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I have a question about a sum when calculating moment generating function. The question is : "Find the moment generating function for each of these two random variables. (i) $X$ = outcome a die toss, $p(x) = P[X= x] = \frac{1}{6}$ for $x = 1,2,3,4,5,6$. I don't understand how the sum in the answer $e^t \cdot \frac{1}{6} + e^{2t} \cdot \frac{1}{6} +...+ e^{6t}\cdot \frac{1}{6} = \frac{1}{6}e^t\frac{e^{6t}-1}{e^t-1}$. Why doesn't it work to have this as the sum of a finite geometric series with the common ratio being $e^t$?

quietkid
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  • The standard geometric series goes like $1+x+\cdots$. You have something like $x+x^2+\cdots=x(1+x+\cdots)$. – Hamed Oct 23 '18 at 20:47
  • thank you! I see, so I need to factor out $e^t$. Also I think I just got confused because in the end you can factor out -1 of the final fraction from $\frac{1-e^{6t}}{1-e^t}$, I think. – quietkid Oct 23 '18 at 20:54

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I think it doesn't work with finite geometric series because $e$ is not rational number, but I am not sure

kr3mn1
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