X^2)]& =1+\sum _{p=1}^{\mathrm{\infty }}\frac{{\lambda }^p\mathrm{E}[X^{2p}]}{p!}\\ & \le 1+\sum _{p=1}^{\mathrm{\infty }}\frac{2{\lambda }^p{K}_3^{2p}\mathrm{\Gamma }(p+1)}{p!}\\ & =1+2\sum _{p=1}^{\mathrm{\infty }}{\lambda }^p{K}_3^{2p}\\ & =2\sum _{p=0}^{\mathrm{\infty }}{\lambda }^p{K}_3^{2p}-1\\ & =\frac{2}{1-\lambda K_3^2}-1\text{for }\lambda K_3^2<1,\end{array}$$which is less than or equal to $2$ for $\lambda \le \frac{1}{3K_3^2}$. Let $K_2\ge 3^{\frac{1}{2}}{K}_3$, then $\mathrm{E}[\exp (X^2/K_2^2)]\le 2.$

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

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

• If $\mathrm{P}(|X|\ge t)\le 2\exp (-t^2/K^2)$, then $\mathrm{E}|X{|}^p\le 2K^p\mathrm{\Gamma }(\frac{p}{2}+1)$ for all $p\ge 1$.
• If $\mathrm{E}|X{|}^p\le 2K^p\mathrm{\Gamma }(\frac{p}{2}+1)$ for all $p\ge 1$, then $\Vert X{\Vert }_{{\psi }_2}\le 3^{\frac{1}{2}}K$.
• If $\Vert X{\Vert }_{{\psi }_2}\le K$, then $\mathrm{P}(|X|\ge t)\le 2\exp (-t^2/K^2)$.

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

Basic properties

Basic properties — * If $X$ is subgaussian, and $k>0$, then $\Vert kX{\Vert }_{{\psi }_2}=k\Vert X{\Vert }_{{\psi }_2}$ and $\Vert kX{\Vert }_{vp}=k\Vert X{\Vert }_{vp}$.

• (Triangle inequality) If $X,Y$ are subgaussian, then $$\Vert X+Y{\Vert }_{vp}^2\le (\Vert X{\Vert }_{vp}+\Vert Y{\Vert }_{vp}{)}^2$$

• (Chernoff bound) If $X$ is subgaussian, then $$Pr(X\ge t)\le e^{-\frac{t^2}{2\Vert X{\Vert }_{vp}^2}}$$ for all $t\ge 0$

$X\lesssim X^{\prime }$ means that $X\le C{X}^{\prime }$, where the positive constant $C$ is independent of $X$ and $X^{\prime }$.

Subgaussian deviation bound — If $X$ is subgaussian, then $$\Vert X-E[X]{\Vert }_{{\psi }_2}\lesssim \Vert X{\Vert }_{{\psi }_2}$$

Proof

By triangle inequality, $\Vert X-E[X]{\Vert }_{{\psi }_2}\le \Vert X{\Vert }_{{\psi }_2}+\Vert E[X]{\Vert }_{{\psi }_2}$. Now we have $\Vert E[X]{\Vert }_{{\psi }_2}=\sqrt{\ln 2}|E[X]|\le \sqrt{\ln 2}E[|X|]\sim E[|X|]$. By the equivalence of definitions (2) and (4) of subgaussianity, we have $E[|X|]\lesssim \Vert X{\Vert }_{{\psi }_2}$.

Independent subgaussian sum bound — If $X,Y$ are subgaussian and independent, then $$\Vert X+Y{\Vert }_{vp}^2\le \Vert X{\Vert }_{vp}^2+\Vert Y{\Vert }_{vp}^2$$

Proof

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

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

Corollary — Linear sums of subgaussian random variables are subgaussian.

Partial converse (Matoušek 2008, Lemma 2.4) — If $E[X]=0,E[X^2]=1$, and $-\ln Pr(X\ge t)\ge \frac{1}{2}a{t}^2$ for all $t>0$, then $$\ln E[e^{tX}]\le C_a{t}^2$$ where $C_a>0$ depends on $a$ only.

Proof

Proof

Let $F(x)$ be the CDF of $X$. The proof splits the integral of MGF to two halves, one with $tX>1$ and one with $tX\le 1$, and bound each one respectively.

$$\begin{array}{rl}E[e^{tX}]& =\underset{\mathbb{ℝ}}{\int }{e}^{tx}dF(x)\\ & =\underset{-\mathrm{\infty }}{\overset{1/t}{\int }}{e}^{tx}dF(x)+\underset{1/t}{\overset{+\mathrm{\infty }}{\int }}{e}^{tx}dF(x)\\ \end{array}$$ Since $e^x\le 1+x+x^2$ for $x\le 1$, $$\begin{array}{rl}\underset{-\mathrm{\infty }}{\overset{1/t}{\int }}{e}^{tx}dF(x)& \le \underset{-\mathrm{\infty }}{\overset{1/t}{\int }}(1+tx+t^2{x}^2)dF(x)\\ & \le \underset{\mathbb{ℝ}}{\int }(1+tx+t^2{x}^2)dF(x)\\ & =1+tE[X]+t^{2}E[X^2]\\ & =1+t^2\\ & \le e^{t^2}\end{array}$$ For the second term, upper bound it by a summation: $$\begin{array}{rl}\underset{1/t}{\overset{+\mathrm{\infty }}{\int }}{e}^{tx}dF(x)& \le e^{2}Pr(X\in [1/t,2/t])+e^{3}Pr(X\in [1/t,2/t])+\dots \\ & \le \sum _{k=1}^{\mathrm{\infty }}{e}^{k+1}Pr(X\ge k/t)\\ & \le \sum _{k=1}^{\mathrm{\infty }}{e}^{k(2-\frac{1}{2}ak/t^2)}\end{array}$$ When $t\le \sqrt{a/8}$, for any $k\ge 1$, $2k-\frac{a{k}^2}{2t^2}\le -\frac{ak}{4t^2}$, so

$$\le \frac{1}{e^{\frac{a}{4t^2}}-1}\le \frac{4}{a}{t}^2$$ When $t>\sqrt{a/8}$, by drawing out the curve of $f(x)=e^{-\frac{a}{2t^2}{x}^2+2x}$, and plotting out the summation, we find that $$\sum _{k=1}^{\mathrm{\infty }}{e}^{k(2-\frac{1}{2}ak/t^2)}\le \underset{\mathbb{ℝ}}{\int }f(x)dx+2\underset{x}{\max }f(x)=e^{\frac{2t^2}{a}}(\sqrt{\frac{2\pi t^2}{a}}+2)<10\sqrt{t^2/a}{e}^{\frac{2t^2}{a}}$$ Now verify that $\ln 10+\frac{1}{2}\ln (t^2/a)+\frac{2}{a}{t}^2<C_a{t}^2$, where $C_a$ depends on $a$ only.

Corollary (Matoušek 2008, Lemma 2.2) — $X_1,\dots ,X_n$ are independent random variables with the same upper subgaussian tail: $$-\ln Pr(X_i\ge t)\ge \frac{1}{2}a{t}^2$$ for all $t>0$. Also, $E[X_i]=0,E[X_i^2]=1$, then for any unit vector $v\in {\mathbb{ℝ}}^n$, the linear sum $\sum _i{v}_i{X}_i$ has a subgaussian tail:
$$-\ln Pr(\sum _i{v}_i{X}_i\ge t)\ge C_a{t}^2$$ where $C_a>0$ depends only on $a$.

Concentration

Gaussian concentration inequality for Lipschitz functions (Tao 2012, Theorem 2.1.12.) — If $f:{\mathbb{ℝ}}^n\to \mathbb{ℝ}$ is $L$-Lipschitz, and $X\sim N(0,I)$ is a standard gaussian vector, then $f(X)$ concentrates around its expectation at a rate $$Pr(f(X)-E[f(X)]\ge t)\le e^{-\frac{2}{{\pi }^2}\frac{t^2}{L^2}}$$ and similarly for the other tail.

Proof

Proof

By shifting and scaling, it suffices to prove the case where $L=1$, and $E[f(X)]=0$.

Since every 1-Lipschitz function is uniformly approximable by 1-Lipschitz smooth functions (by convolving with a mollifier), it suffices to prove it for 1-Lipschitz smooth functions.

Now it remains to bound the cumulant generating function.

To exploit the Lipschitzness, we introduce $Y$, an independent copy of $X$, then by Jensen, $$E[e^{t(X-Y)}]=E[e^{tX}]E[e^{-tY}]\ge E[e^{tX}]e^{-tE[Y]}=E[e^{tX}]$$

By the circular symmetry of gaussian variables, we introduce $X_{\theta }:=Y\cos \theta +X\sin \theta$. This has the benefit that its derivative $X^{\prime }=-Y\sin \theta +X\cos \theta$ is independent of it.

$$\begin{array}{rl}{e}^{t(f(X)-f(Y))}& =e^{t(f(X_{\pi /2})-f(X_0))}\\ & =e^{t\underset{0}{\overset{\pi /2}{\int }}\mathrm{\nabla }f(X_{\theta })\cdot {X_{\theta }}^{\prime }d\theta }\\ & =e^{\pi t/2\underset{0}{\overset{\pi /2}{\int }}\mathrm{\nabla }f(X_{\theta })\cdot {X_{\theta }}^{\prime }\frac{d\theta }{\pi /2}}\\ & \le \underset{0}{\overset{\pi /2}{\int }}{e}^{\pi t/2\mathrm{\nabla }f(X_{\theta })\cdot {X_{\theta }}^{\prime }}\frac{d\theta }{\pi /2}\\ \end{array}$$

Now take its expectation, $$E[e^{t(f(X)-f(Y))}]\le \underset{0}{\overset{\pi /2}{\int }}E[e^{\pi t/2\mathrm{\nabla }f(X_{\theta })\cdot {X_{\theta }}^{\prime }}]\frac{d\theta }{\pi /2}$$ The expectation within the integral is over the joint distribution of $X,Y$, but since the joint distribution of $X_{\theta },{X_{\theta }}^{\prime }$ is exactly the same, we have

$$=E_X[E_Y[e^{\pi t/2\mathrm{\nabla }f(X)\cdot Y}]]$$

Conditional on $X$, the quantity $\mathrm{\nabla }f(X)\cdot Y$ is normally distributed, with variance $\le 1$, so $$\le e^{\frac{1}{2}(\pi t/2{)}^2}=e^{\frac{{\pi }^2}{8}{t}^2}$$

Thus, we have $$\ln E[e^{tf(X)}]\le \frac{{\pi }^2}{8}{t}^2$$

Strictly subgaussian

Expanding the cumulant generating function:$$\frac{1}{2}{s}^2{t}^2\ge \ln \mathrm{E}[e^{tX}]=\frac{1}{2}\mathrm{V}\mathrm{a}\mathrm{r}[X]t^2+{\kappa }_3{t}^3+\cdots$$we find that $\mathrm{V}\mathrm{a}\mathrm{r}[X]\le \Vert X{\Vert }_{\mathrm{v}\mathrm{p}}^2$. At the edge of possibility, we define that a random variable $X$ satisfying $\mathrm{V}\mathrm{a}\mathrm{r}[X]=\Vert X{\Vert }_{\mathrm{v}\mathrm{p}}^2$ 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

	$\Vert X{\Vert }_{{\psi }_2}$	$\Vert X{\Vert }_{vp}^2$	strictly subgaussian?
gaussian distribution ${𝒩}(0,1)$	$\sqrt{8/3}$	$1$	Yes
mean-zero Bernoulli distribution $p{\delta }_q+q{\delta }_{-p}$	solution to $p{e}^{(q/t{)}^2}+q{e}^{(p/t{)}^2}=2$	$\frac{p-q}{2(\log p-\log q)}$	Iff $p=0,1/2,1$
symmetric Bernoulli distribution $\frac{1}{2}{\delta }_{1/2}+\frac{1}{2}{\delta }_{-1/2}$	$\frac{1}{\sqrt{\ln 2}}$	$1$	Yes
uniform distribution $U(0,1)$	solution to $\underset{0}{\overset{1}{\int }}{e}^{x^2/t^2}dx=2$, approximately 0.7727	$1/3$	Yes
arbitrary distribution on interval $[a,b]$		$\le {\left(\frac{b-a}{2}\right)}^2$

The optimal variance proxy $\Vert X{\Vert }_{\mathrm{v}\mathrm{p}}^2$ 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 $\Vert X{\Vert }_{\mathrm{v}\mathrm{p}}^2=\frac{p-q}{2(\log p-\log q)}$. Its subgaussian norm is $t$ where $t$ is the unique positive solution to $p{e}^{(q/t{)}^2}+q{e}^{(p/t{)}^2}=2$.

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$$\Vert X{\Vert }_{{\psi }_2}=\inf \{c>0:\mathrm{E}[\exp \left(\frac{X^2}{c^2}\right)]\le 2\}=\inf \{c>0:\exp \left(\frac{1}{c^2}\right)\le 2\}=\frac{1}{\sqrt{\ln 2}},$$and hence $X$ is a subgaussian random variable.

Bounded distributions

Some commonly used 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 $\Vert X{\Vert }_{\mathrm{v}\mathrm{p}}^2\le {\left(\frac{b-a}{2}\right)}^2$. Hoeffding's inequality is the Chernoff bound obtained using this fact.

Convolutions

Density of a mixture of three normal distributions (μ = 5, 10, 15, σ = 2) with equal weights. Each component is shown as a weighted density (each integrating to 1/3)

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 $\mathrm{E}[\exp \left(\frac{X^2}{c^2}\right)]=\sum _i{p}_i\mathrm{E}[\exp \left(\frac{X_i^2}{c^2}\right)]$we have $\Vert X{\Vert }_{{\psi }_2}\le \underset{i}{\max }\Vert X_i{\Vert }_{{\psi }_2}$, 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: $\Vert X{\Vert }_{{\psi }_2}\le \underset{i}{\sup }\Vert X_i{\Vert }_{{\psi }_2}$.

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 ${\mathbb{ℝ}}^n$.

Define.

• $\Vert X{\Vert }_{{\psi }_2}:=\underset{v\in S^{n-1}}{\sup }\Vert v^{T}X{\Vert }_{{\psi }_2}$, where $S^{n-1}$ is the unit sphere in ${\mathbb{ℝ}}^n$. Similarly for the variance proxy $\Vert X{\Vert }_{vp}:=\underset{v\in S^{n-1}}{\sup }\Vert v^{T}X{\Vert }_{vp}$
• $X$ is subgaussian iff $\Vert X{\Vert }_{{\psi }_2}<\mathrm{\infty }$.

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 $\Vert X{\Vert }_{{\psi }_2}\lesssim 1$.

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

Maximum inequalities

Theorem — If $X_1,\dots ,X_n$ are mean-zero subgaussians, with $\Vert X_i{\Vert }_{vp}^2\le {\sigma }^2$, then for any $\delta >0$, we have $\max (X_1,\dots ,X_n)\le \sigma \sqrt{2\ln \frac{n}{\delta }}$ with probability $\ge 1-\delta$.

Proof

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

Theorem (Exercise 2.5.10 ) — If $X_1,X_2,\dots$ are subgaussians, with $\Vert X_i{\Vert }_{{\psi }_2}\le K$, then $$E\left[\underset{n}{\sup }\frac{|X_n|}{\sqrt{1+\ln n}}\right]\lesssim K,E[\underset{1\le n\le N}{\max }|X_n|]\lesssim K\sqrt{\ln N}$$Further, the bound is sharp, since when $X_1,X_2,\dots$ are IID samples of ${𝒩}(0,1)$ we have $E[\underset{1\le n\le N}{\max }|X_n|]\gtrsim \sqrt{\ln N}$.

Theorem (over a finite set) — If $X_1,\dots ,X_n$ are subgaussian, with $\Vert X_i{\Vert }_{vp}^2\le {\sigma }^2$, then$$\begin{array}{rl}E[\underset{i}{\max }(X_i-E[X_i])]\le \sigma \sqrt{2\ln n},& \mathrm{Pr}(\underset{i}{\max }(X_i-E[X_i])>t)\le n{e}^{-\frac{t^2}{2{\sigma }^2}},\\ E[\underset{i}{\max }|X_i-E[X_i]|]\le \sigma \sqrt{2\ln (2n)},& \mathrm{Pr}(\underset{i}{\max }|X_i-E[X_i]|>t)\le 2n{e}^{-\frac{t^2}{2{\sigma }^2}}\end{array}$$

Proof

For any t>0:$$\begin{array}{rl}E[\underset{1\le i\le n}{\max }(X_i-E[X_i])]& =\frac{1}{t}E[\ln \underset{i}{\max }{e}^{t(X_i-E[X_i])}]\\ & \le \frac{1}{t}\ln E\left[\underset{i}{\max }{e}^{t(X_i-E[X_i])}\right]\text{by Jensen}\\ & \le \frac{1}{t}\ln \sum _{i=1}^{n}E{e}^{t(X_i-E[X_i])}\\ & \le \frac{1}{t}\ln \sum _{i=1}^n{e}^{{\sigma }^2{t}^2/2}\text{by def of }\Vert \cdot {\Vert }_{vp}\\ & =\frac{\ln n}{t}+\frac{{\sigma }^{2}t}{2}\\ & \stackrel{t=\sqrt{2\ln n}/\sigma }{=}\sigma \sqrt{2\ln n},\end{array}$$This is a standard proof structure for proving Chernoff-like bounds for sub-Gaussian variables. For the second equation, it suffices to prove the case with one variable and zero mean, then use the union bound. First by Markov, $\mathrm{Pr}(X>t)\le \mathrm{Pr}(e^{sX}>e^{st})\le e^{-st}E[e^{sX}]$, then by definition of variance proxy, $\le e^{-st}{e}^{{\sigma }^2{s}^2/2}$, and then optimize at $s=-t^2/2{\sigma }^2$.

Corollary (over a convex polytope) — Fix a finite set of vectors $v_1,\dots ,v_n$. If $X$ is a random vector, such that each $\Vert v_i^{T}X{\Vert }_{vp}^2\le {\sigma }^2$, then the above 4 inequalities hold, with $\underset{v\in \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{v}(v_1,\dots ,v_n)}{\max }(v^{T}X-E[v^{T}X])$ replacing $\underset{i}{\max }(X_i-E[X_i])$. Here, $\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{v}(v_1,\dots ,v_n)$ is the convex polytope hulled by the vectors $v_1,\dots ,v_n$.

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

Inequalities

Theorem. (Theorem 2.6.1 ) There exists a positive constant $C$ such that given any number of independent mean-zero subgaussian random variables $X_1,\dots ,X_n$, $${\Vert \sum _{i=1}^n{X}_i\Vert }_{{\psi }_2}^2\le C\sum _{i=1}^n{\Vert X_i\Vert }_{{\psi }_2}^2$$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 $X_1,\dots ,X_N$,$$\mathbb{\mathbb{P}}(\left|\sum _{i=1}^N{X}_i\right|\ge t)\le 2\exp (-\frac{c{t}^2}{\sum _{i=1}^N{\Vert X_i\Vert }_{{\psi }_2}^2})\mathrm{\forall }t>0$$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 $X_1,\dots ,X_N$,$$\mathbb{\mathbb{P}}(\left|\sum _{i=1}^N{X}_i\right|\ge t)\le 2\exp (-c\min (\frac{t^2}{\sum _{i=1}^N{\Vert X_i\Vert }_{{\psi }_1}^2},\frac{t}{\underset{i}{\max }{\Vert X_i\Vert }_{{\psi }_1}}))$$ 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 $p\ge 2$, and any $a_1,\dots ,a_N\in \mathbb{ℝ}$,$${\left(\sum _{i=1}^N{a}_i^2\right)}^{1/2}\le {\Vert \sum _{i=1}^N{a}_i{X}_i\Vert }_{L^p}\le CK\sqrt{p}{\left(\sum _{i=1}^N{a}_i^2\right)}^{1/2}$$

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, $X^{T}AX$, is also subgaussian/subexponential. Further, the upper bound on the tail of $X^{T}AX$, 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$$P(|X^{T}AX-E[X^{T}AX]|>t)\le \max (2e^{-\frac{c{t}^2}{K^4\Vert A{\Vert }_F^2}},2e^{-\frac{ct}{K^2\Vert A\Vert }})=2\exp [-c\min (\frac{t^2}{K^4\Vert A{\Vert }_F^2},\frac{t}{K^2\Vert A\Vert })]$$where $K=\underset{i}{\max }\Vert X_i{\Vert }_{{\psi }_2}$, and $\Vert A{\Vert }_F=\sqrt{\sum _{ij}{A}_{ij}^2}$ is the Frobenius norm of the matrix, and $\Vert A\Vert =\underset{\Vert x{\Vert }_2=1}{\max }\Vert Ax{\Vert }_2$ is the operator norm of the matrix.

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

In the statement of the theorem, the constant $c$ 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$$P(|\Vert AX{\Vert }_2-\Vert A{\Vert }_F|>t)\le 2e^{-\frac{c{t}^2}{K^4\Vert A{\Vert }^2}}$$In words, the random vector $AX$ is concentrated on a spherical shell of radius $\Vert A{\Vert }_F$, such that $\Vert AX{\Vert }_2-\Vert A{\Vert }_F$ is subgaussian, with subgaussian norm $\le \sqrt{3/c}\Vert A\Vert K^2$.

See also

• Platykurtic distribution

References

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