By definition, the weak law states that for a specified large $n$, the average is likely to be near $\mu$. Thus, it leaves open the possibility that $|\bar{X_n}-\mu| \gt \eta$ happens an infinite number of times, although at infrequent intervals.

The strong law shows that this almost surely will not occur. In particular, it implies that with probability 1, we have that for any $\eta > 0$ the inequality $|\bar{X_n}-\mu| \lt \eta$ holds for all large enough $n$.

Now my question is application of these laws. How do I know which distribution satisfies the strong law vs the weak law. For example, consider a distribution $X_n$ be iid with finite variances and zero means. Does the mean $\frac{\sum_{k=1}^{n} X_k}{n}$ converge to $0$ almost surely (strong law of large numbers) or only in probability (weak law of large numbers)?

  • $\begingroup$ Both laws always hold. Also finite variance isn't necessary (but it's easier to prove the laws with finite variance.) $\endgroup$ Dec 7, 2010 at 21:15

2 Answers 2


From section 7.4 of Grimmett and Stirzaker's Probability and Random Processes (3rd edition).

The independent and identically distributed sequence $(X_n)$, with common distribution function $F$, satisfies $${1\over n} \sum_{i=1}^n X_i\to \mu$$ in probability for some constant $\mu$ if and only if the characteristic function $\phi$ of $X_n$ is differentiable at $t=0$ and $\phi^\prime(0)=i \mu$.

For instance, the weak law holds but the strong law fails for $\mu=0$ and symmetric random variables with $1-F(x)\sim 1/(x\log(x))$ as $x\to\infty$.

  • $\begingroup$ Perhaps worth adding for new readers that the final example does not have a finite mean, but almost does. $\endgroup$
    – Henry
    Oct 14 at 17:00

If $X_1,X_2,\ldots$ is a sequence of i.i.d. random variables with finite mean $\mu$ (in your example, $\mu = 0$), then by the strong law of large numbers, $\frac{{\sum\nolimits_{i = 1}^n {X_i } }}{n}$ converges to $\mu$ almost surely. In particular, $\frac{{\sum\nolimits_{i = 1}^n {X_i } }}{n}$ converges to $\mu$ in probability. So, you actually don't have to assume finite variance.

  • $\begingroup$ Can you give an example where weak law holds but strong law does not hold? $\endgroup$
    – user957
    Dec 7, 2010 at 21:18
  • $\begingroup$ @user957: what hypothesis do you want to drop? $\endgroup$ Dec 7, 2010 at 21:52
  • $\begingroup$ I have tried to find a simple example, but with no success. $\endgroup$
    – Shai Covo
    Dec 8, 2010 at 3:08
  • $\begingroup$ You can get away with just pairwise independence as well. (Etemadi's law) $\endgroup$
    – Jens
    Feb 25, 2011 at 14:00
  • $\begingroup$ @Jens: Indeed. Thank you. $\endgroup$
    – Shai Covo
    Feb 27, 2011 at 6:28

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