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The expected number of returns of a symmetric random walk is given by

$\sum_{k=0}^n \binom{2k}{k} / 2^{2k} -1$

The exercise is to compute an explicit form for this. I tried to do this in the following way:

$\sum_{k=0}^n \binom{2k}{k} / 2^{2k}= \sum_{k=0}^n 1/(2 \pi i) \int_{|z|=0.5} 1/2^{2k}(1+z)^{2k}/ z^{k+1} dz $

$=1/(2 \pi i) \int_{|z|=0.5}\sum_{k=0}^n 1/2^{2k}((1+z)^2)^k/ z^{k+1}dz $

From here one can use geometric series and simplifying the double ratio to arrive at $1/(2 \pi i) \int_{|z|=0.5} 1/2^{2n}((1+z)^{2n+2}- 2^{2n+2} z^{n+1})/ ((z-1)^2z^{n+1})dz$

The monomial in the numerator with power n has a coefficient of 2n+2 choose n. All other terms vanish when applying the residue theore. Hence the result should be $\binom{2n+2}{n}$. But I know from the lecture that it is not (also from the asymptotic behavior) I cant figure out what I did wrong.

uniquehorn
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1 Answers1

2

Here is an answer that repeats some of the material from the question but typeset slightly differently.

We seek to compute $$\sum_{k=0}^n {2k\choose k} 2^{-2k}.$$

Introduce the integral representation $${2k\choose k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2k}}{z^{k+1}} \; dz.$$

This gives for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z} \sum_{k=0}^n \frac{(1+z)^{2k}}{z^k} 2^{-2k} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z} \sum_{k=0}^n \left(\frac{(1+z)^2}{4z}\right)^k \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z} \frac{((1+z)^2/4/z)^{n+1}-1}{(1+z)^2/4/z-1} \; dz \\ = \frac{4}{2\pi i} \int_{|z|=\epsilon} \frac{((1+z)^2/4/z)^{n+1}-1}{(1-z)^2} \; dz.$$

The second term in the difference does not contribute and we are left with $$\frac{4}{2\pi i} \int_{|z|=\epsilon} \frac{1}{(1-z)^2} ((1+z)^2/4/z)^{n+1} \; dz \\ = \frac{4}{4^{n+1}}\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{(1-z)^2} (1+z)^{2n+2} \; dz .$$

Extracting coefficients from this we obtain $$\frac{1}{2^{2n}} \sum_{q=0}^n {2n+2\choose q} (n-q+1).$$

Here we have calculated $$[z^n] \frac{1}{(1-z)^2} (1+z)^{2n+2}$$ by matching the first $n+1$ powers $z^q$ up to $z^n$ from the polynomial with the corresponding power $z^{n-q}$ from the rational term.

There are two pieces here, the first is: $$\frac{n+1}{2^{2n}} \sum_{q=0}^n {2n+2\choose q} = \frac{n+1}{2^{2n}} \frac{1}{2} \left(2^{2n+2} - {2n+2\choose n+1}\right).$$

The second piece is: $$- \frac{1}{2^{2n}} \sum_{q=0}^n {2n+2\choose q} \times q = - \frac{1}{2^{2n}} \sum_{q=1}^n {2n+2\choose q} \times q \\ = - \frac{2n+2}{2^{2n}} \sum_{q=1}^n {2n+1\choose q-1} = - \frac{2n+2}{2^{2n}} \sum_{q=0}^{n-1} {2n+1\choose q} \\ = - \frac{2n+2}{2^{2n}} \frac{1}{2} \left(2^{2n+1} - 2 {2n+1\choose n}\right).$$

The nonbinomial term in the first piece is $$\frac{n+1}{2^{2n}} \frac{1}{2} 2^{2n+2} = 2n+2$$ and in the second piece $$- \frac{2n+2}{2^{2n}} \frac{1}{2} 2^{2n+1} = -(2n+2)$$ so these two cancel.

This leaves $$\frac{n+1}{2^{2n}} \left(2{2n+1\choose n} - \frac{1}{2}{2n+2\choose n+1}\right)$$ which is $$\frac{n+1}{2^{2n}} \left(2 \frac{n+1}{2n+2} {2n+2\choose n+1} - \frac{1}{2}{2n+2\choose n+1}\right) \\= \frac{n+1}{2^{2n}} \frac{1}{2} {2n+2\choose n+1} = \frac{n+1}{2^{2n+1}} {2n+2\choose n+1}.$$

Apparently this method is due to Egorychev.

Marko Riedel
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