Answer on Question #79236 – Math – Statistics and Probability
Question
Let e=(i,j) represent an arbitrary resulting from two rolls of the four-sided die of example 2.1.1. Tabulate the discrete pdf and sketch the graph of the CDF for the
(a) Y(e)=i+j
(b) Z(e)=i−j
(c) W(e)=(i−j)2
Solution
The set of all possible outcomes is
(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),
(4,3), (4,4). Each of the outcomes has probability 1/16.
a) Y(e)=i+j
The set of possible values is {2,3,4,5,6,7,8}. The probabilities are:
P(Y=2)=P((1,1))=1/16P(Y=3)=P((1,2))+P((2,1))=2/16P(Y=4)=P((1,3))+P((2,2))+P((3,1))=3/16P(Y=5)=P((1,4))+P((2,3))+P((3,2))+P((4,1))=4/16P(Y=6)=P((2,4))+P((3,3))+P((4,2))=3/16P(Y=7)=P((3,4))+P((4,3))=2/16P(Y=8)=P((4,4))=1/16
b) Z(e)=i−j
The set of possible values is {−3,−2,−1,0,1,2,3}
P(Z=−3)=P((1,4))=1/16P(Z=−2)=P((1,3))+P((2,4))=2/16P(Z=−1)=P((1,2))+P((2,3))+P((3,4))=3/16P(Z=0)=P((1,1))+P((2,2))+P((3,3))+P((4,4))=4/16P(Z=1)=P((2,1))+P((3,2))+P((4,3))=3/16P(Z=2)=P((3,1))+P((4,2))=2/16P(Z=3)=P((4,1))=1/16
c) W(e)=(i−j)2
The set of possible values is {0,1,4,9}
P(Z=0)=P((1,1))+P((2,2))+P((3,3))+P((4,4))=4/16P(Z=1)=P((1,2))+P((2,3))+P((3,4))+P((2,1))+P((3,2))+P((4,3))=6/16P(Z=4)=P((1,3))+P((2,4))+P((3,1))+P((4,2))=4/16P(Z=9)=P((1,4))+P((4,1))=2/16
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