x x 1 0 1/(N 1) (N 2)/(N 1)

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1 Please simplify your answers to the extent reasonable without a calculator, show your work, and explain your answers, concisely. If you set up an integral or a sum that you cannot evaluate, leave it as it is; and if the result is needed for the next part, say how you would use the result if you had it. 1. Suppose Bob is trying to guess a specific natural number x {1,...,N}. On his first guess he chooses a number X 0 uniformly at random. For t N, if X t x the game ends; if X t x, he guesses X t+1 uniformly at random from among the numbers different from X t. a. [5 points] What is the expected number of guesses it takes Bob to find x? WecandescribethisasaMarkovprocessontwostatesx and x {1,...,N}\{x }, with transition probability matrix: x ( x ) x 1 0 x. 1/(N 1) (N 2)/(N 1) Let ν be the expected number of steps it takes to reach x from x. Then from P we have ν 1+ N 2 N 1 ν, which we can solve to find ν N 1. Then the expected number of guesses it takes Bob to find x is 1 N (0+1)+ N 1 N (N 1+1) N 1+ 1 N. b. [5 points]supposebobhasabadmemoryandcan trememberthenumber heguessed previously, so that he guesses X t+1 uniformly at random from among {1,...,N}. In this case what is the expected number of guesses it takes him to find x? Bob s bad memory changes the transition probability matrix to: Now ( x x ) x 1 0 x. 1/N (N 1)/N ν 1+ N 1 N ν, so ν N and the expected number of guesses is 1 N (0+1)+ N 1 N (N +1) N.

2 c. [5 points] Suppose Bob has a good memory, and at each step guesses uniformly at random among the numbers he has not guessed at any previous step. Now what is the expected number of guesses it takes him to find x? Bob has equal probability 1/N of guessing correctly on any of guesses 1 through N, so the expected number of guesses in this case is N k1 k 1 N N(N 1) 2 1 N N [15 points] 2 n N is a prime number if its only divisors are 1 and itself. The Prime Number Theorem says that the primes are distributed approximately as if they came from an inhomogeneous Poisson process P(x) with intensity λ(x) 1/lnx. [We have to say approximately since (1) the primes are integers, not general real numbers, and (2) the primes take determined, not random, values.] Use this theorem to estimate the number of primes in the interval [2,N]. Since P(x) is an inhomogeneous Poisson process, the expected number of events in the interval [2,N] is N 1 E[P(N)] E[P(2)] lnx dx. This estimates the number of primes in this interval, and is approximately N/lnN. 3. Suppose we flip a fair coin repeatedly. a. [5 points] What is the expected number of flips until we see Head followed by Tail? We can think of this a Markov process on three states: H, meaning the most recent flip was Head; HT meaning the most recent two flips were Head then Tail; and 0 meaning anything else. Then the transition probability matrix is 2 0 H HT 0 1/2 1/2 0 H 0 1/2 1/2. HT Using ν HT E[#flips HT] 0, from P we have ν 0 E[#flips 0] ν ν H ν H E[#flips H] ν H Solving this system of linear equations gives ν 0 4.

3 b. [5 points] What is the expected number of flips until we see Head followed by Head? Again this a Markov process on three states; now HH, meaning the most recent two flip were both Heads; H meaning only the most recent flips was Head; and 0 meaning anything else. Then the transition probability matrix is 0 H HH 0 1/2 1/2 0 H 1/2 0 1/2. HH Using ν HH E[#flips HH] 0, from P we have ν 0 E[#flips 0] ν ν H ν H E[#flips H] ν 0 Solving this system of linear equations gives ν 0 6. c. [5 points] Give an intuitive explanation for why your answers in (a) and (b) are the same or different. In the Markov chain in case (a), once it reaches state H it cannot return to 0, but in case (b), it can. This makes the expected number of steps to reach the absorbing state longer in case (b). 4. [15 points] Here is a list of the 23 prime numbers between 3 and 100, together with their remainders when divided by 3: In this list 1 follows 1 three times; 2 follows 1 seven times; 1 follows 2 eight times; and 2 follows 2 four times, so we might imagine that the sequence of remainders of prime numbers divided by 3 are the outcome of a Markov process with transition matrix ( 1 2 ) 1 3/10 7/ /3 1/3 If this were true, what fraction of all prime numbers would we expect to have remainder 1 when divided by 3? P is regular, so the limiting distribution for this Markov process is its stationary state, i.e., its left eigenvector with eigenvalue 1: (p 1 p) (p 1 p), so 3 10 p+ 2 (1 p) p, 3 which implies p 20/41; this is the fraction of all primes we would expect to have remainder 1 when divided by 3, if this were a good model for the prime numbers.

4 In fact, a generalization of the theorem mentioned in problem 2, the Prime Number Theorem for arithmetic sequences, says that in the limit as N, 1/2 of prime numbers less than N have remainder 1, and half have remainder 2, when divided by 3. But a recent paper by Robert J. Lemke Oliver and Kannan Soundararajan, Unexpected biases in the distribution of consecutive primes, arxiv: [math.nt], shows that even for very large N, successive primes more often have different than the same remainders when divided by 3. They show that a conjecture of Hardy and Littlewood implies that this bias goes away, slowly, as N, but that conjecture remains unproved. 5. Let X(t) be a Poisson process on R 0, with intensity λ. Suppose Y i Poisson(µ) are independent of one another and of X(t). Let Y(t) X(t) i1 a. [6 points] What is E[Y(t)]? What is Var[Y(t)]? E[Y(t)] λt µ. Var[Y(t)] λt µ+µ 2 λt. b. [6 points] What is E[Y(t) X(t) n]? What is Var[Y(t) X(t) n]? E[Y(t) X(t) n] nµ. Var[Y(t) X(t) n] nµ. c. [4 points] Is Y(t) a Poisson process? Why or why not? No. One way to see this is that E[Y(t)] Var[Y(t)], so Y(t is not a Poisson random variable. Another way to see it is that Y(t) does not make jumps of size 1 the way a Poisson process does. d. [4 points] Give a different probability function for Y i that makes Y(t) a Poisson process. The only possibility is Y i Bernoulli(p). Y i.

5 6. There is a 500m. 1000m. plot of land on Barro Colorado Island in Panama where a careful census has been made of the trees. The locations of trees of one common species, Alseis blackiana, are indicated by dots in the graphic below, where the size of each dot represents the size of the tree. a. [10 points] Do you think the locations of these trees should be modeled as arising from a Poisson process? Why or why not? I think not. At best it appears the Poisson process would be quite inhomogeneous. Also, since a tree cannot grow within another tree, the number of trees within the radius of a tree of its center does not follow a pre-specified, even inhomogeneous, Poisson distribution. Finally, although it is maybe hard to see this without doing some calculations, the numbers of trees in disjoint regions seem not to be independent. b. [10 points] There are 7599 dots in this graphic. If we partition the plot into four congruent rectangles, by dividing it in half vertically and horizontally, the number of dots in each rectangle is 2115, 1950, 1708 and 1826, moving counterclockwise from the northeast rectangle. What is the probability of observing this distribution if the tree locations arise from a homogeneous Poisson process? Conditioning on there being 7599 dots, a homogeneous Poisson process restricted to this rectangle becomes 7599 samples of a uniform process on the rectangle. Since there is probability 1/4 for each sample to fall into each subrectangle, the probability of seeing this distribution is 7599! (1) ! 1950! 1708! 1826! 4 Even without a calculator, we can see that this will be a very small number because the numbers of trees in each subrectangle are so different; if this were really 7599 samples from a uniform distribution it is much more probable that they all would be close to 1900.

6 7. Let Y 1,Y 2 {0,1} be Bernoulli random variables. Suppose and Pr(Y 1 1) p. Pr(Y 2 0 Y 1 0) r Pr(Y 2 1 Y 1 1) s a. [4 points] What is the joint probability function for Y 1 and Y 2? Pr(Y 1 0 and Y 2 0) Pr(Y 2 0 Y 1 0)Pr(Y 1 0) r(1 p) Pr(Y 1 0 and Y 2 1) Pr(Y 2 1 Y 1 0)Pr(Y 1 0) (1 r)(1 p) Pr(Y 1 1 and Y 2 0) Pr(Y 2 0 Y 1 1)Pr(Y 1 1) (1 s)p Pr(Y 1 1 and Y 2 1) Pr(Y 2 1 Y 1 1)Pr(Y 1 1) s(1 p) With the values of Y 1 and Y 2 labeling the rows and columns of a matrix, respectively, this joint probability function is: 0 1 ( ) 0 r(1 p) (1 r)(1 p). 1 (1 s)p ps b. [4 points] For what values of r, s and p are Y 1 and Y 2 independent? To be independent, the joint probability matrix above must be singular, i.e., have nonzero determinant because one row is a multiple of the other since the distribution is a product distribution. Its determinant is rsp(1 p) (1 r)(1 s)p(1 p) (r+s 1)p(1 p) which vanishes if p {0,1} or if r +s 1. c. [4 points] What is Pr(Y 2 1)? Pr(Y 2 1) (1 r)(1 p)+sp. d. [4 points] For what values of r, s and p is Pr(Y 2 1) Pr(Y 1 1)? Solving p (1 r)(1 p)+sp for p gives as long as r +s 2. p 1 r 2 r s, 8. Consider a sequence of customers entering a store. Let Y i {0,1} denote the number of items the i th customer buys, for 0 < i N. Suppose every customer, after the first, sees what the previous customer does; if the previous customer bought nothing, the current customer also buys nothing, with probability r, and if the previous customer bought an item, the current customer does too, with probability s. a. [4 points] Suppose 0 < r s < 1. Without doing any calculation, explain what is the fraction of customers who buy an item in the infinite number of customers limit.

7 In this case buying and not-buying are completely symmetrical, so in the infinite number of customer limit, half of the customers buy an item. b. [4 points] For general 0 < r,s < 1, what is lim n Pr(Y n 1)? The transition probability matrix is ( r 1 r 1 1 s s and we want to find π (1 π 1 π 1 ) that solves π π. Multiplying out gives us the equation r(1 π 1 )+(1 s)π 1 1 π 1, which we solve to get ), lim Pr(Y n 1) π 1 1 r n 2 r s. (Notice that this is the same as the answer to problem 1.d, i.e., that the probability of each customer buying an item is the same.) c. [8 points] Still for general 0 < r,s < 1, what is lim n Pr(Y n 1 and Y n+1 1)? The probability transition matrix for the states of consecutive customers is r 1 r s s 10 r 1 r s s Solving π (π 0 π 1 π 2 π 3 ) πp requires solving: rπ 0 +rπ 2 π 0 (1 r)π 0 +(1 r)π 2 π 1. (1 s)π 1 +(1 s)π 3 π 2 sπ 1 +sπ 3 π 2 The first two of these equations imply π 0 (r/(1 r))π 1 and the last two imply π 3 (s/(1 s))π 2. Plugging these back in to the first and last equations gives π 1 π 2. Requiring that the sum of the π i be 1 gives π 1 (1 r)(1 s) 2 r s π 2,

8 which implies that lim Pr(Y n 1 and Y n+1 1) π 3 s(1 r) n 2 r s. An alternate, cleverer, solution is to notice that lim Pr(Y n 1 and Y n+1 1) lim Pr(Y n+1 1 Y n 1)Pr(Y n 1) n n s lim Pr(Y n 1) n 1 r s 2 r s, where we have used the answer to 2.b for the last step. 9. [16 points] Now suppose people enter a store according to a Poisson process with rate λ. The store owner wants to know if someone buying something makes the next person more likely to buy something (as it did in problem 2 for s > 1/2). To help answer this question, suppose each person who enters the store buys something with probability p, independently of what anyone else does. Let C(t) be the number of people who buy something and are followed by the next customer also buying something, both before time t. What is E[C(t)]? E[C(t)] E[C(t) X(t) n]pr ( X(t) n ) n2 n 1 E[Y i 1 and Y i+1 1]Pr ( X(t) n ) n2 i1 (n 1)p 2(λt)n e λt n! n2 p 2 e (λt λt (λt) n 1 (n 1)! (λt) n ) n! n2 n2 p 2 e λt( λt(e λt 1) (e λt 1 λt) ) p 2 (e λt 1+λt). Extra Credit. [5 points] Suppose instead of the people s purchasing decisions being independent, they are dependent as in problem 2. Now what is E[C(t)]? Since we don t know what the first customer does, or even what the probability distribution for the first customer s action is, we have to make some assumption. If we assume that customer probabilities are stationary, the only change in the calculation above is that the p 2 is replaced by lim n Pr(Y n 1 and Y n+1 1) s(1 r) 2 r s

9 from problem 2.c. 10. Let {X(t) t R 0 } be a Poisson point process with rate λ on R 0. For each point 0 < i N of the process, let D i be the distance to its nearest neighbor. a. [3 points] Write D i in terms of the sojourn times {S j j N}. D 1 S 1 and for i > 1, D i min{s i 1,S i }. b. [3 points] Are the {D i 0 < i N} independent? No, since D 1 and D 2, for example, both depend upon S 1. For example, let 0 s < t; then while Pr(D 1 > s,d 2 > t) Pr(S 1 > s,s 1 > t,s 2 > t) Pr(S 1 > t,s 2 > t) Pr(S 1 > t)pr(s 2 > t), Pr(D 1 > s)pr(d 2 > t) Pr(S 1 > s)pr(s 1 > t,s 2 > t) Pr(S 1 > s)pr(s 1 > t)pr(s 2 > t). c. [10 points] What is the probability density function for each D i? f D1 (d) f S1 (d) λe λd. For i > 1, Pr(d < D i d+ d) Pr(d < distance to nearest point d+ d) Pr(d < distance to point on left d+ d and d+ d < distance to point on right) +Pr(d < distance to point on right d+ d and d+ d < distance to point on left) 2e λd λ de λ d e λd, which implies f Di (d) 2λe 2λd, i.e., just the probability distribution for a sojourn time with double the rate. Alternatively, Pr(D i > d) Pr(S i 1 > d,s i > d) Pr(S i 1 > d)pr(s i > d) e λd e λd e 2λd. From this we can compute the probability density function: for d 0. f Di (d) d dd Pr(D i > d) d dd e 2λd 2λe 2λd, 11. [20 points] An ant starts at one vertex of a cube and at each time step walks along an edge to an adjacent vertex, choosing each possible edge with equal probability 1/3.

10 What is the probability the ant returns to its original vertex before reaching the opposite vertex? Let the state of the ant be described by how many edges away from the corner at which it starts it is, e.g., the opposite corner is state 3. Making states 0 and 3 absorbing states, the probability transition matrix is /3 0 2/ /3 0 1/ Sinceafterthefirststeptheantisatstate1,wemustcomputeν i Pr(absorbed at state 0 from state i) for i 1. Considering the next step the ant takes, we get Solving these equations gives ν 1 3/5. ν ν 2 ν ν Let X,Y be random variables with a bivariate normal distribution such that E[Y] 0 and Var[X] 1 Var[Y]. Suppose E[X Y y] 2 y/3. a. [6 points] What is E[X]? E[X] b. [10 points] What is Cov[X,Y]? E[X Y y]f Y (y)dy (2 y/3)f Y (y)dy E[2 Y/3] 2 E[Y]/3 2. Cov[X,Y] E[XY] E[X]E[Y] E[XY] E[XY Y y]f Y (y)dy ye[x Y y]f Y (y)dy (2y y 2 /3)f Y (y)dy E[2Y Y 2 /3] 2E[Y] E[Y 2 ]/3 1/3, because E[Y] 0 and thus E[Y 2 ] Var[Y] 1. Notice that we did not need to use the fact that the joint distribution of X and Y is bivariate normal.