In their classical paper on fluctuations in coin tossing [On Fluctuations in Coin-Tossing](https://www.pnas.org/content/35/10/605), Chung and Feller give a precise formula for the conditional probability of the number of positive “sides” of a random walk with an even number of steps, given a particular outcome for the endpoint.
$$
\mathbf{P}\left(\sum\_{j=1}^{2n} \mathbf{1}\_{(0,\infty)}\left(\frac{\mathsf{X}\_{j-1}+\mathsf{X}\_j}{2}\right)=2k\, \Bigg|\, \mathsf{X}\_{2n}=2\ell\right)
$$
is equal to
$$
\frac{\ell}{\binom{2n}{n-\ell}}\, \sum\_{i=\ell}^{k} \binom{2i}{i-\ell} \binom{2n-2i}{n-i} \cdot \frac{1}{i(n-i+1)}
$$
for a simple random walk $(\mathsf{X}\_0,\mathsf{X}\_1,\dotsc,\mathsf{X}\_{2n})$.


For their Theorem 1
$$
\mathbf{P}\left(\sum\_{j=1}^{2n} \mathbf{1}\_{(0,\infty)}\left(\frac{\mathsf{X}\_{j-1}+\mathsf{X}\_j}{2}\right)=2k\right)\,
=\, u\_{2k} u\_{2n-2k}\, ,
$$
for
$$
u\_0=1\, ,\ \text{ and }\ u\_{2k}\, =\, \mathbf{P}(\mathsf{X}\_{2k}=0)\, =\, \frac{1}{2^{2k}} \binom{2k}{k}\, ,\ \text{ for $k=1,2,\dots$,}
$$
the generalization to odd times was performed somewhat recently by Gessel in slides


[Chung–Feller Theorems](https://people.brandeis.edu/%7Egessel/homepage/slides/chung-feller-slides.pdf) last few slides,


and by Grünbaum in an article


[A Feynman–Kac approach to a paper of Chung and Feller on fluctuations in the coin-tossing game](https://arxiv.org/abs/1810.06092v1)


published in [Proceedings of the American Mathematical Society](https://doi.org/10.1090/proc/14758). The answer for that generalization is
$$
\mathbf{P}\left(\sum\_{j=1}^{2n+1} \mathbf{1}\_{(0,\infty)}\left(\frac{\mathsf{X}\_{j-1}+\mathsf{X}\_j}{2}\right)=2k+1\right)\,
=\, u\_{2k} u\_{2n-2k} \cdot \frac{2k+1}{2(n+1)}\, .
$$
Incidentally, the problem was apparently also stated as an exercise in McKean's textbook:


[Probability: The classical limit theorems](https://www.cambridge.org/us/academic/subjects/mathematics/abstract-analysis/probability-classical-limit-theorems?format=HB&isbn=9781107053212) Exercise 3.4.2 on p139.


But to the best of my knowledge nobody has stated a generalization of Chung and Feller's conditional distribution results in Theorem 2 to odd numbers of steps. (I might have missed it somewhere.)


*Note the analog of their Theorem 2a was the limit/specialization of their Theorem 2 to $\ell=0$ giving a uniform distribution. It makes sense from their Theorem 2a formula if you cancel the $\ell$ in the numerator outside their sum with the $i$ (which is necessarily $0$ in the single-summand sum) in the denominator inside the sum.*


As a secondary question, I wonder why combinatorialists sometimes derive useful formulas and advertise them as Gessel (and possibly McKean?) did, but then do not publish them. I have asked that question specifically over at AcademiaSE with more references for an example related to this question:


[Is it common for combinatorialists to not publish all their results? If so, why?](https://academia.stackexchange.com/questions/180689/is-it-common-for-combinatorialists-to-not-publish-all-their-results-if-so-why)



What is the generalization of the formula for Chung and Feller's Theorem 2 to odd numbers of steps?
Speaking only for myself, the reason that I don't publish all my results is that I write very slowly. (It has nothing to do with the nature of combinatorics.) It takes a long time for me to arrange my work into something publishable; it's much quicker to put some results into a slide presentation for a talk.


But I'd like to thank you for the references. I do intend to publish this result some day and it's good to know that there are other references to it.


For the benefit of readers who, like myself, think in terms of counting lattice paths rather than sums of random variables associating with random walks, it may be hepful to restate these formulas in terms of lattice paths. We consider paths made up of up steps $(1,1)$ and down steps $(1,-1)$, starting at the origin. (The horizontal components are irrelevant and are included only for convenience in visualization.) The result of Chung and Feller is that among the $4^n$ paths of length $2n$, the number with $2k$ steps above the $x$-axis is $\binom{2k}{k}\binom{2n-2k}{n-k}$. The analogous result for paths of odd length is that the number of paths of length $2j + 2k − 1$ with $2j − 1$ steps above the $x$-axis and $2k$ steps below is
$\frac{j}{2(j+k)}\binom{2j}{j}\binom{2k}{k}$. (A path of odd length cannot end on the $x$-axis; if it ends above the $x$-axis it must have an odd number of steps above the $x$-axis. Switching $j$ and $k$ allows us to count paths with an even number of steps above the $x$-axis.)


