I am reading Walker's book Codes and Curves and am having trouble proving this Lemma regarding the Asymptotic Gilbert-Varshamov bound.

Suppose that $q$ is a prime power and we define \begin{align*} V_q(n,r) &:= \sum\limits_{i=0}^r {n\choose r}(q-1)^i \end{align*}

We define the Hilbert entropy function as \begin{align*} H_q(x) &:= \cases{0, & x= 0\\ x\log_q(q-1)-x\log_q x - (1-x)log_q(1-x), & $0 < x \leq 1-\frac{1}{q}$} \end{align*}

There is a lemma that states if $0\leq\lambda\leq 1-\frac{1}{q}$ then \begin{align*} \lim\limits_{n\to\infty}\frac{1}{n} \log_q V_q(n,\lfloor \lambda n\rfloor) &= H_q(\lambda) \end{align*}

Walker suggests using Stirling's approximation to get this limit. Here is what I have so far: First, I find that if $0<\lambda \leq 1-\frac{1}{q}$ then \begin{align*} H_q(\lambda) &= \lambda\log_q(q-1)-\lambda\log_q \lambda - (1-\lambda)log_q(1-\lambda)\\ &= \log_q\left(\frac{(q-1)^\lambda}{\lambda^\lambda(1-\lambda)^{1-\lambda}}\right) \end{align*}

Then, try to calculate $\lim\limits_{n\to\infty} \frac{1}{n}\log_q V_q(n,\lfloor \lambda n\rfloor)$. \begin{align*} \lim\limits_{n\to\infty} \frac{1}{n}\log_q V_q(n,\lfloor \lambda n\rfloor) &= \lim\limits_{n\to\infty} \log_q\left(\left(\sum\limits_{i=0}^{\lfloor \lambda n\rfloor} {n\choose i}(q-1)^i\right)^\frac{1}{n}\right)\\ &= \log_q\left(\lim\limits_{n\to\infty} \left(\sum\limits_{i=0}^{\lfloor \lambda n\rfloor} {n\choose i}(q-1)^i\right)^\frac{1}{n} \right) \end{align*}

Looking only at the terms inside the logarithm, I would like to show that \begin{align*} \lim\limits_{n\to\infty} \left(\sum\limits_{i=0}^{\lfloor \lambda n\rfloor} {n\choose i}(q-1)^i\right)^\frac{1}{n} &= \frac{(q-1)^\lambda}{\lambda^\lambda(1-\lambda)^{1-\lambda}} \end{align*}

Unfortunately, I get stuck here. This stackexchange post pointed me to this resource which essentially shows the case for $q=2$ in exercise 9.42. It looks easy to generalize to this problem using the provided method. However, I do not quite understand this crucial step:

If we let $m = \lfloor\lambda n\rfloor$, then we get that \begin{align*} {n\choose m}\sum\limits_{i=0}^m \left(\frac{\alpha}{1-\alpha}\right)^i = {n\choose m}\frac{1-\alpha}{1-2\alpha} \end{align*} This step seems so simple based off of geometric series, but I cannot get my calculations into the provided form.

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    $\begingroup$ This might help math.stackexchange.com/questions/136996/… and math.stackexchange.com/questions/160248/… $\endgroup$ – leonbloy Aug 22 at 19:39
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    $\begingroup$ The sum $V_q(n,r)$ is identical to $F_{n,r}(q-1)$ in the paper 'Decoding Generalised Hyperoctahedral Groups...' by R Bailey and T. Prellberg, Contrib. Discrete Mathematcs vol. 7 #1, pgs 1-14. The $F_{n,r}(x)$ is turned into a contour sum and estimated asymptotically as $n \to \infty$ by the saddle point method. There is a function that looks like your entropy function. I have not tried to map the answer to your form, but if the statement is correct, there should exist a mapping. $\endgroup$ – skbmoore Aug 22 at 23:31
  • $\begingroup$ I don't quite get why you equate $ {n\choose i}$ with $n^i$ in the penultimate equation? $\endgroup$ – leonbloy Sep 1 at 17:25
  • $\begingroup$ Sorry for not looking at this post in awhile. It appears that I believed that ${n\choose i}\sim \frac{n^i}{i!}$ through some usage of Stirling's approximation. However, I cannot find the reasoning I used to get here, nor am I able to rework out this calculation. Let me try to fix this. $\endgroup$ – J. Pistachio Sep 1 at 17:32
  • $\begingroup$ @leonbloy I have removed my mistake and added some extra resources and information that I have found. $\endgroup$ – J. Pistachio Sep 1 at 18:14

Here I show that

$$\lim_{t\to \infty} \left(\sum\limits_{k=0}^{at} \frac{t^k}{k!} \right)^{1/t}= \left(\frac{e}{a}\right)^a $$

Letting $n(q-1) = t$ and $a = \frac{\lambda}{q-1}$

$$ \begin{align} \lim\limits_{n\to\infty}\left(\sum\limits_{i=0}^{\lambda n}\frac{n^i}{i!}(q-1)^i \right)^\frac{1}{n}&= \lim\limits_{t\to\infty}\left(\sum\limits_{i=0}^{at}\frac{t^i}{i!}\right)^\frac{q-1}{t}\\ &=\left(\frac{e}{a}\right)^{a(q-1)} \\ &= \left(\frac{q-1}{\lambda}\right)^\lambda e^\lambda \end{align} $$

This does not quite agree with your desired answer. Perhaps the discrepancy is due to an error in your penultimate equation, which looks wrong to me.


The trick in this is to first upper and lower bound $V_q$ by respectively $n$ and $1$ times the max term in the sum, and then take $\log$. Then the game becomes controlling this max term, which is much easier to handle. A key result needed for this is the following lemma, which can be shown using Stirling's approximation:

For any $k \in [1:n-1],$ $$ \frac{1}{n} \ln\binom{n}{k} = (1 + o_n(1)) h\left(\frac{k}{n}\right),$$ where $h(x) := -x\ln x - (1-x) \ln (1-x)$ is the binary entropy function.

You should take a pass at showing this, but see, for intance, this for both a proof and other nice asymptotics of the binomial coefficients. More precise, non-asymptotic statements are also straightforward to get. For instance, this also only uses Stirling's approximation.

Now, let $K:= \lfloor \lambda n \rfloor,$ and $$\varphi := \max_{i \in [1:K]} \binom{n}{i} (q-1)^i.$$ I'll consider the $\lambda > 0$ case, and work with $n$ large enough so that $K \ge 2.$ We have $$ \varphi \le V_q \le K \varphi \le n \varphi,$$ which implies that $$\frac{1}{n} \ln V_q = \frac{1}{n} \ln \varphi + o_n(1).$$ At this point the argument is straightforward. I urge you to take a pass yourself before reading on.

Next, it follows that \begin{align} \frac{1}{n} \ln \varphi &= \max_{i \in [0:K]} \frac{1}{n} \ln \binom{n}{i} + \frac{i}{n} \ln (q-1) \\ &= (1 + o_n(1)) \left\{\max_{i \in [0:K]} h(i/n) + (i/n)\ln (q-1) \right\}, \end{align} where the second line uses the quoted asymptotic equality.

Now note that treated as a function for any real $0 \le x \le 1-1/q$, quantity $$ \rho(x) := h(x) + x \ln(q-1)$$ is non-decreasing in $x$. Indeed, $$\rho' = \ln(q-1) + \ln(1-x/x) \ge \ln(q-1) + \ln(1/q/ (1-1/q) = 0.$$ (Aside: the $H_q$ in your question is the same as $\rho/\ln q$).

This means that $$\frac{1}{n} \ln \varphi = (1 + o_n(1)) \left( h(K/n) + (K/n) \ln(q-1) \right)$$

Finally, $K/n \to \lambda,$ and by continuity $h(K/n) \to h(\lambda)$ finishes the job.


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