Let $\rho \in (0,1)$ and $n \in \mathbb{N}$. I'm trying to show that $$ \frac{n(1+\rho)}{2^n} \sum_{i=0}^n \frac{\binom{n}{i}}{n+(3n-4i)\rho} \to 1 $$ as $n \to \infty$. I have no idea how to show this, I thought perhaps using the standard bounds for the binomial and upper bounding with an exponential would produce something helpful but I don't quite see hwo to make it work. Any ideas?


Using the binomial theorem I managed to write the expression as $$ \frac{n(1+\rho)}{2^n} \sum_{i=0}^n \frac{\binom{n}{i}}{n+(3n-4i)\rho} = \frac{n(1+\rho)}{2^n} \int_0^1 t^{3\rho n + n - 1}(1+t^{-4\rho})^n \, \mathrm{d}t $$

I thought this was perhaps helpful but the final integral evades me. Any ideas?

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    $\begingroup$ Proposed outline: in the usual sum $\sum_{i=0}^n \binom ni = 2^n$, the bulk of the mass in the sum comes from terms very near $i=\frac n2$, say with $|i-\frac n2| \le n^{2/3}$. Separate the contribution of these terms, for which $n+(3n-4i)\rho$ is very close to $n(1+\rho)$, from the rest of the terms. $\endgroup$ – Greg Martin Jun 24 '20 at 8:02
  • $\begingroup$ @GregMartin I think I see the intuition of what you are suggesting and I can get that the equivalent statement for the sum in question would be $|i-n/2| < n^{2/3}/(4\rho)$ but I'm not sure how to deal with each term after splitting. Do you, perhaps, have a reference where a similar problem is solved regarding the bulk of the mass of a standard binomial sum? $\endgroup$ – Lundborg Jun 24 '20 at 9:44

While the hint in @Greg Martin's comment is ok, it's a bit tedious. In fact, most of the heavy lifting is done already in the law of big numbers: if $X_i$ are i.i.d. random variables with $P(X_i=0)=P(X_i=1)=\frac12$ and $\displaystyle S_n=\sum^n_{i=1}X_i,$ we may recognize $$P(S_n=i)=2^{-n}\binom{n}{i},$$ so we have $$\frac{n(1+\rho)}{2^n} \sum_{i=0}^n \frac{\binom{n}{i}}{n+(3n-4i)\rho}=(1+\rho)\,\mathbb{E}\frac1{1+(3-4S_n/n)\rho}.$$

Now $S_n/n\to\frac12$ a.s. as $n\to\infty$ due to the law of large numbers, and from the dominated convergence theorem, we know that with $$Y_n=\frac1{1+(3-4S_n/n)\rho},$$ we have $$\lim_{n\to\infty}\mathbb{E}Y_n=\mathbb{E}\lim_{n\to\infty}Y_n,$$ since $1+(3-4S_n/n)\rho\ge1-\rho,$ i.e. $Y_n\le\frac1{1-\rho}.$ Thus, $$(1+\rho)\,\mathbb{E}\frac1{1+(3-4S_n/n)\rho}\to\frac{1+\rho}{1+(3-4\cdot\frac12)\rho}=1.$$

  • $\begingroup$ I'm sorry but are you sure this is valid? I don't follow your derivation and my immediate reading screams $\mathbb{E}(g(X)) \neq g(\mathbb{E}(X))$ but perhaps I am missing something. $\endgroup$ – Lundborg Jun 24 '20 at 14:30
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    $\begingroup$ @ProfessorVector: your insulting tone is unnecessary and harmful. This is true regardless of whether your answer has mathematical issues or not. It's alarming to see somebody who has been on this StackExchange for over three years demonstrate so little understanding of its basic code of conduct. $\endgroup$ – Greg Martin Jun 24 '20 at 16:57
  • $\begingroup$ After having spent some more time looking at this solution, I now see what you're doing. Its quite a nice solution, however, I would have appreciated more explanation about how you got from the LHS to RHS of the equality. $\endgroup$ – Lundborg Jun 24 '20 at 17:08
  • $\begingroup$ @Greg Martin And the suggestion that I'm unaware of $\mathbb{E}(g(X)) \neq g(\mathbb{E}(X))$ is what? Necessary and harmless, I guess. Some things cut two ways. $\endgroup$ – user436658 Jun 24 '20 at 17:57
  • $\begingroup$ @Lundborg I've edited my answer a bit, I hope it's clearer, now. $\endgroup$ – user436658 Jun 24 '20 at 18:13

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