mgf adding two binomials
We are to solve this problem using mgf. technique (transforms of rv)
x1= b(4,1/2) and x2= b(6,1/3), They are independent.
Let Y = X1+X2
We are looking for the mean and variance.
I assume that we need to proceed as follows:
My(t) = E(e^tY) = ....= E(e^tx1)E(e^tx2).
So I get (1/2 + 1/2 * e^t)^4 * (2/3 + 1/3 * e^t)^6
Now I stuck. Dont i need to create an mgf I recognize (binomial I imagine) and pick out the mean and variance from the mgf parameters?
The sum of two indep binomials with the same parameter p is a binomial.
(With N=n_1+n_2 and same p.)
If the paramters are different I do not believe it's a binomial.
MGFs are unique and since we cannot recognize this one as a binomial, it isn't.
HOWEVER, you can get the mean and the second moment by differentiating with respect to t and setting t=0.
And from that you can get the variance, but I don't know why you would use MGFs to find the mean and variance in this problem.
This is a MOMENT generating function, so you can clearly get the MOMENTS by differentiating and setting t=0.
That's how it got its name.
Thanks for the reply - I did forget about trying to differentiate the mgf.
I ended up expanding the terms (1/2 + 1/2 * e^t)^4 * (2/3 + 1/3 * e^t)^6 and then looking at the "weights" on the e^tX terms to create a pmf which I then used to find the mean and variance.
Thanks again for the other tip!