Chapter 6: Q. 6.2 (page 277)

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

Expert verified

(a) $E [X Y Z] = 6$

(b) $E [X Y + X Z + Y Z] = 10$

Step by step solution

Given information (part a)

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

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

Explanation (part a)

The random variable XYZ can assume the following values:

$1 \times 2 \times 3 = 6 2 \times 1 \times 1 = 2 2 \times 2 \times 1 = 4 2 \times 3 \times 2 = 12$

Use the formula: $E [X] = \sum x P (X = x)$

Also,

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

So,

$E [X Y Z] = \frac{1}{4} (6) + \frac{1}{4} (2) + \frac{1}{4} (4) + \frac{1}{4} (12)$

$= \frac{24}{4}$

$= 6$

Given information (part b)

The random variable $X Y + X Z + Y Z$ can assume the following values:

$(1 \times 2) + (2 \times 3) + (1 \times 3) = 2 + 6 + 3 = 11 (2 \times 1) + (1 \times 1) + (2 \times 1) = 2 + 1 + 2 = 5 (2 \times 2) + (2 \times 1) + (2 \times 1) = 4 + 2 + 2 = 8 (2 \times 3) + (2 \times 2) + (3 \times 2) = 6 + 4 + 6 = 16$

Use the formula: $E [X] = \sum x P (X = x)$

Explanation (part b)

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

So,

$E [X Y + Y Z + X Z] = \frac{1}{4} (11) + \frac{1}{4} (5) + \frac{1}{4} (8) + \frac{1}{4} (16)$

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

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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

Step by step solution

Given information (part a)

Explanation (part a)

Given information (part b)

Explanation (part b)

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The joint probability mass function of the random variables X, Y, Z isp(1,2,3)=p(2,1,1)=p(2,2,1)=p(2,3,2)=14Find (a) E[XYZ], and (b) E[XY + XZ + YZ].

Short Answer

Step by step solution

Given information (part a)

Explanation (part a)

Given information (part b)

Explanation (part b)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Theoretical and Mathematical Physics

Pure Maths

Mechanics Maths

Geometry

Statistics

Study anywhere. Anytime. Across all devices.

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].