Exercises 6.10: 18, 21

Also:

1. Suppose $X$ is uniform on $(a, b)$. Show that the moment generating function of $X$ is $M_X(t) = \frac{e^{tb} - e^{ta}}{t(b - a)}$.
2. Suppose $X$ is geometric with probability of success $p$ and let $q = 1 - p$. Show that the moment generating function of $X$ is $M_X(t) = \frac{p}{1 - qe^t}$ for $qe^t < 1$.
3. Suppose $X$ has moment generating function $M_X(t) = \frac{1}{(1 - 2t)^3}$ for $t < \frac{1}{2}$. Find the mean and variance of $X$.
4. Suppose $X \sim \text{Expo}(\lambda)$. Use the moment generating function of $X$ to show that the $k$th moment of $X$ is $\mu_k = \frac{k!}{\lambda^k}$.

For Problem Set due 2 November: 18

Answers:

1. $M_X(t) = \frac{1}{b - a}\int_a^b e^{tx}dx = \frac{e^{tb} - e^{ta}}{t(b - a)}$
2. $M_X(t) = \sum_{k=0}^\infty e^{tk}q^kp = p\sum_{k=0}^\infty (qe^t)^k = \frac{p}{1 - qe^t}$ provided $qe^t < 1$.
3. $M_X'(t) = \frac{6}{(1 - 2t)^4}$ and $M_X''(t) = \frac{48}{(1 - 2t)^5}$, so $E[X] = M_X'(0) = 6$, $E[X^2] = M_X''(0) = 48$, and $\text{Var}(X) = 48 - 36 = 12$
4. $M_X(t) = \frac{\lambda}{\lambda - t} = \frac{1}{1 - \frac{t}{\lambda}} = \sum_{k=0}^\infty \frac{t^k}{\lambda^k}$. Hence $\frac{\mu_k}{k!} = \frac{1}{\lambda^k}$.