1. ## mgf

$P(X = k) = \left(\begin{array}{cc}n\\k\end{array}\right)p^kq^ {n-k}$ k=0,...,n.
Where n is a positive integer, 0 < p < 1 , q = 1 - p.

How would I show the mgf of X is

$M_{x}(t) = (pe^t +q)^n\ \ -\infty < t < \infty$

Also, if $Y = \Sigma_{1\le i \le m} X_{i}$

how would I derive the mgf of Y?

2. Hello,

The mgf of a discrete distribution is :

$M_X(t)=\sum_{k=0}^\infty \mathbb{P}(X=k)e^{kt}$

So here, we have :

\begin{aligned} M(t)
&=\sum_{k=0}^n {n\choose k} e^{kt}p^k q^{n-k} \\
&=\sum_{k=0}^n {n\choose k} (pe^t)^k q^{n-k}\end{aligned}

Now just recall Newton's binomial formula :

$(a+b)^n=\sum_{k=0}^n a^k b^{n-k}$

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 $[M(t)]^m$

3. Originally Posted by bigdoggy
$P(X = k) = \left(\begin{array}{cc}n\\k\end{array}\right)p^kq^ {n-k}$ k=0,...,n.
Where n is a positive integer, 0 < p < 1 , q = 1 - p.

How would I show the mgf of X is

$M_{x}(t) = (pe^t +q)^n\ \ -\infty < t < \infty$

Also, if $Y = \Sigma_{1\le i \le m} X_{i}$

how would I derive the mgf of Y?
See here: http://www.mathhelpforum.com/math-he...tribution.html

4. So if the mgf is

$M_{x}(t) = (pe^t +q)^n\ \ -\infty < t < \infty$
then the mgf of Y would be $(pe^t +q)^{mn}\ \mbox{where}\ Y = \Sigma_{1\le i \le m}X_{i} \mbox{ with }\ X_{i}$ independent??

How would I find the distribution of Y,and does E[Y]= M'(0) and V[Y]= M''(0)??

5. Originally Posted by bigdoggy
So if the mgf is

$M_{x}(t) = (pe^t +q)^n\ \ -\infty < t < \infty$
then the mgf of Y would be $(pe^t +q)^{mn}\ \mbox{where}\ Y = \Sigma_{1\le i \le m}X_{i} \mbox{ with }\ X_{i}$ independent??

How would I find the distribution of Y,and does E[Y]= M'(0) and V[Y]= M''(0)??

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

$(pe^t+q)^n(pe^t+q)^m=(pe^t+q)^{n+m}$

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

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

$X_1+\cdots +X_m\sim Bin(nm,p)$

6. Why have you mentioned the mgf of X + Y?

I came with the mgf of Y to be $(pe^t +q)^{mn}$ by using the post from Moo, which used $M_Y(t) = (pe^t +q)^{mn}$

Is this correct??

Also is E[Y]= M'(0) and V[Y]= M''(0) correct??

7. Originally Posted by bigdoggy
Why have you mentioned the mgf of X + Y?

I came with the mgf of Y to be $(pe^t +q)^{mn}$ by using the post from Moo, which used $M_Y(t) = (pe^t +q)^{mn}$

Is this correct??

Also is E[Y]= M'(0) and V[Y]= M''(0) correct??
NO, E[Y]= M'(0)=nmp BUT E[Y^2]= M''(0) SO V(Y)=nmpq
AS I wrote 30 minutes ago.

8. Originally Posted by matheagle
NO, E[Y]= M'(0)=nmp BUT E[Y^2]= M''(0) SO V(Y)=nmpq
AS I wrote 30 minutes ago.
Do you mean no, the mgf of Y isn't $(pe^t +q)^{mn}$??
What is it here??

9. It's $(pe^t+q)^{mn}$.. matheagle didn't deny it.

The problem is your formula for the variance...

We know that $\mathbb{E}(X^k)=M^{(k)}(0)$ (k-th derivative)

But $V(X)=\mathbb{E}(X^2){\color{red}-[\mathbb{E}(X)]^2}=M''(0)-[M'(0)]^2$, and not what you wrote.

10. Thanks again Moo!!