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Question

The joint probability mass function of the random variables X, Y, Z is

p(1,2,3)=p(2,1,1)=p(2,2,1)=p(2,3,2)=14p ( 1,2,3 ) = p ( 2,1,1 ) = p ( 2,2,1 ) = p ( 2,3,2 ) = \frac { 1 } { 4 }

Find (a) E[XYZ], and (b) E[XY+XZ+YZ].

Solution

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(a)

Observe that the random variable XYZXYZ can assume values 2,4,6,122,4,6,12 with the same probability 14\frac{1}{4}. Hence, the required expectation is

E(XYZ)=14(2+4+6+12)=6\begin{align*} E(XYZ) = \frac{1}{4} (2+4+6+12) = 6 \end{align*}

(b)

Observe that the random variable XY+XZ+YZXY + XZ + YZ can assume values 11,5,8,1611,5,8,16 with the same probability 14\frac{1}{4}. Hence, the required expectation is

E(XY+XZ+YZ)=14(11+5+8+16)=10\begin{align*} E(XY+XZ+YZ) = \frac{1}{4} (11+5+8+16) = 10 \end{align*}

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