I am trying to interpret the following definition:

A non-degenerate distribution is a stable distribution if it satisfies the  

following property:
Let X1 and X2 be independent copies of a random variable X. Then X is said to 

be stable if for any constants a>0 and b>0 the random variable aX1 + bX2 has 

the same distribution as cX + d for some constants c>0 and d. The distribution      

is said to be strictly stable if this holds with d = 0 (Nolan 2009).

I have searched the internet for more intuitive explanations but couldn't find anything. Can someone help?

I couple of questions:

  1. what does independent copies of a random variable mean?

  2. is the random variable X already a distribution?

So does this say basically that if we add for example two normal distributions then it will also be normal?

  • 2
    $\begingroup$ Any random variable, by definition, has a distribution, which might be discrete or continuous based on the set of values it admits and independent copies mean independent draws from the distribution of $X$, i.e. $X_1,X_2 \sim F_X$. Stable distributions are called so because a sum such as the one you described, leaves the shape of the distribution unchanged. Also, the example with Normal is correct. i.e. $ X \sim N(\mu,\sigma^2) \Rightarrow aX_1+bX_2 \sim N((a+b)\mu, (a^2+b^2)\sigma^2)$ and $ cX+d \sim N(c\mu+d, c^2)$, so, the definition holds for $c^2 = a^2+b^2$ and $d = (a+b-c)\mu$. $\endgroup$ Commented Jul 17, 2012 at 22:08

1 Answer 1


The central limit theorem and infinite divisibility are properties associated with the normal distribution. The stable distributions extend family of distributions to other distributions that have the inifinite divisibility property and and a limit theroem called the stable law. Convergence of avergaes to the normal distribution requires certain properties of the random variables. We shall just consider the simple case where the random variables are independent and identically distributed. in that case the central limit theorem holds under mild conditions. (1) the variance must exist and (2) any moment slightly higher than the 2nd moment must exist (one form of the CLT). Now distributions that do not have a finite variance and even some like the Cauchy that don't have a finite mean will still have their independent sums properly normalized converge to one of the stable distributions. The stable distributions are indexed by a parameter alpha that can be >0 but <=2. The index 2 corresponds to the normal distribution. A more thorough account of the properties of stable distributions can be found in Wikipedia at this link:


The other questions were answered correctly by VitalStatistix in his comments.

  • $\begingroup$ Just an aside: If our random variables are iid, we only need finite nonzero variance to get the CLT; the second moment condition listed is unnecessary. :) $\endgroup$
    – cardinal
    Commented Jul 18, 2012 at 10:16
  • $\begingroup$ @cardinal Yes the CLT comes about in various forms with the Lindeberg-Feller conditions for example are more general than the version I mentioned. That is why I mentioned "one form" and my point was just that there are random variables that don't obey the CLT but do have a stable limit. But I was not aware or forgot about examples where finite variance was not needed in the IID case. Can you give me an example where the variance is infinite? $\endgroup$ Commented Jul 18, 2012 at 10:28
  • $\begingroup$ Sorry I was not clear. I meant that requiring a finite moment strictly larger than the second to exist in order to apply the CLT is not necessary in the iid case. We just need the first condition you list (i.e., nontrivial finite variance). :) $\endgroup$
    – cardinal
    Commented Jul 18, 2012 at 10:47
  • $\begingroup$ @cardinal Sure that was the reason for my remark. Something more than second moments is needed and the Lyaponov condition is just one simple way to get it. This wikipedia article covers this well. en.wikipedia.org/wiki/Central_limit_theorem $\endgroup$ Commented Jul 18, 2012 at 11:01
  • 1
    $\begingroup$ Referring to your link, look under the "Classical CLT" heading or the Lindeberg-Levy CLT. This is the situation you describe. No additional moment finiteness is needed beyond the second moment. :) The Lyapunov variant drops the identical distribution assumption. Since we give something up, we have to add back some structure to retain the result. One way to do this is to require a slightly higher moment (though this is just a special case of full Lindeberg-Feller). :) $\endgroup$
    – cardinal
    Commented Jul 18, 2012 at 11:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.