- #1
azdang
- 84
- 0
Let X be a Binomial B([tex]\frac{1}{2}[/tex],n), where n=2m.
Let a(m,k) = [tex]\frac{4^m}{(\stackrel{2m}{m})}P(X = m + k)[/tex].
Show that [tex]lim_{m->\infty}(a(m,k))^2 = e^{-k^2}[/tex].
So far, I've found that P(X = m+k) = [tex](\stackrel{2m}{m+k}) \frac{1}{4^m}[/tex]
Then, a(m,k)=[tex]\frac{m!m!}{(m+k)!(m-k)!}.[/tex]
But I have no idea how to show that the limit of [tex]a^2[/tex] will be equal to [tex]e^{-k^2}[/tex].
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!
Let a(m,k) = [tex]\frac{4^m}{(\stackrel{2m}{m})}P(X = m + k)[/tex].
Show that [tex]lim_{m->\infty}(a(m,k))^2 = e^{-k^2}[/tex].
So far, I've found that P(X = m+k) = [tex](\stackrel{2m}{m+k}) \frac{1}{4^m}[/tex]
Then, a(m,k)=[tex]\frac{m!m!}{(m+k)!(m-k)!}.[/tex]
But I have no idea how to show that the limit of [tex]a^2[/tex] will be equal to [tex]e^{-k^2}[/tex].
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!