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We know that $$\sum_{j=0}^N\binom{N}{j}2^{-N}=(1/2+1/2)^N=1\qquad\text{ and}\qquad\sum_{j=0}^N\binom{N}{j}(-1)^j2^{-N}=(1/2-1/2)^N=0,$$ hence $\sum_{j:0\leq 2j\leq N}\binom{N}{2j}2^{-N}=1/2$. More generally, given $p\in(0,1)$, it seem plusible to me that we can have $$\sum_{j:0\leq n+jk\leq N}\binom{N}{n+jk}p^{n+jk}(1-p)^{N-n-jk}\to 1/k\text{ as }N\to\infty\text{ for any given }k\in\mathbb N\text{ and }n=0,1,\cdots,k-1.$$ But I have no idea how to prove this. Is his true? If so how can one prove it? Can we also give a rate of convergence in terms of $N$?

Equivalent Triangle
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  • I am not quite sure if I understand the summation. It seems to me, you want to only add the multiples of $k$ in the summation, is that correct? In any case, you can first try simulating the results and see if the pattern you are guessing holds. If it does, then it is worth trying to prove, otherwise you might end up trying to prove something that is not true. – sudeep5221 Apr 03 '20 at 13:15
  • Yes, I am summing multiples of $k$. It seems to be obviously true to me because it corresponds to the probability of a random walk with probability $p$ ending up on position $n$ after $N$ steps on $\mathbb Z/k\mathbb Z$, at least when $k$ is odd – Equivalent Triangle Apr 03 '20 at 13:24

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Let's rewrite the sum as $S_N(k,n,p):=\sum\limits_{j=0}^{N}\binom{N}{j}[j\equiv n\ (k)]p^j(1-p)^{N-j}$, where we use $$[j\equiv n\ (k)]:=\begin{cases}1,&j\equiv n\pmod{k}\\0,&\hfill\text{otherwise}\hfill\end{cases}=\frac1k\sum_{s=0}^{k-1}\color{blue}{\zeta_k}^{(j-n)s},$$ where, in turn, $\zeta_k=\exp(2\pi i/k)$ is a primitive $k$-th root of unity. Summing over $j$, we obtain $$S_N(k,n,p)=\frac1k\sum_{s=0}^{k-1}\zeta_k^{-ns}(1-p+p\zeta_k^s)^N.$$ The absolute value of the $s$-th term is $r_s^N$, where $r_s=\sqrt{1-4p(1-p)\sin^2(s\pi/k)}$.

Thus, when $N\to\infty$, only the term with $s=0$ "survives", giving the desired limit of $1/k$.

metamorphy
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