Question

A fair die is rolled twice, with outcomes X for the first roll and Y for the second roll. Find the moment generating function MX`Y ptq of X ` Y . Note that your answer should be a function of t and can contain unsimplified finite sums.

Answer 1

Answer:

$\mathbf{\dfrac{e^{2t}}{36} + \dfrac{e^{3t}}{18} + \dfrac{e^{4t}}{12} +\dfrac{e^{5t}}{9} + \dfrac{5e^{6t}}{36} + \dfrac{7e^{7t}}{6} + \dfrac{5e^{8t}}{36} + \dfrac{e^{9t}}{9} + \dfrac{e^{10t}}{12} + \dfrac{e^{11t}}{18} + \dfrac{e^{12t}}{36} }$

Step-by-step explanation:

The objective is to find the moment generating  function of  $M_{X+Y}(t) \ of \ X+Y$.

We are being informed that the fair die is rolled twice;

So; X to be the value for the  first roll

Y to be the value of the second roll

The outcomes  of X are:  X = {1,2,3,4,5,6}

Where ;

$P (X=x) = \dfrac{1}{6}$

The outcomes  of Y are:  y = {1,2,3,4,5,6}

Where ;

$P (Y=y) = \dfrac{1}{6}$

The outcome of Z = X+Y

$= \left[\begin{array}{cccccc}(1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\ (2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\ (3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6) \\ (4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6) \\ (5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6) \\ (6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6) \end{array}\right]$

= [2,3,4,5,6,7,8,9,10,11,12]

Here;

$P (Z=z) = \dfrac{1}{36}$

∴ the moment generating function $M_{X+Y}(t) \ of \ X+Y$is as follows:

$M_{X+Y}(t) \ of \ X+Y$ = $E(e^{t(X+Y)}) = E(e^{tz})$

⇒ $\sum \limits^{12}_ {z=2 } et ^z \ P(Z=z)$

= $\mathbf{\dfrac{e^{2t}}{36} + \dfrac{e^{3t}}{18} + \dfrac{e^{4t}}{12} +\dfrac{e^{5t}}{9} + \dfrac{5e^{6t}}{36} + \dfrac{7e^{7t}}{6} + \dfrac{5e^{8t}}{36} + \dfrac{e^{9t}}{9} + \dfrac{e^{10t}}{12} + \dfrac{e^{11t}}{18} + \dfrac{e^{12t}}{36} }$