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What is the relationship between a random variable obeying the subexponential distribution defined here and a random variable $X$ satisfying $P\left(\left|X\right|>t\right)\le\alpha e^{-\beta t}$ for all $t>0$ and some $\alpha,\beta>0$?

Thanks a lot for any helpful answers.

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There is no such thing as THE subexponential distribution. You might wish to explain which distribution you have in mind. – Did Sep 15 '12 at 9:48
@did is it true that when $\lim_{t\rightarrow\infty}$, $P(t>|x|)\sim 1-e^{-\beta t}$? – Seyhmus Güngören Sep 15 '12 at 9:53
@SeyhmusGüngören I do not understand your question. – Did Sep 15 '12 at 10:19
@did can we say that any $X$ satisfying $P\left(\left|X\right|>t\right)\le\alpha e^{-\beta t}$ has an exponential distribution? then it will be also subexponential. – Seyhmus Güngören Sep 15 '12 at 10:29
@SeyhmusGüngören The usual meaning of the term subexponential distribution for distributions on $[0,+\infty)$ (Teugels, Pitman, others) is that $P(X+X'\gt x)\sim2P(X\gt x)$ when $x\to+\infty$, for $X$ and $X'$ i.i.d. following this distribution. In particular, $e^{cx}P(X\gt x)\to\infty$ for every $c\gt0$. Hence, no, exponential distributions are not subexponential. – Did Sep 15 '12 at 12:18

As explained in the comments, if the distribution of a nonnegative random variable $Y$ is subexponential, then $\mathrm P(Y\gt x)\gg\mathrm e^{-cx}$ when $x\to+\infty$, for every positive $c$. Hence, if $\mathrm P(X\gt x)\leqslant \alpha\mathrm e^{-\beta x}$ when $x\to+\infty$ for some positive $\beta$, then $\mathrm P(X\gt x)\ll\mathrm P(Y\gt x)$ when $x\to+\infty$. And?

Edit The fact that every nonnegative random variable $X$ with subexponential distribution is such that $G(x)\gg\mathrm e^{-cx}$ when $x\to+\infty$, for every positive $c$, where $G(x)=\mathrm P(X\gt x)$, is a basic result of the field. Let us give a hands-on proof of a very restricted version of this fact, namely, that $G(x)\sim a\mathrm e^{-bx}$ is impossible.

By hypothesis, $\mathrm P(X+X'\gt x)\sim2G(x)$ when $x\to+\infty$, where $X'$ is independent of $X$ with the same distribution. Note that $$ [X+X'\gt x]=[X\gt x\vee X'\gt x]\cup A_x,\quad A_x=[X+X'\gt x\geqslant X,x\geqslant X'], $$ and that the two events on the RHS are disjoint. In particular, $$ \mathrm P(X+X'\gt x)=\mathrm P(A_x)+1-\mathrm P(X\leqslant x,X'\leqslant x). $$ By independence, $\mathrm P(X\leqslant x,X'\leqslant x)=\mathrm P(X\leqslant x)^2=(1-G(x))^2$, hence $$ \mathrm P(X+X'\gt x)=\mathrm P(A_x)+2G(x)-G(x)^2. $$ Since $G(x)^2\ll G(x)$, the property of being subexponential is equivalent to $$ G(x)\gg\mathrm P(A_x). $$ Now, $A_{2x}\supset[x\lt X\leqslant 2x,x\lt X'\leqslant 2x]$ hence, if the distribution is subexponential, then $$ G(2x)\gg\mathrm P(x\lt X\leqslant 2x)^2=G(x)^2-2G(x)G(2x)+G(2x)^2. $$ Since $G(x)G(2x)\ll G(2x)$ and $G(2x)^2\ll G(2x)$, one gets $G(2x)\gg G(x)^2$. But if $G(x)\sim a\mathrm e^{-bx}$, then $G(2x)\sim a^2\mathrm e^{-2bx}\sim a G(x)^2$, hence $G(2x)\gg G(x)^2$ is impossible.

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Thank you for the answer, but how did you get $\mathrm P(Y\gt x)\gg\mathrm e^{-cx}$ when $x\to+\infty$? Thanks. – maq Sep 15 '12 at 16:54

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