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 bounded from above (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}

The subgaussian norm of ${\displaystyle X}$, denoted as ${\displaystyle \Vert X\Vert _{\psi _{2}}}$, is$${\displaystyle \Vert X\Vert _{\psi _{2}}=\inf \left\{c>0:\operatorname {E} \left[\exp {\left({\frac {X^{2}}{c^{2}}}\right)}\right]\leq 2\right\}.}$$In other words, it is the Orlicz norm of ${\displaystyle X}$ 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.

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.

Let ${\displaystyle X}$ be a random variable. Let ${\displaystyle K_{1},K_{2},K_{3},\dots }$ be positive constants. 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 ${\displaystyle t\geq 0}$;
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 ${\displaystyle \Gamma }$ 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 ${\displaystyle t}$;
6. Moment-generating function (of ${\displaystyle X^{2}}$): ${\displaystyle \operatorname {E} [e^{X^{2}t^{2}}]\leq e^{K^{2}t^{2}}}$ 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.

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

Proof. ${\displaystyle (1)\implies (3)}$ By the layer cake representation,$${\displaystyle {\begin{aligned}\operatorname {E} |X|^{p}&=\int _{0}^{\infty }\operatorname {P} (|X|^{p}\geq t)dt\\&=\int _{0}^{\infty }pt^{p-1}\operatorname {P} (|X|\geq t)dt\\&\leq 2\int _{0}^{\infty }pt^{p-1}\exp \left(-{\frac {t^{2}}{K_{1}^{2}}}\right)dt\\\end{aligned}}}$$

After a change of variables ${\displaystyle u=t^{2}/K_{1}^{2}}$, we find that$${\displaystyle {\begin{aligned}\operatorname {E} |X|^{p}&\leq 2K_{1}^{p}{\frac {p}{2}}\int _{0}^{\infty }u^{{\frac {p}{2}}-1}e^{-u}du\\&=2K_{1}^{p}{\frac {p}{2}}\Gamma \left({\frac {p}{2}}\right)\\&=2K_{1}^{p}\Gamma \left({\frac {p}{2}}+1\right).\end{aligned}}}$$${\displaystyle (3)\implies (2)}$ By the Taylor series ${\textstyle e^{x}=1+\sum _{p=1}^{\infty }{\frac {x^{p}}{p!}},}$$${\displaystyle {\begin{aligned}\operatorname {E} [\exp {(\lambda X^{2})}]&=1+\sum _{p=1}^{\infty }{\frac {\lambda ^{p}\operatorname {E} {[X^{2p}]}}{p!}}\\&\leq 1+\sum _{p=1}^{\infty }{\frac {2\lambda ^{p}K_{3}^{2p}\Gamma \left(p+1\right)}{p!}}\\&=1+2\sum _{p=1}^{\infty }\lambda ^{p}K_{3}^{2p}\\&=2\sum _{p=0}^{\infty }\lambda ^{p}K_{3}^{2p}-1\\&={\frac {2}{1-\lambda K_{3}^{2}}}-1\quad {\text{for }}\lambda K_{3}^{2}<1,\end{aligned}}}$$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,$${\displaystyle \operatorname {P} (|X|\geq t)=\operatorname {P} \left(\exp \left({\frac {X^{2}}{K_{2}^{2}}}\right)\geq \exp \left({\frac {t^{2}}{K_{2}^{2}}}\right)\right)\leq {\frac {\operatorname {E} [\exp {(X^{2}/K_{2}^{2})}]}{\exp \left({\frac {t^{2}}{K_{2}^{2}}}\right)}}\leq 2\exp \left(-{\frac {t^{2}}{K_{2}^{2}}}\right).}$$${\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.

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

Expanding the cumulant generating function:$${\displaystyle {\frac {1}{2}}s^{2}t^{2}\geq \ln \operatorname {E} [e^{tX}]={\frac {1}{2}}\mathrm {Var} [X]t^{2}+\kappa _{3}t^{3}+\cdots }$$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.

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.

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.

	${\displaystyle \|X\|_{\psi _{2}}}$	${\displaystyle \|X\|_{vp}^{2}}$	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 ${\displaystyle U(0,1)}$	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]

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 ${\displaystyle \Vert X\Vert _{\mathrm {vp} }^{2}={\frac {p-q}{2(\log p-\log q)}}}$.^[5] 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$${\displaystyle \Vert X\Vert _{\psi _{2}}=\inf \left\{c>0:\operatorname {E} \left[\exp {\left({\frac {X^{2}}{c^{2}}}\right)}\right]\leq 2\right\}=\inf \left\{c>0:\exp {\left({\frac {1}{c^{2}}}\right)}\leq 2\right\}={\frac {1}{\sqrt {\ln 2}}},}$$and hence ${\displaystyle X}$ is a subgaussian random variable.