Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

share|improve this question
    
Both laws always hold. Also finite variance isn't necessary (but it's easier to prove the laws with finite variance.) –  trutheality Dec 7 '10 at 21:15

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
    
Can you give an example where weak law holds but strong law does not hold? –  user957 Dec 7 '10 at 21:18
    
@user957: what hypothesis do you want to drop? –  Qiaochu Yuan Dec 7 '10 at 21:52
    
I have tried to find a simple example, but with no success. –  Shai Covo Dec 8 '10 at 3:08
    
You can get away with just pairwise independence as well. (Etemadi's law) –  Jens Feb 25 '11 at 14:00
    
@Jens: Indeed. Thank you. –  Shai Covo Feb 27 '11 at 6:28

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 $x\to\infty$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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