Probability & Statistics Ch.2 Section 1 Question 7

· Mohammad-Ali Bandzar

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={1,2,3}S=\{1,2,3\}, X=I{1}X=\textit{I}_{\{1\}}, Y=I{2,3}Y=\textit{I}_{\{2,3\}}, Z=I{1,2}Z=\textit{I}_{\{1,2\}}. Let W=XY+ZW=X-Y+Z.

a) Compute W(1)W(1) b) Compute W(2)W(2) c) Compute W(3)W(3) d) Determine whether or not WZW \geq Z

Solution

(a) Compute W(1)W(1)

W(1)=X(1)Y(1)+Z(1)=10+1=2W(1)=X(1)-Y(1)+Z(1) = 1-0+1 = 2

(b) Compute W(2)W(2)

W(2)=X(2)Y(2)+Z(2)=01+1=0W(2)=X(2)-Y(2)+Z(2) = 0-1+1 = 0

(c) Compute W(3)W(3)

W(3)=X(3)Y(3)+Z(3)=01+0=1W(3)=X(3)-Y(3)+Z(3) = 0-1+0 = -1

(d) Determine whether or not WZW \geq Z

We compare W to Z for every element in our sample space:

  • W(1)Z(1)21W(1) \geq Z(1) \rightarrow 2 \geq 1 ✓ True
  • W(2)Z(2)01W(2) \geq Z(2) \rightarrow 0 \geq 1 ✗ FALSE
  • W(3)Z(3)10W(3) \geq Z(3) \rightarrow -1 \geq 0 ✗ FALSE

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

Alternatively, we could have concluded that the range of W is [1,2][-1,2] and the range of Z is [0,1][0,1], therefore WZW \ngeq Z.