A Simple Note on the Basic Properties of
Subgaussian Random Variables

Let us begin the main body with the adjusted definition of subgaussian variables, followed by Theorem 1, which states the equivalence of various characterizations.

(2)$\implies$(3) (according to Ref. [2], with adjustments): Given property (2), we find that the moment-generating function of $X^{2}$ satisfies

$$\begin{split}\mathbb{E}\left[\exp(\lambda X^{2}/\sigma_{2}^{2})\right]&=\sum_{k=0}^{\infty}\frac{\lambda^{k}\,\mathbb{E}[X^{2k}]}{k!\,\sigma_{2}^{2k}}\,\\[-3.0pt] &\leq 1+\rho_{2}\sum_{k=1}^{\infty}\lambda^{k}=1+\frac{\lambda\rho_{2}}{1-\lambda},\end{split}$$

for all $\lambda\in[0,1)$. We observe that the rightmost side of the above expression increases with $\lambda$; we can freely select any $\lambda\in(0,1)$ and accordingly set $C_{32}=\sqrt{1/\lambda}$, $\varphi_{32}(\rho)=1+\lambda\rho/(1-\lambda)$, with which property (3) holds.

(3)$\implies$(4) (according to [3], with adjustments): From property (3) and the inequality $\lambda X\leq\lambda^{2}\sigma_{3}^{2}/4+X^{2}/\sigma_{3}^{2}$, we find

$$\begin{split}\mathbb{E}\left[\exp(\lambda X)\right]&\leq\mathbb{E}\left[\exp(\lambda^{2}\sigma_{3}^{2}/4+X^{2}/\sigma_{3}^{2})\right]\\ &=\exp(\lambda^{2}\sigma_{3}^{2}/4)\cdot\mathbb{E}\left[\exp(X^{2}/\sigma_{3}^{2})\right]\\ &\leq\rho_{3}\exp\big{(}\lambda^{2}\sigma_{3}^{2}/4\big{)}\end{split}$$

for all $\lambda\in\mathbb{R}$. Therefore, property (4) holds for $\sigma_{4}\geq\sqrt{2}\sigma_{3}/2$ and $\rho_{4}\geq\rho_{3}$, i.e., $C_{43}=\sqrt{1/2}$ and $\varphi_{43}(\rho)=\rho$.

(4)$\implies$(5) (generic Chernoff bound, see [2, 3]): Given property (4), Markov’s inequality implies

$$\begin{split}\mathbb{P}\left[X\geq\lambda\sigma_{5}\right]&=\mathbb{P}\left[\exp(xX)\geq\exp(x\lambda\sigma_{5})\right]\\ &\leq\mathbb{E}\left[\exp(xX)\right]\exp(-x\lambda\sigma_{5})\\ &\leq\rho_{4}\exp\left(\sigma_{4}^{2}x^{2}/2-\lambda\sigma_{5}x\right)\end{split}$$

for all $\lambda\geq 0$ and $x>0$. Note that $\mathbb{P}\left[X\geq\lambda\sigma_{5}\right]\leq\rho_{4}\exp(\sigma_{4}^{2}x^{2}/2-\lambda\sigma_{5}x)$ holds trivially for $x=0$. Minimizing the right-hand side with respect to $x$, we obtain

$$\begin{split}\mathbb{P}\left[X\geq\lambda\sigma_{5}\right]&\leq\rho_{4}\inf_{x\geq 0}\exp\left(\sigma_{4}^{2}x^{2}/2-\lambda\sigma_{5}x\right)\\ &=\rho_{4}\exp\left[-{\lambda^{2}\sigma_{5}^{2}}/2\sigma_{4}^{2}\right],\end{split}$$

where the infimum is attained at $x=\lambda\sigma_{5}/\sigma_{4}^{2}$. Since property (4) is invariant when replacing $X$ with $-X$, we similarly find $\mathbb{P}[-X\geq\lambda\sigma_{5}]\leq\rho_{4}\exp[-{\lambda^{2}\sigma_{5}^{2}}/2\sigma_{4}^{2}]$ for all $\lambda\geq 0$. Therefore, property (5) holds whenever $\sigma_{5}\geq\sigma_{4}$ and $\rho_{5}\geq\rho_{4}$, i.e., $C_{54}=1$ and $\varphi_{54}(\rho)=\rho$.

(5)$\implies$(1): This is trivial, with $C_{15}=1$ and $\varphi_{15}(\rho)=2\rho$.

So far, we have demonstrated the equivalence of all characterizations; any $C_{ij}$ and $\varphi_{ij}$ not directly addressed can be determined using the transitivity of implications. Furthermore, improvements can be made to certain $\varphi_{ij}$ with the following additional implications.

(3)$\implies$(1) [2]: This follows immediately from Markov’s inequality, since

$$\begin{split}\mathbb{P}\left[\lvert X\rvert\geq\lambda\sigma_{3}/\sqrt{2}\right]&=\mathbb{P}\left[X^{2}/\sigma_{3}^{2}\geq\lambda^{2}/2\right]\\ &\leq\mathbb{E}\left[\exp(X^{2}/\sigma_{3}^{2})\right]\exp(-\lambda^{2}/2)\\ &\leq\rho_{3}\exp(-\lambda^{2}/2)\end{split}$$

for any $\lambda\geq 0$, given property (3). Therefore, property (1) holds for all $\sigma_{1}\geq\sqrt{2}\sigma_{3}/2$ and $\rho_{1}\geq\rho_{3}$, i.e., $C_{13}=\sqrt{1/2}$ and $\varphi_{13}(\rho)=\rho$. The value $C_{13}=\sqrt{1/2}$ obtained here equals that derived via chain of implications, $C_{15}C_{54}C_{43}$, while $\varphi_{15}(\rho)=\rho$ obtained here is slightly better than the composition $\varphi_{15}\circ\varphi_{54}\circ\varphi_{43}(\rho)=2\rho$.

(3)$\implies$(2) [6]: By expanding property (3) into a power series, we obtain

$$\mathbb{E}\left[\exp(X^{2}/\sigma_{3}^{2})\right]=1+\sum_{k=1}^{\infty}\frac{\mathbb{E}\left[X^{2k}\right]}{k!\,\sigma_{3}^{2k}}\leq\rho_{3}.$$

Since every term in the sum is non-negative, it follows that

$$\mathbb{E}\,\big{[}X^{2k}\big{]}\leq(\rho_{3}-1)\cdot\sigma_{3}^{2k}\cdot k!$$

for all positive integers $k$. Therefore, property (2) holds, with $C_{23}=1$ and $\varphi_{23}(\rho)=\rho-1$, where $\varphi_{23}(\rho)$ is slightly better than the composition $\varphi_{21}\circ\varphi_{13}(\rho)=\rho$.

(1)$\implies$(3): This can be derived from the chain (1)$\implies$(2)$\implies$(3) or directly as shown in Refs. [6, 7]. However, it should be noted that property (1) may lead to a slightly better conclusion than property (2). Given property (1), we have

$$\begin{split}\mathbb{E}\left[\lvert X\rvert^{n}\right]&=n\int_{0}^{\infty}\mathbb{P}\left[\lvert X\rvert\geq\lambda\sigma_{1}\right]\,(\lambda\sigma_{1})^{n-1}\,\mathrm{d}(\lambda\sigma_{1})\\ &\leq n\sigma_{1}^{n}\int_{0}^{\infty}\min\big{\{}\rho_{1}\exp(-\lambda^{2}/2),1\big{\}}\,\lambda^{n-1}\,\mathrm{d}\lambda\\ &=\frac{1}{2}n\big{(}\sqrt{2}\sigma_{1}\big{)}^{n}\int_{0}^{\infty}\min\big{\{}\rho_{1}\exp(-x),1\big{\}}\,x^{\tfrac{n-2}{2}}\,\mathrm{d}x\\ &=\frac{1}{2}n\big{(}\sqrt{2}\sigma_{1}\big{)}^{n}\bigg{(}\int_{0}^{\ln\rho_{1}}\mkern-5.0mux^{\tfrac{n-2}{2}}\,\mathrm{d}x+\rho_{1}\int_{\ln\rho_{1}}^{\infty}\mkern-5.0mu\exp(-x)\,x^{\tfrac{n-2}{2}}\,\mathrm{d}x\bigg{)}\\ &=\rho_{1}\cdot\big{(}\sqrt{2}\sigma_{1}\big{)}^{n}\cdot\Gamma\left(\frac{n}{2}+1,\,\ln\rho_{1}\right)\end{split}$$

for any $n\geq 1$, where $\Gamma(\cdot\,,\cdot)$ denotes the upper incomplete gamma function. When $n=2k$ with $k=1,2,3\dots$, we have

$$\mathbb{E}\big{[}X^{2k}\big{]}\leq\rho_{1}\cdot 2^{k}\sigma_{1}^{2k}\cdot\Gamma\left(k+1,\,\ln\rho_{1}\right)=\sum_{i=0}^{k}\frac{(\ln\rho_{1})^{i}}{i!}\cdot 2^{k}\sigma_{1}^{2k}\cdot k!,$$

which is slightly better than the result $\mathbb{E}\big{[}X^{2k}\big{]}\leq\rho_{1}\cdot 2^{k}\sigma_{1}^{2k}\cdot k!$ in the derivation for implication (1)$\implies$(2). Furthermore, we find

$$\begin{split}\mathbb{E}\left[\exp(\lambda X^{2}/2\sigma_{1}^{2})\right]&=\sum_{k=0}^{\infty}\frac{\lambda^{k}\,\mathbb{E}[X^{2k}]}{k!\,2^{k}\sigma_{1}^{2k}}\leq\sum_{k=0}^{\infty}\lambda^{k}\sum_{i=0}^{k}\frac{(\ln\rho_{1})^{i}}{i!}\\[-3.0pt] &=\sum_{i=0}^{\infty}\frac{(\ln\rho_{1})^{i}}{i!}\sum_{k=i}^{\infty}\lambda^{k}=\frac{1}{1-\lambda}\sum_{i=0}^{\infty}\frac{(\lambda\ln\rho_{1})^{i}}{i!}=\frac{\rho_{1}^{\lambda}}{1-\lambda},\end{split}$$

for all $\lambda\in[0,1)$. Therefore, we can freely select any $\lambda\in(0,1)$, and property (3) holds with $C_{31}=\sqrt{2/\lambda}$ and $\varphi_{31}(\rho)=\rho^{\lambda}/(1-\lambda)$, where $\varphi_{31}(\rho)$ is slightly better than the composition $\varphi_{32}\circ\varphi_{21}(\rho)=1+\lambda\rho/(1-\lambda)$.

(4)$\implies$(3) (according to [3]): Starting from property (4), we have

$$\mathbb{E}\bigg{[}\exp\Big{(}xX-\frac{1}{2\lambda}\sigma_{4}^{2}\,x^{2}\Big{)}\bigg{]}\leq\rho_{4}\exp\bigg{(}\frac{1}{2}\Big{(}1-\frac{1}{\lambda}\Big{)}\sigma_{4}^{2}x^{2}\bigg{)}$$

for all $x\in\mathbb{R}$ and $\lambda\in(0,1)$. By integrating both sides with respect to $x$ on $(-\infty,+\infty)$ and using Fubini’s theorem, we obtain

$$\frac{\sqrt{2\lambda\pi}}{\sigma_{4}}\mathbb{E}\left[\exp\Big{(}\frac{\lambda X^{2}}{2\sigma_{4}^{2}}\Big{)}\right]\leq\frac{\rho_{4}}{\sigma_{4}}\cdot\frac{\sqrt{2\lambda\pi}}{\sqrt{1-\lambda}},$$

which simplifies to

$$\mathbb{E}\left[\exp\Big{(}\frac{\lambda X^{2}}{2\sigma_{4}^{2}}\Big{)}\right]\leq\frac{\rho_{4}}{\sqrt{1-\lambda}}$$

for all $\lambda\in(0,1)$. Therefore, property (3) holds with $C_{34}=\sqrt{2/\lambda}$ and $\varphi_{34}(\rho)=\rho/\sqrt{1-\lambda}$. Meanwhile, the implication chain (4)$\implies$(5)$\implies$(1)$\implies$(3) leads to $C_{31}C_{15}C_{54}=\sqrt{2/\lambda}$ and $\varphi_{31}\circ\varphi_{15}\circ\varphi_{54}=(2\rho)^{\lambda}/(1-\lambda)$. We may finally choose

$$\varphi_{34}(\rho)=(1-\lambda)^{-1}\min\left\{\rho\sqrt{1-\lambda},\,(2\rho)^{\lambda}\right\}.$$

(3)

 

Definition 1 and the equivalent characterizations in Theorem 1 do not require subgaussian variables to be centered (by “subgaussian” we mean that any of the properties in Theorem 1 is satisfied). Only $(\sigma,1)$-subgaussian variables, as conventionally defined, are guaranteed to have a zero mean. Conversely, any centered subgaussian variable must be $(\sigma,1)$-subgaussian for some $\sigma$.

Case (b) (adjusted from [3]): The moments of odd orders of a random variable $Y$ can be bounded according to the Cauchy–Schwarz inequality, as

$$\mathbb{E}\left[Y^{2k+1}\right]\leq\sqrt{\mathbb{E}\left[Y^{2k}\right]\,\mathbb{E}\left[Y^{2k+2}\right]}\leq\frac{1}{2}\left(x\mathbb{E}\big{[}Y^{2k}\big{]}+x^{-1}\mathbb{E}\big{[}Y^{2k+2}\big{]}\right)$$

for all $x>0$ and $k=0,1,2,\dots$. Applying this for $\lambda X$ and substituting it into the power series expansion of $\mathbb{E}\left[\exp(\lambda X)\right]$, we find

$$\begin{split}\mathbb{E}\left[\exp(\lambda X)\right]&=1+\lambda\mathbb{E}\left[X\right]+\sum_{k=1}^{\infty}\left\{\frac{\mathbb{E}\left[(\lambda X)^{2k}\right]}{(2k)!}+\frac{\mathbb{E}\left[(\lambda X)^{2k+1}\right]}{(2k+1)!}\right\}\\ &\leq 1+\sum_{k=1}^{\infty}\left\{\frac{\lambda^{2k}\mathbb{E}[X^{2k}]}{(2k)!}+\frac{x_{k}\lambda^{2k}\mathbb{E}[X^{2k}]+x_{k}^{-1}\lambda^{2k+2}\mathbb{E}[X^{2k+2}]}{2\cdot(2k+1)!}\right\}\\[-2.0pt] &\leq 1+\rho_{2}\sum_{k=1}^{\infty}\left\{\frac{\lambda^{2k}\sigma_{2}^{2k}\,k!}{(2k)!}+\frac{x_{k}\lambda^{2k}\sigma_{2}^{2k}\,k!+x_{k}^{-1}\lambda^{2k+2}\sigma_{2}^{2k+2}\,(k+1)!}{2\cdot(2k+1)!}\right\}\\[-2.0pt] &=1+\rho_{2}\cdot\frac{\lambda^{2}\sigma_{2}^{2}}{2!}\left(1+\frac{x_{1}}{6}\right)+\rho_{2}\sum_{k=2}^{\infty}\frac{\lambda^{2k}\sigma_{2}^{2k}k!}{(2k)!}\left(\frac{k}{x_{k-1}}+1+\frac{x_{k}}{4k+2}\right)\\[-2.0pt] &=1+\sum_{k=1}^{\infty}c_{k}\frac{\lambda^{2k}\sigma_{2}^{2k}}{k!},\end{split}$$

where we have used the assumption $\mathbb{E}\left[X\right]=0$ and property (2), and have defined coefficients

$$c_{k}\coloneqq\left\{\begin{aligned} &\frac{1}{2}\left(1+\frac{x_{1}}{6}\right)\rho_{2},~{}&&\textup{for }k=1;\\ &\frac{(k!)^{2}}{(2k)!}\left(\frac{k}{x_{k-1}}+1+\frac{x_{k}}{4k+2}\right)\rho_{2},~{}&&\textup{for }k=2,3,\dots,\end{aligned}\right.$$

where $x_{k}>0$ are to be determined. By choosing $x_{k}=2(k+1)$ for $k\geq 2$, we have

$$c_{k}=\left\{\begin{aligned} &\frac{1}{2}\left(1+\frac{x_{1}}{6}\right)\rho_{2},~{}&&\textup{for }k=1;\\ &\frac{1}{6}\left(\frac{2}{x_{1}}+\frac{8}{5}\right)\rho_{2},~{}&&\textup{for }k=2;\\ &\frac{(k!)^{2}}{(2k)!}\left(2+\frac{1}{4k+2}\right)\rho_{2},~{}&&\textup{for }k=3,4,\dots,\end{aligned}\right.$$

and

$$\left\{\begin{aligned} c_{1}&=\frac{1}{2}\left(1+\frac{x_{1}}{6}\right)\rho_{2};\\ \frac{c_{2}}{c_{1}}&=\frac{1}{3}\cdot\frac{2/x_{1}+8/5}{1+x_{1}/6}=\frac{2}{3x_{1}}\frac{1+\frac{4}{5}x_{1}}{1+\frac{1}{6}x_{1}}\leq\frac{2}{3x_{1}}\left[\frac{19}{120}\Big{(}x_{1}-6\Big{)}+\frac{29}{10}\right],~{}&&\textup{when }x_{1}\in(0,6];\\ \frac{c_{3}}{c_{2}}&=\frac{3}{10}\cdot\frac{2+1/14}{2/x_{1}+8/5}\leq\frac{9}{28}<\frac{29}{90},~{}&&\textup{when }x_{1}\in(0,6];\\ \frac{c_{k+1}}{c_{k}}&=\frac{(k+1)}{4(k+3/2)}\cdot\frac{(8k+13)}{(8k+5)}<0.29,~{}&&\textup{for }k=3,4,\dots.\end{aligned}\right.$$

Now we need to choose the value of $x_{1}$ appropriately and find a common upper bound for $c_{1}$, $c_{2}/c_{1}$, and all $c_{k+1}/c_{k}$ ($k\geq 2$). When $\rho_{2}\leq{29}/{90}$, we can choose $x_{1}=6$, so that $c_{1}\leq 29/90$ and $c_{k+1}/c_{k}\leq 29/90$ for all $k\geq 1$. When $\rho_{2}>{29}/{90}$, we choose $x_{1}=x_{1}^{\ast}\in(0,6)$ such that

$$\frac{1}{2}\left(1+\frac{x_{1}^{\ast}}{6}\right)\rho_{2}=\frac{2}{3x_{1}^{\ast}}\left[\frac{19}{120}\Big{(}x_{1}^{\ast}-6\Big{)}+\frac{29}{10}\right]>\frac{29}{90},$$

(6)

which leads to an explicit solution

$$x_{1}^{\ast}=\frac{3}{\rho_{2}}\bigg{[}\sqrt{\Big{(}\rho_{2}-\frac{19}{90}\Big{)}^{2}+\frac{26}{15}\rho_{2}}-\left(\rho_{2}-\frac{19}{90}\right)\bigg{]}$$

and accordingly

$$c_{1}^{\ast}=\frac{1}{4}\bigg{[}\sqrt{\Big{(}\rho_{2}-\frac{19}{90}\Big{)}^{2}+\frac{26}{15}\rho_{2}}+\left(\rho_{2}+\frac{19}{90}\right)\bigg{]}.$$

With such choices of $x_{k}$ ($k=1,2,3,\dots$), we finally have

$$\begin{split}\mathbb{E}\left[\exp(\lambda X)\right]&\leq 1+\sum_{k=1}^{\infty}c_{k}\frac{\lambda^{2k}\sigma_{2}^{2k}}{k!}\leq 1+\sum_{k=1}^{\infty}\frac{\lambda^{2k}\sigma_{2}^{2k}}{k!}\max\left\{\frac{29}{90},c_{1}^{\ast}\right\}^{k}\\ &\leq\exp\left(\max\Big{\{}\frac{29}{90},c_{1}^{\ast}\Big{\}}\sigma_{2}^{2}\lambda^{2}\right),\end{split}$$

which proves the claim of case (b).