- #1
exitwound
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I haven't taken math for going on 13 years and I am having a hard time following what this all means. I'm not sure if I'm doing it right.
First of all, can anyone explain to me in English what exactly the Moment Generating Function is? The book does a poor job at it and the teacher wasn't very clear either. In fact, the book *never* says what it is at all, just how it's used.
If you toss a fair die with outcome X, p(x) = 1/6 for x=1,2,3,4,5,6. Find Mx(t).
[tex]M_x(t) = E(e^{tx}) = \sum_{x\in D}e^{tx}p(x) [/tex]
If we start with the Expectation equation above:
[tex]\sum_{x\in D}e^{tx}p(x) = \sum_{x=1}^6 e^{tx}(1/6)[/tex]
[tex]= e^{t\cdot 0}(1/6) + e^{t\cdot 1}(1/6) + e^{t\cdot 2}(1/6) + ... e^{t\cdot 6}(1/6)[/tex]
[tex] = 1 + \frac{e^t}{6} + \frac{e^{2t}}{6} + \frac{e^{3t}}{6} + \frac{e^{4t}}{6} + \frac{e^{5t}}{6} + \frac{e^{6t}}{6}[/tex]
First of all, is this correct so far? And if it is, since it's been umteen years since I've had algebra, what happens here? How can this be simplified?
First of all, can anyone explain to me in English what exactly the Moment Generating Function is? The book does a poor job at it and the teacher wasn't very clear either. In fact, the book *never* says what it is at all, just how it's used.
Homework Statement
If you toss a fair die with outcome X, p(x) = 1/6 for x=1,2,3,4,5,6. Find Mx(t).
Homework Equations
[tex]M_x(t) = E(e^{tx}) = \sum_{x\in D}e^{tx}p(x) [/tex]
The Attempt at a Solution
If we start with the Expectation equation above:
[tex]\sum_{x\in D}e^{tx}p(x) = \sum_{x=1}^6 e^{tx}(1/6)[/tex]
[tex]= e^{t\cdot 0}(1/6) + e^{t\cdot 1}(1/6) + e^{t\cdot 2}(1/6) + ... e^{t\cdot 6}(1/6)[/tex]
[tex] = 1 + \frac{e^t}{6} + \frac{e^{2t}}{6} + \frac{e^{3t}}{6} + \frac{e^{4t}}{6} + \frac{e^{5t}}{6} + \frac{e^{6t}}{6}[/tex]
First of all, is this correct so far? And if it is, since it's been umteen years since I've had algebra, what happens here? How can this be simplified?