k=0,...,n.

Where n is a positive integer, 0 < p < 1 , q = 1 - p.

How would I show the mgf of X is

Also, if

how would I derive the mgf of Y?

Results 1 to 10 of 10

- August 19th 2009, 04:01 AM #1

- Joined
- Nov 2008
- Posts
- 121

- August 19th 2009, 04:17 AM #2
Hello,

The mgf of a discrete distribution is :

So here, we have :

Now just recall Newton's binomial formula :

As for the second one, if the Xi's are independent, then the mgf of Y is the product of the mgf of the Xi.

If they're identically distributed, this will just be

- August 19th 2009, 04:29 AM #3

- August 19th 2009, 01:49 PM #4

- Joined
- Nov 2008
- Posts
- 121

- August 19th 2009, 03:24 PM #5

The point of using MGF's is that you hope to recognize the resulting one.

Here you added m INDEPENDENT binomials.

This techniques proves that the result is a binomial with same probability of success (p) and now the number of trials is nm.

The MGF of X+Y where they are indep bin(n,p) and bin(m,p), SAME p, different sample sizes is

proving that X+Y is bin(n+m,p)

So, for you the mean is (nm)p and the variance is (nm)pq.

- August 19th 2009, 04:33 PM #6

- Joined
- Nov 2008
- Posts
- 121

- August 19th 2009, 04:44 PM #7

- August 19th 2009, 04:56 PM #8

- Joined
- Nov 2008
- Posts
- 121

- August 20th 2009, 04:45 AM #9

- August 20th 2009, 05:01 AM #10