1. moment generating function

Hello,

I want to show using moment generating function, that for a very big n, the binomail distribution is approximately normal.

I need to show that for n->inf lim (pe^t+1-p)^n equals the mgf of the normal distribution, can anyone show me how to do it ?

2. I need to show that for n->inf lim (pe^t+1-p)^n equals the mgf of the normal distribution, can anyone show me how to do it ?
I suspect they can't, considering this is false. For t > 0, this goes to infinty, for t = 0 it goes to 1 and for t < 0 it goes to 0. If you know the CLT, that works in this case, and you can mimic the proof of it if you want to use MGFs to solve this.

3. not trying to hijack thread, but if the proposition is false how can the binomial converge to the normal distribution? i thought it was impossible to have to RVs with the same distribution and different MGFs.

Edit:

I had a play with google. Could not find a complete solution but one text suggested trying a taylor expansion of the MGF, then taking limits of that.

http://www.aiaccess.net/English/Glos...m#tut_binomial

4. trying the taylor expansion

binomial mgf
$[(1-p)+pe^t]^n$

do a taloyr expansion with
f(x) = x^n
x=1-p
h=pe^t

$f(x+h) = [1-p + pe^t]^n \approx (1-p)^n + pe^t \times n(1-p)^{n-1}$

hmm...stuck, would need to show that $lim n(1-p)^{n-1} = 1$

5. Originally Posted by SpringFan25
not trying to hijack thread, but if the proposition is false how can the binomial converge to the normal distribution? i thought it was impossible to have to RVs with the same distribution and different MGFs
This is a scaling issue. It is apparent that some sort of convergence is going on, but if you don't scale it right the result is going to either blow up to infinity or go to 0 in distribution. Stating that X converges in distribution to a normal random variable is very different than stating that $\frac{X - np}{\sqrt{np(1 - p)}}$ does. You could try working with the MGF of that if you want, but proving the CLT (assuming that you take for granted that point wise convergence of MGF guarantees convergence in distribution) is quite easy.