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In random walks, a path may return to origin for the $r$-th time in $n$-th step ($r$ is given but $n$ is not given), and under this condition, these $n$ steps can be split into $r$ sections where each section ends with an $i$-th return ($i=0,1,\ldots,r$). We denote the length of each section as $l_1,l_2,\ldots,l_r$ ($l_1+l_2+\ldots+l_r=n$).

What I concern is the independence between $l_1,l_2,\ldots,l_r$.

In the obvious sense the random walk starts from scratch every time when the path returns to the origin. Thus one may draw a conclusion that they are independent of each other. But Feller's book (on page 91) tells us an unexpected different story:

$n$ is to be of the order of magnitude $r^2$.

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If they are independent, $\frac{n}{r}$ should not be relevant to $r$, but it surely is from the above statement. Thus they are not.

My question is: What key point that we ignored led us to the intuitive but wrong conclusion?

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3 Answers 3

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These parts of the path are called excursions and are indeed not independent if $n$ is fixed. Imagine you have only two excursions, and the first one took $n-2$ steps – this describes the other one almost completely (up to the sign).

On the other hand, if $n$ has a geometric distribution (and it is independent from the random walk), then $r$ is also a geometric random variable, and the corresponding excursions are independent. This is the basis of fluctuation theory (and the Wiener–Hopf factorisation) for random walks.

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  • $\begingroup$ I just edited the question and added that "r is given but n is not given", now what if this is the case? $\endgroup$
    – Daniel
    Commented Aug 29, 2017 at 9:25
  • $\begingroup$ If one considers first $r$ excursions, then indeed these excursions are independent. In particular, the total length is the sum of $r$ independent random variables, as Feller clearly states. The point is that the expected length of an excursion is infinite (for the simple random walk); Feller attempts to explain this: "A closer analysis reveals that one among the $r$ waiting times is likely to be of the same order of magnitude as the whole sum". $\endgroup$ Commented Aug 29, 2017 at 10:59
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Feller explains it on p246. In one sentence, the intuition to apply the law of large numbers is incorrect because it only applies to random variables with expectation.

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Thanks to @Especially Lime's reminder of the difference between the median and the mean (in probability, the equivalents are quantile and expectation). And now I think this is exactly the key point leading to the seemingly wrong intuition (also under the misleading of Feller:(). Actually we can see from Feller's first paragraph in the picture and equation (7.7) that what he was really talking about is the quantile instead of the expectation. But in Feller's second paragraph in the picture, it seems he mixed up the two concepts (he used $r^2$ to prove the average increases roughly in proportion to $r$, but this should be some quantile such as median instead of the average). And this can mislead readers (like me).

And when it comes to the average/expectation, the expectation of $\frac{n}{r}$ actually equals the expectation of $l_1$ (in the sense of infinity of the same order), and this is exactly our intuition. But when it comes to the quantile such as the median, actually it is not clear intuitively whether the median of $n$ is proportional to $r$. And thus we need to calculate.

Now we have figured out what's wrong with our first intuition, it is for the expectation/mean/average instead of the quantile such as the median. And by the way, $l_1,l_2,\ldots,l_r$ are truly independent of each other.

Fully understanding a problem is so satisfying, and it feels really great :)

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    $\begingroup$ Well, Feller was correct: the length of the excursion is in the domain of a $1/2$-stable law, and so the total length of $r$ excursions will be distributed roughly as $r^2 \xi$, where $\xi$ is some fixed positive $1/2$-stable random variable. $\endgroup$ Commented Aug 29, 2017 at 22:33

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