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math-340:m340-f15-hw:hw-21 [2015/10/27 18:24]
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math-340:m340-f15-hw:hw-21 [2015/11/03 18:14] (current)
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 **For Problem Set due 2 November**: 18 **For Problem Set due 2 November**: 18
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 +Answers:
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 +  - $M_X(t) = \frac{1}{b - a}\int_a^b e^{tx}dx =  \frac{e^{tb} - e^{ta}}{t(b - a)}$
 +  - $M_X(t) = \sum_{k=0}^\infty e^{tk}q^kp = p\sum_{k=0}^\infty (qe^t)^k = \frac{p}{1 - qe^t}$ provided $qe^t < 1$.
 +  - $M_X'​(t) = \frac{6}{(1 - 2t)^4}$ and $M_X''​(t) = \frac{48}{(1 - 2t)^5}$, so $E[X] = M_X'​(0) = 6$, $E[X^2] = M_X''​(0) = 48$, and $\text{Var}(X) = 48 - 36 = 12$
 +  - $M_X(t) = \frac{\lambda}{\lambda - t} = \frac{1}{1 - \frac{t}{\lambda}} = \sum_{k=0}^\infty \frac{t^k}{\lambda^k}$. Hence $\frac{\mu_k}{k!} = \frac{1}{\lambda^k}$.