HW Solution 2 Due: July 10:39AM

Transcription

1 ECS 35: Probability and Random Processes 200/ HW Solution 2 Due: July 0:39AM Lecturer: Prapun Suksompong, Ph.D. Instructions (a) A part of ONE question will be graded. Of course, you do not know which problem will be selected; so you should work on all of them. (b) Late submission will not be accepted. (c) Write down all the steps that you have done to obtain your answers. You may not get full credit even when your answer is correct without showing how you get your answer. Problem. Binomial theorem: Sometimes, the number ( n r) is called a binomial coefficient because it appears as the coefficient of x r y n r in the expansion of the binomial (x+y) n. More specifically, we have, for any positive integer n, (x + y) n = x r y n r (2.) r To see how we get (2.), let s consider a smaller case of n = 3. The expansion of (x + y) 3 can be found using combinatorial reasoning instead of multiplying the three terms out. When (x+y) 3 = (x+y)(x+y)(x+y) is expanded, all products of a term in the first sum, a term in the second sum, and a term in the third sum are added. Terms of the form x 3, x 2 y, xy 2, andy 3 arise. To obtain a term of the form x 3, an x must be chosen in each of the sums, and this can be done in only one way. Thus, the x 3 term in the product has a coefficient of. To obtain a term of the form x 2 y, an x must be chosen in two of the three sums (and consequently a y in the other sum). Hence. the number of such terms is the number of 2-combinations of three objects, namely, ( 3 2). Similarly, the number of terms of the form xy 2 is the number of ways to pick one of the three sums to obtain an x (and consequently take a y from each of the other two terms). This can be done in ( ) 3 ways. Finally, the only way to obtain a y 3 term is to choose the y for each of the three sums in the product, and this can be done in exactly one way. Consequently. it follows that (x + y) 3 = x 3 + 3x 2 y + 3xy 2 + y 3. Now, let s state a combinatorial proof of the binomial theorem (2.). The terms in the product when it is expanded are of the form x r y n r for r = 0,, 2,..., n. To count the 2-

2 number of terms of the form x r y n r, note that to obtain such a term it is necessary to choose r xs from the n sums (so that the other n r terms in the product are ys). Therefore. the coefficient of x r y n r is ( n r). From (2.), if we let x = y =, then we get another important identity: = 2 n. (2.2) r (a) What is the coefficient of x 2 y 3 in the expansion of (x + y) 25? (b) What is the coefficient of x 2 y 3 in the expansion of (2x 3y) 25? (c) Use the binomial theorem (2.) to evaluate n k=0 ( ) k( n k). (d) Use the binomial theorem (2.) to simplify the following sums (i) (ii) r even ( n r r odd ( n ) r x r ( x) n r ) x r ( x) n r (e) If we differentiate (2.) with respect to x and then multiply by x, we have r x r y n r = nx(x + y) n. r Use similar technique to simplify the sum n r2( n r) x r y n r. Solution: (a) ( 25 2) = 5, 200, 300. (b) ( ) ( 3) 3 = 25! 2!3! = (c) From (2.), set x = and y =, then we have n k=0 ( ) k( n k) = ( + ) n =

3 (d) To deal with the sum involving only the even terms (or only the odd terms), we first use (2.) to expand (x + y) n and (x + ( y)) n. When we add the expanded results, only the even terms in the sum are left. Similarly, when we find the difference between the two expanded results, only the the odd terms are left. More specifically, r even r odd x r y n r = r 2 ((x + y)n + (y x) n ), and x r y n r = r 2 ((x + y)n (y x) n ). If x + y =, then r even r odd x r y n r = r 2 ( + ( 2x)n ), and x r y n r = r 2 ( ( 2x)n ). (2.3a) (2.3b) (e) n r2( n r) x r y n r = nx ( x(n )(x + y) n 2 + (x + y) n ). Problem 2. Prove the following properties of probability measure using only the probability axioms and the properties whose proofs have already been shown in class. (a) P (A c ) = P (A) (b) P (A) [0, ] (c) If P (A) = = 0, then P (A B) = 0 (d) If P (A) = =, then P (A B) = Solution: (a) Because A and A c partition Ω, we have P (Ω) = P (A) + P (A c ). One of the axioms say that P (Ω) =. Applying this fact to the equality above proves the required property. 2-3

4 (b) One of the axioms already specifies that P (A) 0. So, we only need to show that P (A). Now, for any event A, we know that A Ω and hence, from the monotonicity property shown in class, P (A) P (Ω) =. This completes the proof. (c) [Capinski and Zastawniak, 2003, Q4.26] First, we know that P (A B) 0 from one of the axioms. Now, we have seen in class that P (A B) P (A) +. Plugging in P (A) = = 0 gives P (A B) 0 Because P (A B) 0 and P (A B) 0, it must be = 0. (d) [Capinski and Zastawniak, 2003, Q4.27] Because P (A) = =, we know that P (A c ) = P (B c ) = 0. From the previous part of the problem, we can then conclude that P (A c B c ) = 0. The event A B is the complement of A c B c. Hence, P (A B) = P (A c B c ) = 0 =. Problem 3. Suppose that P (A) = and = 2. Find the range of the possible value 2 3 for P (A B). Hint: Smaller than the interval [0, ]. Solution:[Capinski and Zastawniak, 2003, Q4.2] We will first try to bound P (A B). Note that A B A and A B B. Hence, we know that P (A B) P (A) and P (A B). To summarize, we now know that On the other hand, we know that P (A B) min{p (A), }. P (A B) = P (A) + P (A B). Applying the fact that P (A B), we then have P (A B) P (A) +. If the number of the RHS is > 0, then it is a new information. However, if the number on the RHS is negative, it is useless and we will use the fact that P (A B) 0. To summarize, we now know that max{p (A) +, 0} P (A B). In conclusion, max{(p (A) + ), 0} P (A B) min{p (A), }. [ Plugging in the value P (A) = and = 2 gives the range 2 3 6, ]. The upperbound can be obtained by constructing an example which has A B. The lower-bound can 2 be obtained by considering an example where A B = Ω. 2-4

5 Problem 4. Suppose that P (A) = and =. Find the range of the possible value 2 3 for P (A B). Hint: Smaller than the interval [0, ]. Solution:[Capinski and Zastawniak, 2003, Q4.22] By monotonicity we must have On the other hand, we know that P (A B) max{p (A), }. P (A B) P (A) +. If the RHS is >, then the inequality is useless and we simply use the fact that it must be. To summarize, we have In conclusion, P (A B) min{(p (A) + ), }. max{p (A), } P (A B) min{(p (A) + ), }. Plugging in the value P (A) = and =, we have 2 3 P (A B) [ 2, 5 ]. 6 The upper-bound can be obtained by making A B. The lower-bound is achieved when B A. Problem 5. Let A and B be events for which P (A),, and P (A B) are known. Express the following in terms of these probabilities: (a) P (A B) (b) P (A B c ) (c) P (B (A B c )) (d) P (A c B c ) Solution: (a) P (A B) = P (A) + P (A B). This property is shown in class. 2-5

6 (b) We have seen in class that P (A B c ) = P (A) P (A B). Plugging in the expression for P (A B) from the previous part, we have P (A B c ) = P (A) (P (A) + P (A B)) = P (A B). Alternatively, we can start from scratch with the set identity A B = B (A B c ) whose union is a disjoint union. Hence, P (A B) = + P (A B c ). Moving to the LHS finishes the proof. (c) P (B (A B c )) = P (A B) because A B = B (A B c ). (d) P (A c B c ) = P (A B) because A c B c = (A B) c. Problem 6. Prove the following properties of conditional probability. (a) P (A Ω) = P (A) (b) If B A and 0, then P (A B) =. (c) If A B = and 0, then P (A B) = 0 (d) P (A c B) = P (A B) Solution: (a) By definition P (A Ω) = P (A Ω) P (Ω). Because A Ω = A and P (Ω) =, P (A Ω) = P (A) = P (A). (b) By definition, we have P (A B) = P (A B). This quantity is only defined when 0. Because B A, we have A B = B. Hence, P (A B) = =. (c) By definition, we have P (A B) = P (A B). This quantity is only defined when 0. Because A B =, we have P (A B) = P ( ) = 0 =

7 (d) By definition we have P (A c B) = P (Ac B). Note that P (A c B) = P (B \ A) = P (A B). Therefore, P (A c B) = P (Ac B) = P (A B) = P (A B) = P (A B). Problem 7. Due to an Internet configuration error, packets sent from New York to Los Angeles are routed through El Paso, Texas with probability 3/4. Given that a packet is routed through El Paso, suppose it has conditional probability /3 of being dropped. Given that a packet is not routed through El Paso, suppose it has conditional probability /4 of being dropped. (a) Find the probability that a packet is dropped. (b) Find the conditional probability that a packet is routed through El Paso given that it is not dropped. Solution: [Gubner, 2006, Ex..20] To solve this problem, we use the notation E = {routed through El Paso} and D = {packet is dropped}. With this notation, it is easy to interpret the problem as telling us that P (D E) = /3, P (D E c ) = /4, and P (E) = 3/4. (a) By the law of total probability, P (D) = P (D E)P (E) + P (D E c )P (E c ) = (/3)(3/4) + (/4)( 3/4) = /4 + /6 = 5/6. (b) P (E D c ) = P (E Dc ) P (D c ) = P (Dc E)P (E) P (D c ) = ( /3)(3/4) 5/6 = 8. Problem 8. You have two coins, a fair one with probability of heads and an unfair one 2 with probability of heads, but otherwise identical. A coin is selected at random and tossed, 3 falling heads up. How likely is it that it is the fair one? Solution: [Capinski and Zastawniak, 2003, Q7.28] Let F, U, and H be the events that the selected coin is fair, the selected coin is unfair, and the coin lands heads up, respectively. Because the coin is selected at random, the probability P (F ) of selecting the fair coin is P (F ) =. For fair coin, the conditional probability P (H F ) of heads is For the unfair 2 2 coin, P (U) = P (F ) = and P (U F ) =

8 By the Bayes formula, the probability that the fair coin has been selected given that it lands heads up is P (F H) = P (H F )P (F ) P (H F )P (F ) + P (H U)P (U) = = = = 3 5. Problem 9. You have three coins in your pocket, two fair ones but the third biased with probability of heads p and tails p. One coin selected at random drops to the floor, landing heads up. How likely is it that it is one of the fair coins? Solution: [Capinski and Zastawniak, 2003, Q7.29] Let F, U, and H be the events that the selected coin is fair, the selected coin is unfair, and the coin lands heads up, respectively. We are given that P (F ) = 2 3, P (U) = 3, P (H F ) =, P (H U) = p. 2 By the Bayes formula, the probability that the fair coin has been selected given that it lands heads up is P (F H) = P (H F )P (F ) P (H F )P (F ) + P (H U)P (U) = p = + p. 2-8