Let [math]X_1, X_2, X_3[/math] be a random sample from a discrete distribution with probability function

[[math]] p(x) = \begin{cases} 1/3, \, x=0 \\ 2/3, \, x=1 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Determine the moment generating function, [math]M(t)[/math], of [math]Y = X_1X_2X_3 [/math].

• [math]\frac{19}{27} + \frac{8}{27}e^t[/math]
• [math]1 + 2e^t[/math]
• [math](\frac{1}{3} + \frac{2}{3}e^t)^3[/math]
• [math]\frac{1}{27} + \frac{8}{27}e^{3t}[/math]
• [math]\frac{1}{3} + \frac{2}{3}e^{3t}[/math]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: A

Let g(y) be the probability function for [math]Y = X_1X_2X_3. [/math] Note that [math]Y = 1 [/math] if and only if

[[math]] X_1 = X_2 = X_3 = 1. [[/math]]

Otherwise, [math]Y = 0 [/math]. Since

[[math]] \begin{align*} \operatorname{P}[Y = 1] &= \operatorname{P}[X_1 = 1 \cap X_2 = 1 \cap X_3 = 1] \\ &= \operatorname{P}[X_1 = 1] \operatorname{P}[X_2 = 1] \operatorname{P}[X_3 = 1] = (2/3)^3 \\ &= 8/27. \end{align*} [[/math]]

We conclude that

[[math]] g(y) = \begin{cases} \frac{19}{27}, \quad y = 0 \\ \frac{8}{27}, \quad y = 1 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

and

[[math]] M(t) = E[e^{y_t}] = \frac{19}{27} + \frac{8}{27}e^t.[[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.