Question

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

$p(1,2,3)=p(2,1,1)=p(2,2,1)=p(2,3,2)=14$

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

Short answer

(a) $E\left[XYZ\right]=6$

(b)$E\left[XY+XZ+YZ\right]=10$

Step-by-step solution

Given information (part a)

The joint probability mass function of the random variables X, Y, Z is $p(1,2,3)=p(2,1,1)=p(2,2,1)=p(2,3,2)=\frac{1}{4}$

$E\left[X\right]=\sum xP(X=x)$where X denotes a random variable.

Explanation (part a)

The random variable XYZ can assume the following values:

$$\begin{gathered} 1\times 2\times 3=6 \\ 2\times 1\times 1=2 \\ 2\times 2\times 1=4 \\ 2\times 3\times 2=12 \end{gathered}$$

Use the formula: $E\left[X\right]=\sum xP(X=x)$

Also,

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

So,

$E\left[XYZ\right]=\frac{1}{4}\left(6\right)+\frac{1}{4}\left(2\right)+\frac{1}{4}\left(4\right)+\frac{1}{4}\left(12\right)$

$=\frac{24}{4}$

$=6$

Given information (part b)

The random variable $XY+XZ+YZ$can assume the following values:

$$\begin{gathered} \left(1\times 2\right)+\left(2\times 3\right)+\left(1\times 3\right)=2+6+3=11 \\ \left(2\times 1\right)+\left(1\times 1\right)+\left(2\times 1\right)=2+1+2=5 \\ \left(2\times 2\right)+\left(2\times 1\right)+\left(2\times 1\right)=4+2+2=8 \\ \left(2\times 3\right)+\left(2\times 2\right)+\left(3\times 2\right)=6+4+6=16 \end{gathered}$$

Use the formula:$E\left[X\right]=\sum xP(X=x)$

Explanation (part b)

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

So,

$E[XY+YZ+XZ]=\frac{1}{4}\left(11\right)+\frac{1}{4}\left(5\right)+\frac{1}{4}\left(8\right)+\frac{1}{4}\left(16\right)$

$$\begin{gathered} =\frac{11+5+8+16}{4} \\ =\frac{40}{4} \\ =10 \end{gathered}$$