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

Let $S$ be the set of $2^n$ binary $n$-bit strings. For every $x\in S$, let $f(x)$ is the maximal chain of bits 1 in $x$. So Can we find a good upper bound of $$F(n)=\frac{\sum_{x\in S}f(x)}{2^n}$$ Of course, $O(1)\le F(n) \le O(n)$. I think the upper bound is a constant or $O(\log n)$. Can anyone help me?

share|improve this question
    
Well $1$ is an upper bound, but it is a pretty silly one. A slightly better but still bad one is $\frac{1}{2}$. What do you mean by "the" upper bound? –  Noah Stein Mar 28 '13 at 11:31
    
Actually, we even have $F \le n \, 2^{-n}$, don't we? –  gerw Mar 28 '13 at 11:44
1  
One can easily write down the length generating function for strings where f(x)<=k, and from this the one for your quantity F(n). From the expression thus obtained, one may probably perform some asymptotic analysis (cf. works of Odlyzko or Flajolet/sedgewick), and obtain precise upper bounds or even asymptotic equivalents. –  Philippe Nadeau Mar 28 '13 at 12:14
1  
Consider an interpretation. Think of, say a basketball team whose wins and losses are determined by the toss of a coin. The function $f$ is the length of the longest win streak in an $n$-game season, so $F$ is the expected length of the longest win streak. It seems implausible to expect $F$ to be bounded by a constant. –  Barry Cipra Mar 28 '13 at 13:56
2  
Some relevant papers are Mark Schilling's papers on long runs, csun.edu/~hcmth031/research.html. –  Ira Gessel Mar 28 '13 at 14:39
show 5 more comments

1 Answer

There is an easy way to get a good upper bound.

The probability that there is a streak of length $k$ is at most the expected number of streaks of $1$s length $k$, which is at most the expected number of all-$1$ substrings of length $k$ (which may overlap). It is easy to get the last expected value. There are $n-k+1$ possible substrings of length $k$, so the expected number of all-$1$ substrings of length $k$ is $(n-k+1)2^{-k} \lt n /2^k$. For $k = \lceil \log_2 n \rceil + c$ this gives us an upper bound of $1/2^c$ for the probability that there is a streak of length $\lceil \log_2 n \rceil + c$. So, the average excess over $\lceil \log_2 n \rceil$ is at most $1$, and the average length of the longest streak is at most $\lceil \log_2 \rceil + 1$.

Of course, it's not clear that this upper bound is good until you get a lower bound which is close to this. I think the value should be something like $(\log_2 n) -1$. (Actually, my guess is that the difference from $\log_2 n$ is not asymptotically constant, but fluctuates depending on the fractional part of $\log_2 n$.)

share|improve this answer
    
Interesting. Do you have a feel for an intuitive reason that the fractional part of $\log_2 n$ should have anything to do with it for large $n$? –  Noah Stein Mar 28 '13 at 19:25
    
@Noah Stein: There is often some dependence on the fractional part of $\log_2 n$ in problems where $n$ might behave like $n/2$, but there could be some sort of transition between $n/2$ and $n$. The fluctuations between $n/2$ and $n$ may not die out. This happens in trie-theory (some restricted binary trees, see Flajolet and Sedgewick) and in problems like mathoverflow.net/questions/11255/…. Schilling's papers confirm the dependence on the fractional part of $\log_2 n$ both for the mean and variance. –  Douglas Zare Mar 28 '13 at 23:07
    
Thanks! I would never have guessed that. –  Noah Stein Mar 29 '13 at 13:53
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.