Probability & Statistics Ch.2 Section 1 Question 2

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Let $S=\{high, middle, low\}$. Define random variables X, Y, and Z by:

• X(high) = -12, X(middle) = -2, X(low) = 3
• Y(high) = 0, Y(middle) = 0, Y(low) = 1
• Z(high) = 6, Z(middle) = 0, Z(low) = 4

Determine whether each of the following relations is true or false:

• a) $X < Y$
• b) $X \leq Y$
• c) $Y < Z$
• d) $Y \leq Z$
• e) $XY < Z$
• f) $XY \leq Z$

Solution

Example 2.1.7 from the textbook tells us: we write $X < Y$ to mean that $X(s) < Y(s)$ for all $s\in S$.

(a) $X < Y$

Since our sample space is small, we can manually try every item:

• $X(high) < Y(high) \rightarrow -12 < 0$ ✓ True
• $X(middle) < Y(middle) \rightarrow -2 < 0$ ✓ True
• $X(low) < Y(low) \rightarrow 3 < 1$ ✗ FALSE

Since we have shown that the relation does not hold for at least one element of our sample space, we can conclude that the relation is False.

(b) $X \leq Y$

• $X(high) \leq Y(high) \rightarrow -12 \leq 0$ ✓ True
• $X(middle) \leq Y(middle) \rightarrow -2 \leq 0$ ✓ True
• $X(low) \leq Y(low) \rightarrow 3 \leq 1$ ✗ FALSE

The relation is False.

(c) $Y < Z$

• $Y(high) < Z(high) \rightarrow 0 < 6$ ✓ True
• $Y(middle) < Z(middle) \rightarrow 0 < 0$ ✗ FALSE

The relation is False.

(d) $Y \leq Z$

• $Y(high) \leq Z(high) \rightarrow 0 \leq 6$ ✓ True
• $Y(middle) \leq Z(middle) \rightarrow 0 \leq 0$ ✓ True
• $Y(low) \leq Z(low) \rightarrow 1 \leq 4$ ✓ True

Since our relation holds true for all elements of our sample space, the relation is True.

(e) $XY < Z$

• $X(high) \times Y(high) < Z(high) \rightarrow (-12 \times 0) < 6 = 0 < 6$ ✓ True
• $X(middle) \times Y(middle) < Z(middle) \rightarrow (-2 \times 0) < 0 = 0 < 0$ ✗ FALSE

The relation is False.

(f) $XY \leq Z$

• $X(high) \times Y(high) \leq Z(high) \rightarrow (-12 \times 0) \leq 6 = 0 \leq 6$ ✓ True
• $X(middle) \times Y(middle) \leq Z(middle) \rightarrow (-2 \times 0) \leq 0 = 0 \leq 0$ ✓ True
• $X(low) \times Y(low) \leq Z(low) \rightarrow (3 \times 1) \leq 4 = 3 \leq 4$ ✓ True

Since our relation holds true for all elements of our sample space, the relation is True.