Calculating $\lim_{n \to \infty}(a(m,k))^m$ where $a(m,k) = \frac{4^m}{{2m \choose m}}P(X = m+k)$ and $X \sim B(n,1/2)$

Question

I have a question about an exercise from the book Probability Essential, 2nd edition (Jacod, J. & Protter, P., 2004, p. 34). It is from a chapter 5 called "Random Variables on a Countable Space". The question is formulated in the following way: "Let $X$ be Binomial $B(1/2,n)$ where $n = 2m$. Let $$a(m,k) = \frac{4^m}{{2m \choose m}}P(X = m+k)$$ Show that $\lim_{n \to \infty}(a(m,k))^m = e^{-k^2}$." But what is $B(1/2, n)$? or maybe they mean $B(n,1/2)$? Assume that it should be $B(n, 1/2)$, how can I prove this limit? So this is my approach: \begin{align*} a(m,k) &= \frac{4^m}{{2m \choose m}}P(X = m+k)\\ &= \frac{4^m}{{2m \choose m}}{2m \choose m+k}\frac{1}{2^{m+k}}\frac{1}{2^{2m - m - k}}\\ &= \frac{4^m}{\frac{(2m)!}{m!m!}}\frac{(2m)!}{(m+k)!(m-k)!}\frac{1}{2^{m+k}}\frac{1}{2^{m - k}}\\ &= \frac{4^m}{(m!)^2}\frac{(2m)!}{(m+k)!(m-k)!}\frac{1}{2^{2m}}\\ &= \frac{4^m}{(m!)^2}\frac{(2m)!}{(m+k)!(m-k)!}\frac{1}{4^m}\\ &= \frac{1}{(m!)^2}\frac{(2m)!}{(m+k)!(m-k)!} \end{align*} Can I further simplify this? I don't see quiet yet how $(a(m,k))^m$ converges to $e^{-k^2}$ as $m \to \infty$. Should I find an upper bound such as $$\left(1-\frac{k^2}{m}\right)^m$$ since it converges to $e^{-k^2}$?

Answer 1

You put $m!^2$ in the denominator instead of the numerator, and you cancelled one factor of $(2m)!$ but not the other. Without these mistakes, you get

\begin{eqnarray} \frac{m!}{(m-k)!}\frac{m!}{(m+k)!} &=& \frac{m(m-1)\cdots(m-k+1)}{(m+k)(m+k-1)\cdots(m+1)} \\ &=& \frac{\left(1-\frac1m\right)\cdots\left(1-\frac{k-1}m\right)}{\left(1+\frac km\right)\cdots\left(1+\frac1m\right)} \\ &=& 1-\sum_{j=1}^{k-1}\frac jm-\sum_{j=1}^k\frac jm+O\left(m^{-2}\right) \\ &=& 1-\frac{(k-1)k}2\cdot\frac1m-\frac{k(k+1)}2\cdot\frac1m+O\left(m^{-2}\right) \\ &=&1-\frac{k^2}m+O\left(m^{-2}\right)\;, \end{eqnarray}

and then raising to the $m$-th power and taking the limit $m\to\infty$ yields $\mathrm e^{-k^2}$.

Answer 2

Alright, so it appears that I have made a mistake in my own approach. Thanks to the comments I think I know how to prove the limit now. We have \begin{align*} a(m,k) &= \frac{4^m}{{2m \choose m}}P(X = m+k)\\ &= \frac{4^m}{{2m \choose m}}{2m \choose m+k}\frac{1}{2^{m+k}}\frac{1}{2^{2m - m - k}}\\ &= \frac{4^m}{\frac{(2m)!}{m!m!}}\frac{(2m)!}{(m+k)!(m-k)!}\frac{1}{2^{m+k}}\frac{1}{2^{m - k}}\\ &= 4^m{(m!)^2}\frac{1}{(m+k)!(m-k)!}\frac{1}{2^{2m}}\\ &= 4^m{(m!)^2}\frac{1}{(m+k)!(m-k)!}\frac{1}{4^m}\\ &= \frac{m!}{(m+k)!}\frac{m!}{(m-k)!}\\ &= \frac{m(m-1)\cdots(m-k+1)}{(m+k)(m+k-1)\cdots(m+1)} = \frac{\prod_{j=1}^{k-1}(1 - j/m)}{\prod_{j=1}^k(1 + j/m)} \end{align*} Therefore, as $m \to \infty$, $$(a(m,k))^m \to \frac{\prod_{j=1}^{k-1}e^{-j}}{\prod_{j=1}^ke^j}.$$ Simplifying its limit: \begin{align*} \frac{\prod_{j=1}^{k-1}e^{-j}}{\prod_{j=1}^ke^j} &= \frac{e^{-1}e^{-2}\cdots e^{-k+1}}{e^ke^{k-1}\cdots e^2e^1}\\ &= \frac{e^{-1 - k}e^{-2 - k + 1}e^{-3-k +2}\cdots e^{-k + 1 - 2}}{e^1}\\ &= \frac{e^{-1 - k}e^{-1 - k} \cdots e^{-1 - k}}{e^1}\\ &= \frac{e^{(k-1)(-1 - k)}}{e^1} \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{there are} \ k - 1 \ \text{terms})\\ &= \frac{e^{-k^2 + 1}}{e^1} = e^{-k^2}. \end{align*} We conclude that $$\lim_{m \to \infty}(a(m, k)^m) = e^{-k^2}.$$