Solutions to Selected Problems In: Optimal Statistical Decisions by William DeGroot John L. Weatherwax∗ May 22, 1997

∗

[email protected]

1

Chapter 2 (Experiments, Sample Spaces, and Probability) Problem Solutions Problem 5 (lemmas of probability distributions) Part (d): This results is known as Boole’s inequality. We begin by decomposing the countable union of events Ai A1 ∪ A2 ∪ A3 . . . into a countable union of disjoint events Cj . Define these disjoint events as C1 C2 C3 C4 Cj

= = = = .. . =

A1 A2 \A1 A3 \(A1 ∪ A2 ) A4 \(A1 ∪ A2 ∪ A3 ) Aj \(A1 ∪ A2 ∪ A3 ∪ · · · ∪ Aj−1 )

Then by construction A1 ∪ A2 ∪ A3 · · · = C1 ∪ C2 ∪ C3 · · · , and the Cj ’s are disjoint events, so that we have (by part (a) of this problem) X Pr(A1 ∪ A2 ∪ A3 ∪ · · ·) = Pr(C1 ∪ C2 ∪ C3 ∪ · · ·) = Pr(Cj ) . j

Since Pr(Cj ) ≤ Pr(Aj ) (by part (c) of this problem), for each j, this sum is bounded above by X Pr(Aj ) , j

and Boole’s inequality is proven.

Problem 6 (the probability that at least one will fail) Let p denote the probability of failing so that from the given problem we have that p = 0.01. Then the probability that at least one component fails is n X n pk (1 − p)n−k = 1 − (1 − p)n = 1 − 0.99n , k k=1

or the complement of the probability that all components are functional.

Problem 7 (a secretary with letters) Let E1 , E2 , E3 , E4 , and E5 be the events that letter 1, 2, 3, 4 and 5 are placed in their correct envelope. Then we are asked about the probability that no letter is placed in its correct envelope or complement of the probability of the event E1 ∪ E2 ∪ E3 ∪ E4 ∪ E5 . Thus our probability is given by 1 − P (E1 ∪ E2 ∪ E3 ∪ E4 ∪ E5 ) . To evaluate P (E1 ∪ E2 ∪ E3 ∪ E4 ∪ E5 ) we will use the inclusion/exclusion identity which in this case is given by P (E1 ∪ E2 ∪ E3 ∪ E4 ∪ E5 ) = −

5 X i=1

P (Ei) −

X

X

P (EiEj ) +

i

∗

[email protected]

1

Chapter 2 (Experiments, Sample Spaces, and Probability) Problem Solutions Problem 5 (lemmas of probability distributions) Part (d): This results is known as Boole’s inequality. We begin by decomposing the countable union of events Ai A1 ∪ A2 ∪ A3 . . . into a countable union of disjoint events Cj . Define these disjoint events as C1 C2 C3 C4 Cj

= = = = .. . =

A1 A2 \A1 A3 \(A1 ∪ A2 ) A4 \(A1 ∪ A2 ∪ A3 ) Aj \(A1 ∪ A2 ∪ A3 ∪ · · · ∪ Aj−1 )

Then by construction A1 ∪ A2 ∪ A3 · · · = C1 ∪ C2 ∪ C3 · · · , and the Cj ’s are disjoint events, so that we have (by part (a) of this problem) X Pr(A1 ∪ A2 ∪ A3 ∪ · · ·) = Pr(C1 ∪ C2 ∪ C3 ∪ · · ·) = Pr(Cj ) . j

Since Pr(Cj ) ≤ Pr(Aj ) (by part (c) of this problem), for each j, this sum is bounded above by X Pr(Aj ) , j

and Boole’s inequality is proven.

Problem 6 (the probability that at least one will fail) Let p denote the probability of failing so that from the given problem we have that p = 0.01. Then the probability that at least one component fails is n X n pk (1 − p)n−k = 1 − (1 − p)n = 1 − 0.99n , k k=1

or the complement of the probability that all components are functional.

Problem 7 (a secretary with letters) Let E1 , E2 , E3 , E4 , and E5 be the events that letter 1, 2, 3, 4 and 5 are placed in their correct envelope. Then we are asked about the probability that no letter is placed in its correct envelope or complement of the probability of the event E1 ∪ E2 ∪ E3 ∪ E4 ∪ E5 . Thus our probability is given by 1 − P (E1 ∪ E2 ∪ E3 ∪ E4 ∪ E5 ) . To evaluate P (E1 ∪ E2 ∪ E3 ∪ E4 ∪ E5 ) we will use the inclusion/exclusion identity which in this case is given by P (E1 ∪ E2 ∪ E3 ∪ E4 ∪ E5 ) = −

5 X i=1

P (Ei) −

X

X

P (EiEj ) +

i