NOTES FROM MATH 10B DISCUSSION. Contents

Transcription

1 NOTES FROM MATH 10B DISCUSSION MINSEON SHIN Contents 1. Fri, Jan 23, Mon, Jan 26, Wed, Jan 28, Mon, Feb 2, 2015 (12 Fold Way) 5 5. Wed, Feb 4, Fri, Feb 6, Mon, Feb 9, Wed, Feb 11, 2015 (Bayes Theorem) 9 9. Fri, Feb 13, Wed, Feb 18, Fri, Feb 20, Mon, Feb 23, Wed, Feb 25, Mon, Mar 2, 2015 (Statistics) Wed, Mar 4, Mon, Mar 9, Wed, Mar 11, Fri, Mar 20, Mon, Mar 30, Fri, Apr 3, Wed, Apr 15, 2015 (Matrices) Fri, Apr 17, Wed, Apr 22, (Last edited April 22, 2015 at 2:18pm.) I appreciate any comments (e.g. pointing out typos), suggestions, questions. 1

2 2 MINSEON SHIN 1.1. Exercises. 1. Fri, Jan 23, 2015 Exercise 1. How many integers between 1 and 100 are there that are not divisible by 3 and not divisible by 7? Proof. We use the Principle of Inclusion-Exclusion (PIE). Let S = {1, 2,..., 100} be the set of positive integers between 1 and 100. Let A 1 be the set of positive integers between 1 and 100 that are divisible by 3. Let A 2 be the set of positive integers between 1 and 100 that are divisible by 7. We want to compute S \ (A 1 A 2 ). By PIE, we have S \ (A 1 A 2 ) = S A 1 A 2 + A 1 A 2 and S = 100 and A 1 = = 33 and A 2 = = 14 and A 1 A 2 = = 41, which means S \ (A 1 A 2 ) = = 57. Exercise 2. Show that there are at least 2 people in the world with the same number of Facebook friends. Proof using the pigeonhole principle. Let N be the number of Facebook accounts in the world. A person cannot be Facebook friends with himself, so any person can have at most N 1 Facebook friends. On the other hand, a person can have 0 Facebook friends. Suppose there is someone with 0 Facebook friends. If there is someone else with 0 friends, then we re done. Otherwise, assume everyone else has at least one friend. Everyone else can have at most N 2 Facebook friends (since they are not friends with themselves, nor with the person who has 0). Thus there are N 2 possibilities for the number of friends (pigeonholes) and N 1 people (pigeons), so, by the pigeonhole principle, there must be two people who have the same number of friends. Suppose no one has 0 Facebook friends. Then everyone can have between 1 and N 1 friends. Thus there are N 1 possibilities for the number of friends (pigeonholes) and N people (pigeons), so, by the pigeonhole principle, there must be two people who have the same number of friends. Proof without using the pigeonhole principle. Let N be the number of Facebook accounts in the world. A person cannot be Facebook friends with himself, so any person can have at most N 1 Facebook friends. On the other hand, a person can have 0 Facebook friends. Suppose that all N Facebook accounts have a different number of Facebook friends. Then Person #1 has 0 friends, Person #2 has 1 friend, and so on, and Person #N has N 1 friends. There is a contradiction, since Person #1 is friends with no one, whereas Person #N is friends with everyone. Exercise 3. If each point of the plane is colored either red or blue, show that there are two points of the same color that are distance 1 from each other. 1 If the question had replaced 3 with 4 and 7 with 6, then A1 A 2 consists of all integers divisible by 4 and 6, which is the same as being divisible by their least common multiple (LCM) 12.

3 NOTES FROM MATH 10B DISCUSSION 3 Proof. Consider the vertices of any equilateral triangle of side length 1. By the pigeonhole principle (pigeonholes are colors and pigeons are the three points of the triangle), at least two of the vertices have the same color; they also have distance 1 from each other Example from lecture. See also Exercise 4. Suppose 10 people of different heights are standing in a line. Show that there are 4 people whose heights are either increasing or decreasing. Proof. (As you read this proof, it may help to consider Example 5 and Example 6 and Example 7.) Let a 1,..., a 10 be the heights. We want to show that there are integers 1 i 1 < i 2 < i 3 < i 4 10 such that either a i1 < a i2 < a i3 < a i4 or a i1 > a i2 > a i3 > a i4. Let m i be the length of the longest left-to-right chain of increasing heights, starting at the ith person; notice that m i 1 for all i, since the chain a i has length 1. We are done if there exists some i such that m i 4 (for example, see Example 7), since this means we have a left-to-right increasing chain of length at least 4, starting from the ith person. Suppose m i 3 for all i (for example, see Example 5 and Example 6). Then, since there are 10 people (pigeons) and for each person there is a number m i that is either 1, 2, 3 (holes), by the pigeonhole principle there exists some k = 1, 2, 3 and indices i 1 < i 2 < i 3 < i 4 such that m i1 = m i2 = m i3 = m i4 = k. I claim that a i1 > a i2 > a i3 > a i4. Suppose for the sake of contradiction that a i1 < a i2. Since m i2 = k, by definition of m i2 there is a left-to-right sequence of k people, starting at the i 2 th person, whose heights are increasing (say a j1 < < a jk where j 1 = i 2 ). Then we have a left-to-right sequence of (k + 1) people, starting at the i 1 th person, whose heights are increasing, namely a i1 < a j1 < < a jk ; this contradicts the fact that m i1 = k (i.e. the longest increasing left-to-right sequence starting at i 1 has length k). The proof that a i2 > a i3 and a i3 > a i4 are similar. Example 5. Suppose the heights are 5, 2, 6, 4, 7, 8, 1, 0, 3, 9 from left to right. It may help to consider the information in the following table: i a i m i For example, m 3 = 2 since the longest increasing left-to-right sequence starting at the 3nd person is {6, 9} (the sequences {6, 8} and {6, 7} are equally long), and m 6 = 1 since none of 1, 0, 3, 7 are greater than 8. If we apply the above proof to this example, we have k = 3 and i 1 = 1, i 2 = 2, i 3 = 7, i 4 = 8; observe that 5 > 2 > 1 > 0. Example 6. Suppose the heights are 6, 4, 7, 5, 3, 9, 0, 1, 8, 2 from left to right. Then we have i a i m i If we apply the above proof to this example, we have k = 2 and i 1 = 3, i 2 = 4, i 3 = 5, i 4 = 8; observe that 7 > 5 > 3 > 1.

4 4 MINSEON SHIN Example 7. Suppose the heights are 3, 6, 1, 4, 5, 2, 0, 9, 8, 7 from left to right. Then we have i a i m i Mon, Jan 26, 2015 Exercise 8 (AIME 1983 #10). How many 4-digit numbers starting with 1 have exactly two identical digits? Proof. There are two cases; in the first, there are two 1s, and in the second, there is a repeated digit that is not 1. In the first case, there are 3 possibilities for the placement of the 1, namely 11XY, 1X1Y, 1XY 1 (here X and Y represent two distinct digits that are not equal to 1). For each subcase, there are 9 possibilities for X (since it cannot be 1) and 8 possibilities for Y (since it cannot be 1 or X). Thus there are possibilities. In the second case, there are 3 possibilities for the placement of the two repeated digits, namely 1XXY, 1XY X, 1Y XX. For each subcase, we have 9 possibilities for X and 8 possibilities for Y as above. Thus there are possibilities. In total, there are possibilities. Exercise 9 (AHSME 1985 #15). How many integers between 100 and 400 (inclusive) contain the digit 2? Proof. It s equivalent to count the number of integers between 100 and 399 that contain the digit 2, since 400 doesn t contain the digit 2. There are 300 numbers between 100 and ; we count how many contain 2 and subtract that from 300. There are {1, 3} = 2 possibilities for the leftmost digit, and {0, 1, 3,..., 9} = 9 possibilities for each of the other digits. 3 Thus there are = 162 integers between 100 and 399 that do not contain 2. Thus our final answer is = 138. Exercise If the 24 permutations of MATH are written in alphabetical order, what s the 17th permutation appearing in the list? Proof. We determine the letters of the 17th word, starting from left to right. We note that, for each letter A, H, M, T, there are 6 words starting with that letter; the 1st to 6th words in the list start with A, the 7th to 12th words start with H, and so on: 6 A H M T In general, to count the number of integers between A and B, take the number that is 1 less than A, and subtract it from both integers to get 1 and B (A 1). 3 Here S denotes the cardinality of the set S. 4 See also 1986 AHSME Problem 10.

5 NOTES FROM MATH 10B DISCUSSION 5 Since 17 is between 13 and 18, we know that the first letter of the 17th word is M. So now it remains to determine the 5th word in the list of permutations of AHT. For each of the remaining letters A, H, T, there are 2 words that start with that letter: 2 A H T 5 6 Since 5 is between 5 and 6, we know that the second letter of the 17th word is T. So now it remains to determine the 1st word in the list of permutations of AH, which is AH. Thus our final answer is MTAH. 3. Wed, Jan 28, 2015 Exercise 11. How many rectangles are in the following diagram? Proof. A rectangle is determined by choosing two vertical lines and two horizontal lines; vertical lines determine the rectangle s horizontal boundary and horizontal lines determine the rectangle s vertical boundary. There are 7 vertical lines, so there are ( 7 2) ways to choose the horizontal boundary; there are 5 horizontal lines, so there are ( 5 2) ways to choose the vertical boundary. Thus the final answer is ). ( 7 )( Exercise 12. I draw 5 cards randomly from a standard deck of 52 cards. What is the probability that I get a flush? 5 Proof. There are 4 ways to choose the suit; there are ( ) 13 5 ways to choose the 5 cards that will comprise your flush. There are ( ) 52 5 many 5-card hands in total. 4 ( ) 13 5 Thus the probability is ). ( Mon, Feb 2, 2015 (12 Fold Way) 4.1. The 12 Fold Way. In the table below, there is a general formula which depends on the number of balls b and the number of boxes/bins/urns u; below each formula is a brief explanation. Dist stands for distinguishable and Indist stands for indistinguishable. An assignment of balls to boxes corresponds to a function f : {balls} {boxes} which can be injective (i.e. at most one ball in each box) or surjective (i.e. at least one ball in each box). If neither injectivity nor surjectivity is imposed on the 5 A flush consists of 5 cards of the same suit. More poker hand probability computations are available at

6 6 MINSEON SHIN function, we denote the set by All/Arbitrary. I denote by (D,I,I), for example, the entry corresponding to the situation in which balls are distinguishable, boxes are indistinguishable, and we re only counting assignments in which each box contains at most one ball. Balls (b) Boxes (u) All/Arbitrary Injective Surjective u!s(b, u) u b P (u, b) Dist Dist (For each ball, choose a Indist Dist (For each ball, choose a (Same as (D,I,S) but order matters; think of giving ( b+(u 1) ) b (We have b balls and u 1 dividers.) ( u b) (Choose b boxes that should get a ball.) box.) box not already occupied.) each group a group number.) ( (b u)+(u 1) ) b u (Same as (I,D,A) except we start with b u balls, having placed a ball in each box.) Dist Indist u k=1 S(b, k) (# of ways for n people to form at most k groups.) 1 (Are there enough boxes to contain one ball in each box?) S(b, u) (# of ways for n people to form exactly k groups.) Indist Indist u k=1 p k(b) (# of partitions of n into at most k parts.) 1 (Are there enough boxes to contain one ball in each box?) p k (b) (# of partitions of n into exactly k parts.) In the Injective column of the table, the listed formulas are for the case when u b, i.e. when there are more boxes than balls. If u < b, then there are more balls than boxes, so we cannot put the balls into the boxes in such a way that no box has more than 1 ball; thus in this case all the entries in the Injective column are 0. Similarly, in the Surjective column of the table, the listed formulas are for the case when u b, i.e. when there are more balls than boxes. If u > b, then there are more boxes than balls, and there is no way to put at least one ball in every box. The definition of the Stirling number of the second kind 6 S(n, k) is the number of ways for n people to form exactly k groups. I suggest using the following formula to compute S(n, k): S(n, k) = 1 k! k ( ) k ( 1) i (k i) n i i=0 6 See

7 NOTES FROM MATH 10B DISCUSSION 7 (This formula can be proven using the principle of inclusion-exclusion.) For example, we have S(4, 3) = 1 (( ) ( ) ( ) ( ) )0 4 3! = 6. We may also use the recurrence relation S(n + 1, k) = k S(n, k) + S(n, k 1) for computations, using the fact that S(n, 1) = 1 and S(n, n) = 1 for all positive integers n. The expression p k (n) is the number of ways to divide the integer n into k (not necessarily equal) parts. 7 It satisfies the recurrence relation which you can use to compute p k (n). p k (n) = p k (n k) + p k 1 (n 1) 5. Wed, Feb 4, 2015 Exercise 13. In how many ways can we distribute 23 apples and 17 oranges among 5 children so that each child receives 2 apples and 1 orange? Solution. Here apples are indistinguishable from each other, oranges are indistinguishable from each other, but apples are distinguishable from oranges. We may think of giving each child 2 apples and 1 orange first, and distributing the 13 apples and 12 oranges that we re left with. There are ( ) 13+(5 1) 13 ways to distribute the apples (the apples are indistinguishable balls and the children are distinguishable boxes), and there are ( ) 12+(5 1) 12 ways to distribute the oranges (the oranges are indistinguishable balls and the children are distinguishavle boxes). Thus the desired answer is ( 13+(5 2) 13 )( 12+(5 1) 12 ). Exercise 14. A bakery sells 89 different kinds of doughnuts. In how many ways can we make an order of 216 doughnuts if the order must contain at least one of each type? Solution. The balls are 216 doughnuts and the boxes are the 89 different kinds of doughnuts. We can think of the balls as 216 tallies/counters that we have to distribute among the 89 boxes corresponding to how many of each kind we want to order. The balls are indistinguishable, whereas the boxes are distinguishable. Since we require that there must be at least one doughnut of each kind (which corresponds to every box having at least one ball), our desired answer is ( (216 89)+(89 1) Exercise 15. How many terms are there in (x x 13 ) 79? ). 7 See or_number_of_parts

8 8 MINSEON SHIN Solution. Each term is of the form cx e1 1 xe13 where c is a positive integer corresponding to the coefficient and e 1,..., e 13 are nonnegative integers such that e e 13 = 79. Thus we want to count the number of ordered 13-tuples (e 1,..., e 13 ) such that e e 13 = 79; here there are 79 indistinguishable balls, 13 distinguishable boxes, and no condition about injectivity/surjectivity is imposed. Thus there are ( 79+(13 1) 79 ) terms. Exercise In Exercise 15, what is the sum of all the coefficients that appear when the expression is expanded? Solution. We use the following trick: substitute x 1 = = x 13 = 1. In this case each term cx e1 1 xe13 is equal to its coefficient c1e1 1 e13 = c. Thus the answer is ( ) 79 = Fri, Feb 6, 2015 Exercise 17. A bag contains 4 blue marbles and 12 green marbles. Two are randomly picked. What is the probability that both are blue? Proof. The probability of the first marble being blue is 4 16, and the probability of the second marble being blue is Thus the desired probability is Remark 18 (Philosophy). What is the sample space in Exercise 17? Exercise 19. We roll a fair sided die 2 times. What is the probability that the sum is 8? Proof. There are 5 ways for the sum to equal 8, namely 6+2, 5+3, 4+4, 3+5, 2+6. There are 6 2 total outcomes. Thus the desired probability is Exercise 20. We roll a fair sided die 3 times. What is the probability that the sum is 8? Proof. If the first roll is 6, then there is only one way for the sum to be 8, namely If the first roll is 5, there are two ways, 5+2+1, We may continue this argument until the case when the first roll is 1, in which case there are 6 ways for the sum to be 8, namely 1+6+1, 1+5+2, 1+4+3, 1+3+4, 1+2+5, There are 6 3 total outcomes. Thus the desired probability is = Mon, Feb 9, 2015 Exercise 21. Ann and Bob are dealt 2 cards each from a deck of 52 cards. (i) Given that Ann has 2 aces, what s the probability that Bob also has 2 aces? (ii) Given that Ann has 2 hearts, what s the probability that Bob also has 2 hearts? 8 This is just a fun aside; you probably don t need to know this.

9 NOTES FROM MATH 10B DISCUSSION 9 Solution. (i) Let A be the event that Ann has 2 aces, and let B be the event that Bob has 2 aces. We are looking for P (B A) = P (B A) P (A) ; here P (B A) = ( 4 2)( 2 2) ( 52 2 )( 2 ) 9 and P (A) = (4 2) 1, thus P (B A) = ( 52 2 ) ( 52 2 ). (ii) Let A be the event that Ann has 2 hearts, and let B be the event that Bob has 2 hearts. We are looking for P (B A) = P (B A) P (A) ; here P (B A) = ( 13 2 )( 11 2 ) ( 52 2 )( 2 ) 10 and P (A) = (13 ( 52 2 ) 2 ), thus P (B A) = (11 2 ) ( 52 2 ). Exercise I have 3 different coins A, B, C. Coin A shows heads 1 3 of the time, Coin B shows heads 1 2 of the time, Coin C shows heads 2 3 of the time. I select a coin at random (each coin has probability 1 3 of being selected) and flip it two times. I get two heads. What is the probability that I selected Coin A? Solution. Let A be the event that I selected coin A, let B be the event that I selected coin B, let C be the event that I selected coin C, and let D be the event that I got two heads in two flips. We are looking for P (A D), which is equal to P (A D) 1 = 2 = = P (A D) P (D) P (A D) P (A D) + P (B D) + P (C D) 1 3 ( 1 3 )2 1 3 ( 1 3 ) ( 1 2 ) ( 2 3 )2 where equality 1 is by definition of P (A D), equality 2 is since A, B, C are disjoint events whose union is the entire sample space. 8. Wed, Feb 11, 2015 (Bayes Theorem) Here s how to recognize when you should apply Bayes theorem; broadly speaking, you should think of Bayes theorem if you are given events A and B and the probability P (A B) and are being asked to compute P (B A). We are given events such that A, X 1,..., X n X i X j = if i j and X 1 X n = Ω (i.e. the {X 1,..., X n } form a partition of the sample space Ω) and given probabilities P (X 1 ),..., P (X n ), P (A X 1 ),..., P (A X n ) 9 ( 4 ( 2) ways to give 2 aces to Ann, 2 ( 2) ways to give the other 2 aces to Bob; 52 2 deal 2 cards each to Ann and Bob 10 ( 13) ( 2 ways to give 2 aces to Ann, 11 ) ( 2 ways to give the other 2 aces to Bob; 52 2 deal 2 cards each to Ann and Bob 11 This is a preview of Bayes theorem. )( 2 ) ways to )( 2 ) ways to

10 10 MINSEON SHIN and we are asked to compute P (X i A), which Bayes Theorem says is equal to P (X i A) = Notice that the RHS of (1) is also equal to P (A X i )P (X i ) P (A X 1 )P (X 1 ) + + P (A X n )P (X n ). (1) P (A X i ) P (A X 1 ) + + P (A X n ) = P (A X i). P (A) In problems requiring Bayes theorem, usually the hardest part is identifying what A, X 1,..., X n are. Example 23. Assume % of the world s population is male, that 20% of males have brown eyes, and that 15% of females have brown eyes. Given that a person has brown eyes, what s the probability that the person is female? Solution. Let Ω be the set of all people in the world who identify as either male or female, let A be the set of people with brown eyes, let X 1 be the set of female persons, let X 2 be the set of male persons. We are given P (X 1 ) = 0.5, P (X 2 ) = 0.5, P (A X 1 ) = 0.15, P (A X 2 ) = 0.2, and we are asked to compute P (X 1 A), which is P (X 1 A) = P (A X 1 )P (X 1 ) P (A X 1 )P (X 1 ) + P (A X 2 )P (X 2 ) = = Fri, Feb 13, 2015 Example 24. Suppose we roll a die twice 12. Thus the probability space is Ω = {(1, 1), (1, 2),..., (6, 6)} and the cardinality of Ω is Ω = 36. Let A Ω be the event that the sum of the two rolls is 4, and let B Ω be the event that the product of the two rolls is 4. Then A and B are not independent. To see this, we compute P (A) and P (B) and P (A B) and show that P (A B) P (A)P (B). We have P (A) = P ({(1, 3), (2, 2), (3, 1)}) = 3 36 and P (B) = P ({(1, 4), (2, 2), (4, 1)}) = 3 36 and P (A B) = P ({(2, 2)}) = 1 36, and Let s fix a probability space Ω; a random variable X on Ω is a function X : Ω R. 13 The range of a random variable X is the image of X, i.e. the set of values that X takes. Example 25. We flip a coin twice. The probability space is Ω = {HH, HT, T H, T T }. Let X be the random variable associating to each outcome the number of tails in the outcome, i.e. X({HH}) = 0 and X({HT }) = 1 and X({T H}) = 1 and X({T T }) = 2. Example 26. We roll a die twice. The probability space is Ω = {(1, 1), (1, 2),..., (6, 6)}. Let X be the random variable associating to each outcome (a, b) the number a b. Thus, for example, we have X((1, 1)) = 0 and X((4, 2)) = 2 and X((1, 6)) = as opposed to rolling two die at once, so that order matters 13 Notice in particular that X can take negative values.

11 NOTES FROM MATH 10B DISCUSSION 11 Given a probability space Ω, the probability mass function (pmf) associated to X is the function f X : R R 0 defined by f X (r) = P ({ω Ω : X(ω) = r}) for each r R. The event {ω Ω : X(ω) = r} is also commonly denoted X = r. Example 27. We compute the pmf of Example 25. The range of X is {0, 1, 2}. We have f X (0) = P ({HH}) = 1 4 and f X(1) = P ({HT, T H}) = 2 4 and f X(2) = P ({T T }) = 1 4. For any x {0, 1, 2}, we have f X(x) = 0. Example 28. We compute the pmf of Example 26. The range of X is { 5, 4,..., 4, 5}. It is left to the reader to check that f X (n) = 6 n 36 for n { 5, 4,..., 4, 5}, and f X (x) = 0 if x { 5, 4,..., 4, 5}. Example 29. Suppose we choose 13 cards randomly from the standard deck of 52; thus Ω is the set of 13-card hands, and Ω = ( 52 13). Define X : Ω R to be the number of 2s in a 13-card hand. The range of X is {0, 1, 2, 3, 4}. It is left to the reader to check that ( 4 48 ) f X (n) = n)( 13 n ) ( for n {0, 1, 2, 3, 4} and f X (n) = 0 if n {0, 1, 2, 3, 4}. 10. Wed, Feb 18, 2015 Exercise 30. I play the following game: I roll a 12-sided die. If it lands on a prime, then I win 1 dollar and roll again. If lands on a nonprime, then the game is over. What s the probability that I win at least 3 dollars? Solution. This is a geometric distribution. There are 5 primes between 1 and 12, namely 2, 3, 5, 7, 11; thus the probability of landing on a prime is 5 12 and the probability of landing on a nonprime is I lose on the kth roll if I win the first k 1 and lose on the kth roll; the probability of this happening is ( 5 12 )k Thus we want k=4 ( 5 12 )k , or equivalently 1 3 k=1 ( 5 12 )k = ( 5 12 ) Exercise 31. I have a biased coin which lands on heads with probability 0.7. I flip the coin 10 times. What s the probability of getting at least 8 heads? Solution. This is a binomial distribution. The probability of getting exactly k heads is ( ) 10 k (0.7) k (0.3) 10 k. Thus we want ( ) 10 8 (0.7) 8 (0.3) 2 + ( ) 10 9 (0.7) 9 (0.3) 1 + ( 10) (0.7) 10 (0.3) 0. Exercise 32. The number of crimes in NYC per day is a Poisson random variable with average λ = 5. What s the probability of at most 1 crime happening today? Solution. Let X be the random variable. The probability that exactly k crimes occur is P (X = k) = e λ λ k k!, and we want P (X = 0) + P (X = 1) = e ! + e ! = 6e 5.

12 12 MINSEON SHIN 11. Fri, Feb 20, 2015 Exercise 33. Same setup as in Exercise 30. What is the expected value of the game? Solution. This is the expected value of a geometric random variable which has probability p = 7 12 of success (here success is the event that we stop rolling the die, i.e. the game is over), which is 1 p p = = Exercise 34. Same setup as in Exercise 31. What is the expected number of heads? Solution. This is the expected value of a binomial random variable with n = 10 trials and probability p = 0.7 of success, which is np = = 7. Exercise 35. Let Ω = {ω 1, ω 2, ω 3, ω 4 } be the sample space consisting of four outcomes whose probabilities are P (ω 1 ) = 1 2 and P (ω 2) = 1 3 and P (ω 3) = 1 7 and P (ω 4 ) = We define the random variables X, Y : Ω R sending X(ω k) = k 2 + 3k + 1 and Y (ω k ) = k2 2. Compute E[X] and Var[X] and E[Y ] and Var[Y ]. Solution. We have and E[X] = E[X 2 ] = 4 (k 2 + 3k + 1) P (X = k 2 + 3k + 1) k=1 = 5 P (X = 5) + 11 P (X = 11) + 19 P (X = 19) + 29 P (X = 29) = 5 P (ω 1 ) + 11 P (ω 2 ) + 19 P (ω 3 ) + 29 P (ω 4 ) = = (k 2 + 3k + 1) 2 P (X = (k 2 + 3k + 1) 2 ) k=1 = 5 2 P (X = 5 2 ) P (X = 11 2 ) P (X = 19 2 ) P (X = 29 2 ) = 5 2 P (ω 1 ) P (ω 2 ) P (ω 3 ) P (ω 4 ) = = so Var[X] = E[X 2 ] (E[X]) 2 =

13 NOTES FROM MATH 10B DISCUSSION 13 We have and E[Y ] = 4 k=1 k2 k2 2 P (Y = 2 ) = 0 P (Y = 0) + 2 P (Y = 2) + 5 P (Y = 5) + 8 P (Y = 8) = 0 P (ω 1 ) + 2 P (ω 2 ) + 5 P (ω 3 ) + 8 P (ω 4 ) = = 11 7 E[Y 2 ] = 4 k=1 k2 2 2 P (Y = k2 2 2 ) = 0 2 P (Y = 0 2 ) P (Y = 2 2 ) P (Y = 5 2 ) P (Y = 8 2 ) = 0 2 P (ω 1 ) P (ω 2 ) P (ω 3 ) P (ω 4 ) = = 45 7 so Var[Y ] = E[Y 2 ] (E[Y ]) 2 = Mon, Feb 23, 2015 Here are some more details about HW 5 #17. Problem 36 (HW 5 #17). Assume Ω is a finite set, X : Ω R is a random variable with range R, and f : R R is any function. Show that E[f(X)] = x R f(x) P (X = x). (2) Solution. 14 (See Example 37 for an example.) Notice that f(r) = range(f(x)). 15 In the following, the symbol f 1 (y) is defined to be the set {x R : f(x) = y}. For any y R, we have {ω Ω : f(x(ω)) = y} = x f 1 (y) R {ω Ω : X(ω) = x} (3) 14 Recall that the symbol f(x) means the composite function f X : Ω R, i.e. the random variable on Ω sending ω f(x(ω)). 15 Any y f(r) is equal to f(x) where x R, and in turn this x is equal to X(ω) for some ω Ω. Conversely, if ω Ω, then f(x(ω)) is f of something in R, namely X(ω).

14 14 MINSEON SHIN which implies P (f(x) = y) = 1 P ({ω Ω : f(x(ω)) = y}) 2 = P {ω Ω : X(ω) = x} 3 = 4 = x f 1 (y) R x f 1 (y) R x f 1 (y) R P ({ω Ω : X(ω) = x}) P (X = x) where equality 1 follows by the definition of P (f(x) = y), equality 2 is by (3), equality 3 is by additivity of probability for disjoint subsets of Ω, and equality 4 is by definition of P (X = x). We have E[f(X)] = 5 y P (f(x) = y) y f(r) 6 = y f(r) 7 = y x f 1 (y) R y f(r) x f 1 (y) R 8 = x R f(x) P (X = x) P (X = x) f(x) P (X = x) where equality 5 follows from the definition of expected value of the random variable f(x), equality 6 follows from the above computation, equality 7 is because x f 1 (y) implies f(x) = y, and equality 8 is because the indexing sets are equal, i.e. R = y f(r) (f 1 (y) R). Example 37 (Example associated to Problem 36). (This is a special case of HW 5 #14.) We flip a fair coin 3 times; let X be the random variable that counts the number of tails minus the number of heads; for example X(HTH) = 1 2 = 1. The range of X is R = { 3, 1, 1, 3}. Let f : R R be the function f(x) = x 2 ; then f(x) is the random variable that squares X; for example f(x)(hth) = (1 2) 2 = 1. The range of f(x) is f(r) = {1, 9}. The equation (2) is the statement that = ( 3) ( 1) (1) (3)2 1 8 since P (f(x) = 1) = 6 8, P (f(x) = 9) = 2 8, P (X = 3) = 1 8, etc. In case y = 1, the equation (3) is equivalent to the statement that {ω Ω : f(x(ω)) = 1} = {THH, HTH, HHT, HTT, THT, TTH} is equal to the union of the sets {ω Ω : X(ω) = 1} = {THH, HTH, HHT} {ω Ω : X(ω) = 1} = {HTT, THT, TTH}

15 NOTES FROM MATH 10B DISCUSSION 15 since f 1 (1) R = { 1, 1}. Example 38 (A remark about HW 5 #16). Suppose we have a geometric distribution Ω where the probability of success is p. Let X be the random variable assigning, to each outcome ω, the number of failures before the first success. Let Y be the random variable assigning, to each outcome, the number of trials required to obtain the first success. The difference between X and Y is whether we count the success roll or not. We have Y = X + 1, and E[X] = 1 p p and E[Y ] = 1 p. In HW 5 #16, we may either compute E[X] and add 2 at the end (1 for the first roll and 1 for the last success roll) or we may compute E[Y ] and add 1 at the end (1 for the first roll) Practice problems for the midterm. (1) (2) (3) Chapter4.pdf (starting at page 1) (4) (5) solutions.pdf (6) 20Solutions.pdf (7) Wed, Feb 25, 2015 Problem 39. Choose a random string of 9 digits. What s the probability that it has exactly 4 even 16 digits? Solution. There are ( 9 4) ways to choose the locations of the even numbers, 5 4 ways to choose the four even numbers, and 5 5 ways to choose the five odd numbers. In total, the number of 9-digit numbers is Thus the desired probability is ( 9 4) Alternatively, we may view this as a binomial distribution, where there are 9 trials (for each of the 9 digits) and the probability of success is p = 1 2 (since half of the digits are even); then we re looking for P (X = 4) = ( ) 9 4 ( 1 2 )4 (1 1 2 )5. Problem 40. Choose a random 5-card hand. What s the probability that you get one pair? 17 Solution. There are ( ) ( 13 1 ways to choose the value of the pair, 4 2) ways to choose the suits of the pair, ( ) 12 3 ways to choose the values of the three other cards, and 4 3 ways to choose the suits of the three other cards. Thus the desired probability ( is 13 1 )( 4 2)( 12 3 )4 3. ( 52 5 ) 16 Recall that 0 is an even number, i.e. it is divisible by Here I mean exactly one pair, i.e. we don t want the other three cards to be a triple, in which case the hand would constitute a full house.

16 16 MINSEON SHIN Problem 41. How many ways are there to distribute 245 indistinguishable balls to ten boxes labeled 1, 2,..., 10 such that, for all 1 i 10, box i gets at least i balls? 18 Solution. We first give 1 ball to box 1, give 2 balls to box 2, give 3 balls to box 3, etc. We may then distribute the remaining 245 ( ) = 190 balls to the 10 boxes arbitrarily; the number of such ways is ( ) 190+(10 1) 190. Problem 42. I roll 4 fair 6-sided die. What s the probability that at least one die shows a 1, given that the sum is 12? Solution. Let Z d (n, k) be the number of n-tuples (x 1,..., x n ) such that x i {1, 2,..., d} for all 1 i n and x 1 + +x n = k. We have that Z d (n, k) satisfies the recurrence relation d Z d (n, k) = Z d (n 1, k i) i=1 where the ith summand on the right hand side can be interpreted as the number of ordered n-tuples (i, x 2,..., x n ) such that x i {1, 2,..., d} and i+x 2 + +x n = k, or equivalently the number of ordered (n 1)-tuples (x 2,..., x n ) such that x i {1, 2,..., d} and x x n = k i. We are looking for 1 Z5(4,8) Z 6(4,12) ; here Z 5(4, 8) can be interpreted as the number of ordered 4-tuples (x 1, x 2, x 3, x 4 ) such that x i {2, 3,..., 6} and x 1 + x 2 + x 3 + x 4 = 12. We have so and so Z 6 (3, 11) = Z 6 (2, 10) + + Z 6 (2, 5) = = 27 Z 6 (3, 10) = Z 6 (2, 9) + + Z 6 (2, 4) = = 27 Z 6 (3, 9) = Z 6 (2, 8) + + Z 6 (2, 3) = = 25 Z 6 (3, 8) = Z 6 (2, 7) + + Z 6 (2, 2) = = 21 Z 6 (3, 7) = Z 6 (2, 6) + + Z 6 (2, 1) = = 15 Z 6 (3, 6) = Z 6 (2, 5) + + Z 6 (2, 0) = = 10 Z 6 (4, 12) = Z 6 (3, 11) + + Z 6 (3, 6) = = 125 Z 5 (3, 7) = Z 5 (2, 6) + + Z 5 (2, 2) = = 15 Z 5 (3, 6) = Z 5 (2, 5) + + Z 5 (2, 1) = = 10 Z 5 (3, 5) = Z 5 (2, 4) + + Z 5 (2, 0) = = 6 Z 5 (3, 4) = Z 5 (2, 3) + + Z 5 (2, 1) = = 3 Z 5 (3, 3) = Z 5 (2, 2) + + Z 5 (2, 2) = = 1 Z 5 (4, 8) = Z 5 (3, 7) + + Z 5 (3, 3) = = I mean here that box 1 gets at least 1 ball, box 2 gets at least 2 balls, box 3 gets at least 3 balls, etc.

17 NOTES FROM MATH 10B DISCUSSION 17 thus the desired probability is = Problem 43. I roll a fair 6-sided die until I get two 4s in a row. expected number of times I have to roll the die? What s the Solution. 19 Let a be the desired answer, i.e. the expected number of times I have to roll the die until I get two 4s in a row. Let b be the expected number of times I have to roll the die until I get two 4s in a row, assuming that I have just rolled a We have a = (a + 1) (b + 1)1 6 since either the 1st roll is not 4 (with probability 5 6 ) and we are effectively back to square one (so that the expected number of times you have to roll after the 1st roll is again a), or the 1st roll is a 4 (with probability 1 6 ) and the expected number of rolls is after the 1st roll is b (by definition of b). We have b = (a + 1) (1)1 6 since either the 1st roll is not 4 (with probability 5 6 ) and we are effectively back to square one (so that the expected number of times you have to roll after the 1st roll is again a), or the 1st roll is a 4 (with probability 1 6 ) and we stop. Solving the above system of equations gives b = 36 and a = 42. Problem 44. I roll a fair 24-sided die until two of the previous rolls differ by a multiple of 11. What s the probability that I have to roll 15 times? 21 Solution. This is a pigeonhole problem in disguise of a geometric distribution problem. I claim that the game must stop after at most 12 rolls. Suppose I have 12 rolls of the die; if any two of them are equal, then their difference is 0, which is a multiple of 11, hence the game must have stopped. Suppose all 12 rolls are distinct. Let us partition {1, 2,..., 24} into the following 11 groups: {1, 12, 23}, {2, 13, 24}, {3, 14}, {4, 15}, {5, 16}, {6, 17}, {7, 18}, {8, 19}, {9, 20}, {10, 21}, {11, 22} There are 12 pigeons (corresponding to each roll of the die) and 11 pigeonholes (corresponding to the groups above), so by the pigeonhole principle at least one pigeonhole must have two pigeons, i.e. there are two rolls contained in the same group; these rolls differ by a multiple of From 20 Equivalently, b is the expected number of times I have to roll if I stop after either rolling a 4 on the 1st roll or rolling two 4s in a row after not rolling a 4 on the 1st roll. 21 For example, two possible sequence of rolls would be 1, 20, 6, 7, 3, 21, 15, 19, 14 (since 3 14 is a multiple of 11) or 1, 3, 1 (since 1 1 is a multiple of 11).

18 18 MINSEON SHIN 14. Mon, Mar 2, 2015 (Statistics) Definition 45. A random variable is discrete if its range is finite, i.e. finitely many possible values the random variable can take. there are Example 46. I flip a fair coin once; let X be the random variable taking the value 1 if the coin shows heads and 0 if the coin shows tails. Then the range of X is a finite set, namely {0, 1}; thus X is discrete. In Math 10B, we have been working only with discrete random variables, but continuous random variables will be important as well. Definition 47. A random variable is continuous if there exists a continuous nonnegative function f : R R such that P (a < X < b) = b a f(x) dx for all a < b. We call f the probability density function (or pdf) of X. Here is an important example of a continuous random variable. Example 48. A normal random variable (or Gaussian random variable) is a continuous random variable X whose pdf is of the form f(x) = 1 ) ( σ 2π exp (x µ)2 2σ 2 (4) for some µ R and σ R >0. 22 We can verify that, if µ and σ are as above, then E[X] = µ and Var[X] = σ 2. Definition 49. Let X 1 and X 2 be two random variables. (i) If X 1 and X 2 are discrete, we say that they are identically distributed if their probability mass functions (pmf) are the same. (ii) If X 1 and X 2 are continuous, we say that they are identically distributed if their probability density functions (pdf) are the same. Example. I toss a fair coin 4 times. We define the following (discrete) random variables: (i) X 1 is the number of heads (ii) X 2 is the number of tails (iii) X 3 = X 1 X 2 (iv) X 4 is 1 if the first flip is heads and 0 if the first flip is tails (v) X 5 is 1 if the second flip is heads and 0 if the second flip is tails Then X 1 and X 2 are not independent but they are identically distributed; X 1 and X 3 are not independent and not identically distributed; X 4 and X 5 are both independent and identically distributed. The Law of Large Numbers tells you why you need to perform a large number of experiments for your results to be accurate. 22 Here exp(x) means e x.

19 NOTES FROM MATH 10B DISCUSSION 19 Theorem 51 (Law of Large Numbers). Let X 1, X 2, be independent and identically distributed discrete random variables. Then for each ɛ > 0 we have lim P ( Y n µ > ɛ) = 0 n where Y n := X1+ +Xn 24 n and µ := E[X i ] for any i. 25 The Central Limit Theorem tells you why the normal curve appears so often in statistics. Theorem 52 (Central Limit Theorem). Assume the notation of Theorem 51, and suppose in addition that each X i is bounded. Set σ := Var[X i ] for any i and set Z n := Yn µ σ/ n. Then lim P (a < Z n < b) = 1 b n 2π a e x2 /2 dx for all a < b. 26 (Note that the RHS is the integral of (4) in the case µ = 0 and σ = 1.) Parameter estimates. Definition 53. A statistical model is a collection P of pdfs (in case we re dealing with continuous random variables) or pmfs (in case we re dealing with discrete random variables); statistical models are indexed by one or more parameters. 27 Example 54. Assume we have a coin and that it is possibly biased. We flip it n times. A statistical model for the number of heads would be { ( } n P = f p (k) = )p k (1 p) n k k p [0,1] where every f p is a pmf; here the parameter p represents the bias of the coin (i.e. it has probability p of showing heads). Example 55. A statistical model for the heights of Berkeley students, assuming that it approximately has the normal distribution, would be { ( P = f µ,σ (x) = 1 σ 2π exp 1 ( ) )} 2 x µ 2 σ µ R 0,σ R >0 where every f µ,σ is a pdf; here the statistical model is indexed by two parameters, namely µ and σ. Definition 56. A statistic 28 is a rule for guessing the value of a population parameter θ based on a random sample from the population. A statistic is a random variable on the probability space of random samples of the population. More 23 In other words, there is a random variable for each positive integer n. 24 This random variable is also denoted by X, but this notation doesn t make it clear that it depends on n. 25 The assumption that the {Xi } n 1 are identically distributed implies in particular that E[X i ] = E[X j ] for any i, j; thus it doesn t matter which index i is used to define µ. 26 See also ~stark/java/html/normhilite.htm. 27 See Prob-Stat.pdf slide See Prob-Stat.pdf slide 75.

20 20 MINSEON SHIN precisely, given random variables X 1,..., X n, a statistic 29 is a random variable R of the form f(x 1,..., X n ) where f : R n R is some function. Example 57. Let X 1,..., X n be random variables, and let f : R n R be a function. Here are some possible functions f that give familiar statistics: (i) f 1 (x 1,..., x n ) = x1+ +xn n (ii) f 2 (x 1,..., x n ) = max{x 1,..., x n } 1 n (iii) f 3 (x 1,..., x n ) = n 1 k=1 (x k f 1 (x 1,..., x n )) 2, where f 1 is as in (i) Definition 58. A 95% confidence interval 30 for the extimate µ of the expected value is ( µ 2 σ n, µ + 2 σ n ) where µ and σ are the estimates for the expected value and standard error, computed from the observed data. Example 59. Suppose that the true proportion of the U.S. voting population that approves of the President is p; we want to estimate p. We survey 1000 people and find that 523 of them approve. Suppose s 1,..., s 1000 are the people surveyed, and let X i be the random variable that takes value 1 if person s i says they approve of the president, and 0 otherwise. Then we may treat X i as a Bernoulli random variable (on the probability space Ω = {U.S. voting population}) with parameter p. Here µ = and σ = µ(1 µ) = Then we say with 95% confidence that p lies in the interval ( , ) Hypothesis testing. 31 Hypothesis tests are like statistical analogues of proofs by contradiction (at least in spirit). Suppose I am playing Risk with a friend and he suggests using five die, one of which happens to show 6 on five out of seven rolls. I get angry and I want to convince everyone in the world that the die is not fair (i.e. some numbers have higher probability of showing up than other numbers). How would I go about doing such a thing? It would be unreasonable for me to somehow dissect the die and show that one side has more atoms than the others. I would instead devise a hypothesis test; if I assume ( for the sake of contradiction ) that the die is fair, take some measurements of the die (i.e. roll it a lot of times) and show (using some calculations) that the observed outcome is highly improbable (say, has less than 5% 32 chance of happening) given this assumption, then hopefully every reasonable person will conclude that the die is not fair. The first thing to do is to choose a null hypothesis 33 ; in this case, my null hypothesis is H 0 : The die is fair. 29 See Discrete Probability.pdf slide You might be wondering where the number 0.95 comes from; it is an approximation of 1 2π 2 2 e x2 /2 dx; here 1 2π e x2 /2 is the pdf of the normal distribution. Compare with the Central Limit Theorem. 31 See Discrete Probability.pdf slide I take α = 0.05 because I heard it s what statisticians usually do. We could also take α = 0.03; as α becomes smaller, it becomes harder to reject the null hypothesis. 33 The word null in null hypothesis should be interpreted as default or distinguished or the one in question ; it refers to the 0 in H 0. It seems to have little to do with the act of nullifying (e.g. a law).

21 NOTES FROM MATH 10B DISCUSSION 21 Some alternative hypotheses could be H 1 : 6 is twice as likely to show up as any other number. H 2 : 6 is three times as likely to show up as any other number. (Since I am worried about credibility and I am paranoid about saying something wrong, I would rather make H 0 be a laughably ridiculous claim and say H 0 is false than say H 1 is true or H 2 is true. 34 ) I conduct my experiment (e.g. roll the die 300 more times) and record the number of times each number shows up. Value Observed Frequency Expected Frequency In the above table, Expected Frequency is the expected number of times we expect each number to show up when the die is rolled 300 times, assuming that the null hypothesis is true, i.e. that the die is fair. Now I compute what is called the goodness of fit statistic (it is an example of the notion of statistic as defined in Definition 56), which in this case turns out to be r = (47 )2 + (46 )2 + (38 )2 + (41 )2 + (55 )2 + (73 )2 = according to the observed frequencies. The goodness-of-fit statistic r measures the difference between the observed outcomes and the expected outcomes. I define the random variables N 1,..., N 6, where N i counts the number of i s that show up when I roll this die 300 times, and I define the random variable R = (N1 )2 + (N2 )2 + (N3 )2 + (N4 )2 + (N5 )2 + (N6 )2 as well. Since 300 is a large number of trials, a deep theorem of probability theory tells me that 1 P (R 16.08) 2 3 x 2 e x/2 dx Γ(3) which in other words states that the probability of getting a result at least as extreme as the one observed, assuming the null hypothesis is true, is approximately ; and since this is less than 0.05 (or 5%), the international statistics community allows me to rest my case, or reject the null hypothesis. This probability P (R r) is called the p-value, and measures the probability of obtaining a result at least as extreme as the one observed, assuming the null hypothesis. What would have happened if my experiment went as follows? 34 Okay, I m not saying the die is loaded, but...

22 22 MINSEON SHIN Value Observed Frequency Expected Frequency Then the goodness of fit statistic would be r = (47 )2 + (46 )2 + (44 )2 + (43 )2 + (51 )2 + (69 )2 = 9.44 according to the observed frequencies, and the deep theorem of probability theory would tell me that 1 P (R 9.44) 2 3 x 2 e x/2 dx Γ(3) 9.44 and since this is not less than 0.05 (or 5%), I would have failed in trying to prove my case ( fail to reject the null hypothesis ). This is called a Type II error, in other words a false negative. 35 Definition 60. The Gamma function is defined by Γ(x) = 0 t x 1 e t dt for all positive real numbers x. We have Γ(n) = (n 1)! for all positive integers n. Definition 61. Let X be a random variable. We say that X has the chi-squared distribution with k degrees of freedom if the probability density function of X is 1 f X (x) = 2 k/2 Γ(k/2) xk/2 1 e x/2 for all x R. This implies in particular that for all 0 a < b. 36 P (a X b) = 1 2 k/2 Γ(k/2) b 15. Wed, Mar 4, 2015 a x k/2 1 e x/2 dx Problem I flip a biased coin 100 times and get 14 heads. Use this to construct a 95% confidence interval for p, the probability of heads. Solution. The sample mean is µ = = , and since this is a bernoulli distribution, the probability of heads is equal to the expected value of the number of heads on any given flip; thus we set p = µ. We set Var = p(1 p) = ; thus 2 our 95% confidence interval for p is ( / , / ). 35 Max Friedlander, the great art historian, once said, It may be an error to buy a work of art and discover that it is a fake, but it is a sin to call a fake something that is genuine. (John McPhee, Roomful of Hovings ) 36 See also ~stark/java/html/chihilite.htm. 37 See Example 8.2, 8.4, 8.6 in Discrete Probability.pdf.

23 NOTES FROM MATH 10B DISCUSSION 23 Problem Earthquakes in Berkeley follow a Poisson distribution with unknown parameter λ per year. In the last 10 years, there were 9, 13, 8, 10, 9, 4, 1, 11, 2, 6 earthquakes in Berkeley. Use this to construct a 95% confidence interval for λ. Solution. The sample mean is µ = = 7.3, and since this is a Poisson distribution, the expected value and variance are both equal to λ; thus we take Var = λ = µ. Thus our 95% confidence interval for λ is ( , ). Problem 64. I flip a biased coin until it shows heads. I did this 10 times, and it took me 2, 9, 5, 5, 8, 7, 1, 8, 10, 1 flips, respectively (including the last flip). Use this to construct a 95% confidence interval for p, the probability of heads. Solution. The sample mean is µ = = 5.6. The expected value of a geometric random variable (in which we count the success as well) with probability p of success is 1 1 p p, and the variance is p. Thus we can take p to satisfy 2 10 µ = 1 p = 5.6, in which case p = 56, and we can take the sample variance to be Var = 1 p p = 25.76; thus our 95% confidence interval is (5.6 2( ), ( )). 16. Mon, Mar 9, 2015 There were requests for #6 and #12 on Homework 6. For #6, here is a solution without using any statistics. We make the change of variable t = x 2 so that 0 x 2 e x/2 dx = 0 (2t) 2 e (2t)/2 2 dt = 8 I also gave a preview of dynamics and recursion. Example 65. Consider the recurrence relation for n 2. 0 t 2 e t dt = 8 Γ(3) = 16. a n = 2a n 1 a n 2 (5) (1) Suppose the initial conditions are a 0 = 1 and a 1 = 1. The first few terms of the sequence are a 0 = 1, a 1 = 1, a 2 = 1, a 3 = 1, etc. In this case a n = 1 for all nonnegative integers n, since if a n 2 = 1 and a n 1 = 1 then (5) implies a n = 2a n 1 a n 2 = = 1. (2) Suppose the initial conditions are a 0 = 2 and a 1 = 5. The first few terms of the sequence are a 0 = 2, a 1 = 5, a 2 = 8, a 3 = 11, etc; this seems to be an arithmetic sequence with common difference 3, starting from a 0 = 2. So 38 See Example 8.3, 8.5, 8.7 in Discrete Probability.pdf.

24 24 MINSEON SHIN let s show that a n = 3n + 2 for all n We have a 0 = 2 = 3(0) + 2 and a 1 = 5 = 3(1) + 2, so the formula is true for n = 0, 1. Suppose we have shown the formula for 0 n k, and let s show that a k+1 = 3(k + 1) + 2. We have a k+1 = 2a k a k 1 = 2(3k+2) (3(k 1)+2) = 3k+5 = 3(k+1)+2. In general, by writing (5) as a n a n 1 = a n 1 a n 2 we can show that a n = a 0 + (a 1 a 0 )n; here a 0 is the starting term and a 1 a 0 is the common difference. Example 66. Consider the recurrence relation a n = 3a n 1 2a n 2 (6) for n 2 with initial conditions a 0 = 2 and a 1 = 5. Writing (6) as the characteristic polyomial is a n 3a n 1 + 2a n 2 = 0 x 2 3x + 2 which has roots 1 and 2; the general solution to (6) is a n = c 1 1 n + c 2 2 n. We can use the initial conditions to solve for c 1 and c 2 ; for n = 0, we have c 1 + c 2 = c c = a 0 = 2; for n = 1, we have c 1 + 2c 2 = c c = a 1 = 5; so we get a system of linear equations c 1 + c 2 = 2 c 1 + 2c 2 = 5 which implies c 1 = 1 and c 2 = 3. Thus a n = n for all n Wed, Mar 11, 2015 The following is regarding the Theorem on slide 17 of Dynamics.pdf, in particular the case when β 0 (so the recursion is inhomogeneous) and α 1. We want to show that the solution to a n = αa n 1 + β (7) with initial condition a 0 is a n = α n a 0 + β( αn 1 α 1 ). I claim that, by rewriting (7) as a n c = αa n 1 + β c (8) for some carefully chosen c (which will depend only on α and β), we ll reduce to the case when β = 0. I want to choose c such that c 1 = β c α ; for such c, if I define a n := a n c for all n, I can rewrite (7) as which is easy to solve. Solving c 1 = β c α able to rewrite (7) as a n a n = αa n 1 (9) β ( 1 α = α a n 1 β for c gives c = 1 α. This means I am β ) 1 α 39 How did I get an = 3n + 2? Any arithmetic sequence is of the form a n = dn + e where d is the common difference and e is the initial term. We are guessing that d = 3 (by looking at the first few terms), and e = a 0 = 2 by definition. (10)