Binomial Distribution Probability

[azdang]

Let X be a Binomial B($$\frac{1}{2}$$,n), where n=2m.

Let a(m,k) = $$\frac{4^m}{(\stackrel{2m}{m})}P(X = m + k)$$.

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

So far, I've found that P(X = m+k) = $$(\stackrel{2m}{m+k}) \frac{1}{4^m}$$

Then, a(m,k)=$$\frac{m!m!}{(m+k)!(m-k)!}.$$

But I have no idea how to show that the limit of $$a^2$$ will be equal to $$e^{-k^2}$$.

I think my work up to that point seems okay. I got things to cancel out, so that is usually a good sign. Any hints? Thank you!

 

[Billy Bob]

I got what you got, and this is not a good sign. If we are correct, then look at the k=3 case.

a(m,3) simplifies to $$\frac{m(m-1)(m-2)}{(m+3)(m+2)(m+1)}$$ and the limit is 1 as m approaches infinity.

 

[azdang]

Well, we are looking for the limit of $$a^2$$. Does that make a difference?

Also, you said you got the same thing as me, but I'm a little confused how what you wrote (even though I know it's for the specific k=3 case) is the same as what I got. I'll look at it again to see if I can see the connection, but maybe what I got for a was wrong?

 

[Billy Bob]

$$a(m,3)=\frac{m!m!}{(m+3)!(m-3)!}=\frac{m!\cdot m(m-1)(m-2)(m-3)!}{(m+3)(m+2)(m+1)m!\cdot (m-3)!} =\frac{m(m-1)(m-2)}{(m+3)(m+2)(m+1)}$$

and

$$\lim_{m\to\infty}a(m,3)^2=\left(\lim_{m\to\infty}a(m,3)\right)^2=1^2=1$$



I think you should check if the problem is stated correctly. I get the same thing as you did for a(m,k) every time I check.

 

[azdang]

Yep, that's definitely what the problem says so, I'm not sure.

What you've shown definitely seems right and I'm assuming it works for other cases when k is something else, as well. I'll check it out and ask my teacher about it if need be. Thanks!