2
$\begingroup$

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 $e^{-k^2}$?

$\endgroup$
  • $\begingroup$ Please use a more descriptive title for your question. Avoid generic words like "question" and "exercise" that would apply to many or most other questions. Also avoid repeating tag names like "probability theory"; tags are broad categories, and the title should locate the question within the field of the tag. Also, I removed the probability-theory tag; please avail yourself of the tag summaries when choosing tags. $\endgroup$ – joriki Mar 20 '20 at 12:14
  • $\begingroup$ See math.stackexchange.com/q/3261840/321264 $\endgroup$ – StubbornAtom Mar 20 '20 at 12:31
  • $\begingroup$ The title has been changed! I agree that I should have avoided such generic words. $\endgroup$ – Vic Ryan Mar 20 '20 at 13:23
1
$\begingroup$

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 $\mathrm e^{-k^2}$.

$\endgroup$
  • $\begingroup$ About the second term of the 3rd step, the index of the sum should be $j = 1$ right? and we sum this second term up to $k$, I believe. Further, could you please explain how the 3rd step is equal to the 4th step? I have written out the summation but I did not obtain the expression that is stated there. $\endgroup$ – Vic Ryan Mar 20 '20 at 13:19
  • 1
    $\begingroup$ @VicRyan: Yes, I've fixed the summation. About the summations: The sum of the positive integers up to $n$ is $\frac{n(n+1)}2$, so the sums up to $k-1$ and $k$ are $\frac{(k-1)k}2$ and $\frac{k(k+1)}2$, respectively. $\endgroup$ – joriki Mar 20 '20 at 16:13
0
$\begingroup$

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 $$\lim_{m \to \infty}(a(m, k)^m) = e^{-k^2}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.