MGF of a sum = Product of MGFs

Question

So my book says the following:

The moment generating function for the geometric distribution is: $g(t) = \cfrac{pe^t}{1-e^t(1-p)}$ where $|e^t(1-p)| <1$ of course for the geometric series to converge.

The moment generating function for negative binomial distribution is: $$ g(t) = \left(\frac{pe^t}{1-e^t(1-p)}\right)^r, $$ where $r$ is the number of geometric random variables you sum up for the negative binomial.

My books says that you just take the moment generating function for the geometric distribution to the r power to find the moment generating function for the negative binomial distribution as the moment generating function of a sum of independent random variables is equal to the product of each variables moment generating function, but is my question is why is the bolded statement true? Can someone explain this to me?

Thank You!

What have you tried so far? Can you do it for two random variables? — saulspatz, Apr 27 '21 at 16:49
YoGoodSlope, are you sure you understand why $e^{tX}$ and $e^{tY}$ are independent just because $X$ and $Y$ are independent? i'm curious how probability without measure theory is taught in your classes — BCLC, Apr 28 '21 at 00:09

score 3 · Accepted Answer · answered Apr 27 '21 at 16:51

3

HINT

Let $X,Y$ be two independent random variables and $Z=X+Y$. Then, $$ M_Z(t) = \mathbb{E} \left[e^{tZ}\right] = \mathbb{E} \left[e^{t(X+Y)}\right] = \mathbb{E} \left[e^{tX} e^{tY}\right] $$

can you take it from here?

answered Apr 27 '21 at 16:51

gt6989b

54,422

Comments are not for extended discussion; this conversation has been moved to chat. – Xander Henderson Apr 28 '21 at 14:26

MGF of a sum = Product of MGFs

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