# Limit of a binomial probability

I am doing some self-study on measure theory and probability. This question is taken from Jacod and Protter's Essentials of Probability, Chapter 5 (Q15):

Let $X$ be a binomial random variable, $X\sim \mathrm{Binomial}(p=\frac{1}{2}, n)$, where $n=2m$. Let $a(m,k)=\dfrac{4^{m}}{2m \choose m} P(X=m+k)$. Show that $\lim_{m\to \infty} (a(m,k))^{m}=e^{-k^{2}}$.

So far I have been able to get:

\begin{align*} [a(m,k)]^{m}&=\left[ \frac{4^{m}}{2m \choose m} P(X=m+k) \right]^{m}\\ &=\left[ \frac{4^{m}}{2m \choose m} {2m \choose (m+k)}p^{m+k}(1-p)^{2m-(m+k)} \right]^{m}\\ &=\left[ 4^{m} \frac{(m!)^2}{(m-k)!(m+k)!}\left(\frac{1}{2}\right)^{m+k} \left(\frac{1}{2}\right)^{m-k} \right]^{m}\\ &=\left[\frac{(m!)^2}{(m-k)!(m+k)!}\right]^{m}\\ &\approx\left[\left(\frac{m}{\sqrt{(m-k)(m+k)}}\right)\left(\frac{m}{e}\right)^{2m}\left(\frac{e}{m-k}\right)^{m-k}\left(\frac{e}{m+k}\right)^{m+k}\right]^{m}\text{, by stirling}\\ &=\left[\frac{m^{2m+1}}{(m-k)^{m-k+0.5}(m+k)^{m+k+0.5}}\right]^{m}\\ \end{align*} However, I am unable to get the answer from here. I think I am close, but have been working with it for a while and can't get anywhere. I hate moving on without understanding where I am making a mistake (especially with self-study). If anyone feels up to the challenge any help would be greatly appreciated.

\begin{align} \left[\frac{(m!)^2}{(m-k)!(m+k)!}\right]^{m} & = \left[\frac{m!}{(m-k)!}\frac{m!}{(m+k)!}\right]^{m} \\& = \left[\frac{m\times(m-1)\times\cdots\times(m-k+1)}{(m+k)\times(m+k-1)\times\cdots\times(m+1)}\right]^{m} \\& = \left[\left(1+\frac{k}{m}\right) \left(1+\frac{k}{m-1}\right)\cdots\left(1+\frac{k}{m-k+1}\right) \right]^{\Large -m} \\& \end{align}

Let $0 \leq u \leq k-1$.

\begin{align} \left(1+\frac{k}{m-u}\right)^{\Large -m} = \left(\left(1+\frac{k}{m-u}\right)^{\Large \frac{m-u}{k}+\frac{u}{k}}\right)^{\Large -k}, \end{align}

where $$\left(1+\frac{k}{m-u}\right)^{\Large \frac{m-u}{k}}=e \;\text{ as } \; m\to \infty$$ and $$\left(1+\frac{k}{m-u}\right)^{\Large \frac{u}{k}}=\left(1+\frac{\Large \frac{k}{m}}{1- \Large \frac{u}{m}}\right)^{\Large \frac{u}{k}}=1 \;\text{ as } \; m\to \infty.$$

Could you do the rest?

• Yes thank you, this solution is very clear! Excellent. – möbius Jun 25 '15 at 21:10

I got $\dfrac{(m!)^2}{(m-k)!(m+k)!}$ where you have $\left( \dfrac{m!}{(m-k)!} \right)^2$.

• A silly mistake on my part. I have corrected the answer, but am still uncertain how to move any further. – möbius Jun 25 '15 at 15:59