Answer:
[tex]\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} }[/tex]
Step-by-step explanation:
The objective is to find the moment generating function of [tex]M_{X+Y}(t) \ of \ X+Y[/tex].
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 ;
[tex]P (X=x) = \dfrac{1}{6}[/tex]
The outcomes of Y are: y = {1,2,3,4,5,6}
Where ;
[tex]P (Y=y) = \dfrac{1}{6}[/tex]
The outcome of Z = X+Y
[tex]= \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][/tex]
= [2,3,4,5,6,7,8,9,10,11,12]
Here;
[tex]P (Z=z) = \dfrac{1}{36}[/tex]
∴ the moment generating function [tex]M_{X+Y}(t) \ of \ X+Y[/tex]is as follows:
[tex]M_{X+Y}(t) \ of \ X+Y[/tex] = [tex]E(e^{t(X+Y)}) = E(e^{tz})[/tex]
⇒ [tex]\sum \limits^{12}_ {z=2 } et ^z \ P(Z=z)[/tex]
= [tex]\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} }[/tex]
