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 want to estimate the maximum value, which was expected not reacchable, of a time series. the distribution of samples in a interval was unknown. It can be conclude from pdf figures that those samples are in the same distribution with different parameters, that would change slowly. Is there any theorically proved method for that?

share|improve this question
    
What do you mean by "expected not reachable"? If you gathered some data empirically and the plots suggest a trend, it is likely that there is some underlying theory that might explain the behavior. But (if I understand your question correctly) one does not usually derive theories from empirical plots. You might want to look at hypothesis testing. –  sai May 18 '12 at 2:33
    
@sai, maximum value means it is a up boundary, which samples will only smaller than that. the distribution is not ordinary. I find that box plot seems work, but i don't know what is its theoretical basis. –  Readon Shaw May 18 '12 at 6:18
add comment

2 Answers

up vote 0 down vote accepted

You can see P[Mn<=x]=P[X<=x]^n for the iid case that the maximum's distribution can be derived from the distribution for individual observation. Similar formulaes can be derived for the other order statistics. H. A. David's book on orde statistics is a good source for this. In the dependent case there is no easy way. I did research in the case of stationary autoregressive processes and could not find a nice exact formula. To prove limit theorems there are inequalties that can be used and mixing conditions can be verified to prove limit theorem's. Gnedenko's theorem is for extremes what the central limit theorem is for averages. It can be used to justify modeling extremes using an extreme value type since in large samples the distribution for the maximum is close to that extreme type for a large class of of population distributions F. Unfortunately the correlated case is much more difficult. In some dependent cases you can get the limit distirbution to exist and not be an extreme value type (when mixing conditions fail) See Galambos' book on extremes for an example of this.

share|improve this answer
add comment

If you are interested in the limiting distribution of the maximum of a sequence of iid random variables and you know something about the tail behavior of the distribution, Gnedenko's theorem will tell you (if the limit exists) how to normalize the maximum term in the sequence so that it converges to one fo the three extreme value types and which one it converges to. In your case you know there is a finite upper bound on the maximum. So the answer for the limiting distribution would be type III. The theorem extends to stationary time series under special conditions called mixing conditions. This theory can be found for example in "Extremes and Related properties of Random Sequences and Processes by Leadbetter, Lindgren and Rootzen (1983) Springer-Verlag. this theory may help with your problem although not directly. In the iid case let X1, X2,...,Xn be an iid sequence with Mn representing the maximum of these n values. You can use by independence P[Mn<=x]=P[X1<=x] P[X2<=x]...P[Xn<=x] and since they are identically distributed this is equal to P[X1<=x]$^n$. So take for example X1 uniform on [0, A]. Then P[Mn<=x] = x$^n$/A$^n$ for x < A this will go to zero as n tends to infinity showing that Mn is converging to A in probability. So P[Mn>x]=1-x$^n$ /A$^n$ . This tells you exactly probability that Mn is between x and A for any x < A. So for very large n and assuming x is within a small value say e of A (i. e. x=A-e) you have a very high probability that Mn will be larger than A-e.

share|improve this answer
    
It is useful. Is there any tutorial on calculate the maximum value based on this theory? @Michael Chernick –  Readon Shaw May 19 '12 at 8:50
add comment

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.