Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

if we are given a finite number N of points drawn from a probability distribution, expectation can be approximated as a finite sum over these points: E[f]=(1/N)(summation of f(x) over these N points).

comparing this to the actual calculation of E[f]=summation of p(x)f(x), won't the difference between the actual value and approximate value be a lot in cases where p(x) varies a lot?

share|improve this question

closed as not a real question by Nate Eldredge, George Lowther, Qiaochu Yuan, Yemon Choi, Andres Caicedo Jan 20 '11 at 6:39

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

4  
Yes it might. The standard deviation of the sample mean will be large if the underlying distribution has too large standard deviation (if that's what you mean by "varies a lot"). But this is far from a research level question, so not really suitable here. Maybe math.stackexchange would be a better fit? –  George Lowther Jan 20 '11 at 0:24
    
Or stats.stackexchange.com. You are interested in something like the variance of the sample mean. –  Nate Eldredge Jan 20 '11 at 3:57

1 Answer 1

The Strong Law of Large Numbers guarantees almost sure convergence of the sample mean to the population mean. If your distribution has large variance then yes the convergence is slower. However, the probability of being away from the population mean is bounded by:

$P(|s_n-\mu|>\epsilon)<\frac{\sigma^2}{n\epsilon^2}$

Where $\mu$ and $\sigma$ are true mean and standard deviation and $s_n$ is the sample mean from $n$ points.

share|improve this answer
    
It's worth commenting that all bets are off if the population has a distribution that doesn't have a finite variance. –  Brian Borchers Jan 20 '11 at 0:44
1  
@Brian: the SLLN holds with no assumption on the variance. It requires only the assumption of finite mean. –  Qiaochu Yuan Jan 20 '11 at 0:49
    
Are there rate of convergence estimates when dealing with infinite variance? I would guess one would have to truncate the random variables... –  Alex R. Jan 20 '11 at 1:30
    
@Alex Yes, there are long books written on the subject (see, eg, Gnedenko/Kolmogorov, or Feller v II, especially the section on Stable Laws). –  Igor Rivin Jan 20 '11 at 3:26

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