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.

I'm trying to show that given $(X_i)$ i.i.d., $E[X_i^2] < \infty$, $\mu := E[X_i]$ then $P\Big [ \lim_{n \rightarrow \infty} \frac{S_n}{n} = \mu \Big ] = 1$ where $ S_n := \sum_{k=1}^n X_k$.

So far, I have rewritten $P\Big [ \lim_{n \rightarrow \infty} \frac{S_n}{n} = \mu \Big ]$ as

$$ \lim_{k \rightarrow \infty} P \Big [ \cap_{n \geq k} \{ \omega \Big | |\frac{S_n}{n} - \mu| < \varepsilon \} \Big ]$$

But I'm not sure how to proceed from here. I have $$ \sum_{n} P\Big [ |X_n - X|  > \varepsilon \Big] < \infty \Rightarrow P\Big [ \lim_n X_n = X \Big ] = 1$$ which I think I should apply but I don't see how. Can anyone help me with this? Also, I don't see where $E[X_i^2] < \infty$ comes in. Many thanks for your help!

share|improve this question
    
Hint: Note that $P[Y^2 > 1] < E[Y^2]$ for any $Y$. So $P[(X - \mu)^2 > C] = P[(X - \mu)^2/C > 1] < E[(X-\mu)^2/C] = \text{Var}[X]/C$, where $\mu=E[X]$ and $\text{Var}[X]=E[X^2]-\mu^2$, for any $C>0$. What can you prove about $\text{Var}[S_n]$? –  mjqxxxx Feb 10 '11 at 18:22

2 Answers 2

up vote 6 down vote accepted

A proof of the strong law of large numbers can be more or less complicated depending on your hypotheses. In your case, since you assume that $E[X_i^2]<\infty$ there is a straightforward proof. I am taking this from section 7.4 of the third edition of Probability and Stochastic Processes by Grimmett and Stirzaker.

First, by splitting into positive and negative parts we can assume (without loss of generality) that $X_i\geq 0$.

Second, using the positivity, it suffices to prove that $S_{n^2}/n^2\to\mu$ almost surely; that is, we only need convergence along that subsequence.

Next, Chebyshev's inequality gives
$$P(|S_{n^2}/n^2-\mu|>\varepsilon_n)\leq{E[X_i^2]\over n^2\varepsilon_n^2}.$$

Choosing $\varepsilon_n\downarrow 0$ so slowly that the right hand side above is summable, Borel-Cantelli finishes the job since then $$P(|S_{n^2}/n^2-\mu| \leq \varepsilon_n \mbox{ for all but finitely many }n) = 1.$$

In fact, the strong law of large numbers holds under the weaker hypothesis that $E[|X_i|]<\infty$. There are various proofs in the literature, but every student of probability ought to be familiar with Etemadi's tour de force elementary proof. Etemadi uses a clever truncation argument and similar tools to those above, and only needs pairwise independence of the $X_i$'s, not full independence. Some good textbooks like Grimmett and Stirzaker (section 7.5), Billingsley's Probability and Measure (2nd edition), or Durrett's Probability: Theory and Examples (2nd edition) include Etemadi's treatment.

N. Etemadi, An elementary proof of the strong law of large numbers, Z. Wahrscheinlichkeitstheorie verw. Gebeite 55, 119-122 (1981)

share|improve this answer
    
Of the books you recommend, which one is the most useful for analysts? I'm more into stochastic differential equations, but I would like to have a good book on probability theory in my "personal library". –  Jonas Teuwen Feb 11 '11 at 21:50
1  
I think that Billingsley's book is most encyclopedic and would be quite useful in the personal library of any analyst or mathematician. Karl Stromberg's "Probability for analysts" is also worth considering. He explains some of the core theorems of probability from a Fourier analysis point of view. –  Byron Schmuland Feb 11 '11 at 23:32

This is the strong law of large numbers.

If you want a measure theoretic proof, check out these lecture notes: http://staff.science.uva.nl/~spreij/onderwijs/master/mtp.pdf

It is in paragraph 10.4.

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.