# Thread: Help understanding how to use MGF to calculate first moment of Binomial Distribution

1. ## Help understanding how to use MGF to calculate first moment of Binomial Distribution

I have it all in front of me, I'm just having a bit of trouble understanding the math.

$\phi(t)=E[e^{tX}]$

$=\sum_{k=0}^{n}e^{tk}$ ${n}\choose{k}$ $p^{k}(1-p)^{n-k}$
$=\sum_{k=0}^{n}$ ${n}\choose {k}$ $(pe^{t})^{k}(1-p)^{n-k}$

Then all of a sudden the summation and combination are gone and the book has

$=(pe^{t}+1-p)^n$

It goes on to show the differentiation and whatnot, which I understand. I just don't understand what was done with the combination and the summation. Any help would be appreciated.

2. ## Re: Help understanding how to use MGF to calculate first moment of Binomial Distribut

Originally Posted by downthesun01
I have it all in front of me, I'm just having a bit of trouble understanding the math.

$\phi(t)=E[e^{tX}]$

$=\sum_{k=0}^{n}e^{tk}$ ${n}\choose{k}$ $p^{k}(1-p)^{n-k}$
$=\sum_{k=0}^{n}$ ${n}\choose {k}$ $(pe^{t})^{k}(1-p)^{n-k}$

Then all of a sudden the summation and combination are gone and the book has

$=(pe^{t}+1-p)^n$

It goes on to show the differentiation and whatnot, which I understand. I just don't understand what was done with the combination and the summation. Any help would be appreciated.
$(pe^t+1-p)^2=( (pe^t) + (1-p) )^2$

Now consider the binomial expansion of $(a+b)^n$ where $a=pe^t$ and $b=(1-p)$

CB

3. ## Re: Help understanding how to use MGF to calculate first moment of Binomial Distribut

Thank you for responding, but I don't know what you're talking about.

4. ## Re: Help understanding how to use MGF to calculate first moment of Binomial Distribut

Ok, I looked up binomial expansion. I don't understand the math, but apparently:

$(x+y)^{n}=\sum_{k=0}^{n}$ ${n}\choose{k}$ $x^{k}y^{n-k}$

I'll try to see if I can figure this all out now. Thanks