Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

how to prove $$\sum _{|k|\lt\sqrt m}\binom{2m}{m+k}\ge2^{2m-1},\forall m\ge1$$ Thanks in advance .

share|improve this question
add comment

1 Answer

up vote 7 down vote accepted
+50

Consider tossing a fair coin $2m$ time and let $X$ count the number of heads. Then from Chebyshev inequality, we have that $$\mathbb{P}(\vert X - m\vert \geq k \sqrt{m/2}) \leq \dfrac1{k^2}$$ Take $k =\sqrt{2}$, to get that $$\mathbb{P}(\vert X - m\vert \geq \sqrt{m}) \leq \dfrac12, \,\,\, \text{i.e., }\,\,\, \mathbb{P}(\vert X - m\vert < \sqrt{m}) \geq \dfrac12$$ Hence, we have $$\sum_{\vert k \vert <\sqrt{m}} \dfrac{\dbinom{2m}{m+k}}{2^{2m}} \geq \dfrac12$$ which gives us $$\sum_{\vert k \vert <\sqrt{m}} \dbinom{2m}{m+k} \geq 2^{2m-1}$$

share|improve this answer
    
how did you go from the second line to the third line? –  AlanH Mar 24 '13 at 6:36
    
@AlanH Note that $X$ can take values only from $1,2,3,\ldots,2m$. We need $\vert X - m \vert < \sqrt{m} \implies X$ can take values take integers in the domain $(m-\sqrt{m}, m+\sqrt{m})$. Since $X$ is binomial random variables, we have $$P(X = \ell) = \dfrac1{2^{2m}} \dbinom{2m}{\ell}$$ Hence, $$P(\vert X - m \vert < \sqrt{m}) = P(X = m-\lfloor \sqrt{m} \rfloor + 1) + P(X = m-\lfloor \sqrt{m} \rfloor + 2) + \cdots + P(X = m+\lceil \sqrt{m} \rceil - 1)$$ –  user17762 Mar 24 '13 at 6:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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