1 Math 285 Homework Problem List for S2016

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1 1 Math 85 Homework Problem List for S016 Note: solutions to Lawler Problems will appear after all of the Lecture Note Solutions. 1.1 Homework 1. Due Friay, April 8, 016 Look at from lecture note exercises: 1.1 Han in lecture note exercises: 1., 1.3, 1.4, 1.5, 1.6 Han in from Lawler 5.1 on page Homework. Due Friay, April 15, 016 Look at from lecture note exercises: 1.8, 1.9, 1.11 Han in lecture note exercises: 1.7, 1.10, 1.1 Han in from Lawler 1.1, 1.4, Homework 3. Due Friay, April, 016 Look at from Lawler 1.10, Han in from Lawler 1.5, 1.8, 1.9, 1.14, 1.18*,.3 *Hint: show the invariant istribuiton is uniform. 1.4 Homework 4. Due Friay, April 9, 016 Look at lecture note exercises: 1.13 Han in from Lawler problems: 7.6, 7.7, 7.8. Please use the result in 7.8 to verify your numerical approximations foun in 7.7. *Hints for 7.8. Recall that for n N that { } S n = k = k 0,..., k n ) {0, 1} n+1 : k i 1 + k i 1 for 1 i n an your goal is to compute p n m) := # {k S n : k m = 1} # S n ) As Lawler suggests, for i, j {0, 1} an m N, let 1 for 0 m n. r m ij) = # {k = k 0,..., k m ) S m : k 0 = i an k m = j}. Notice that if k = k 0,..., k n ) S n with k m = 1, then 0, k 0,..., k m 1, 1) S m+1 an 1, k m+1,..., k n, 0) S n m+1 an visa versa) where k m 1 an k m+1 must both be zero so that 0, k 0,..., k m, 0) S m an 0, k m+,..., k n, 0) S n m. From these consierations we fin # {k S n : k m = 1} = r m+1 01) r n m+1 10) = r m 00) r n m 00). Similarly k = k 0,..., k m ) S m then 0, k 0,..., k n, 0) S n+ an visa versa from which we learn # S n ) = r n+ 00). Combining these results shows p n m) = r m 00) r n m 00). r n+ 00) A little thought shows this formula is correct at m = 0 an m = n m provie we use the convention that r 0 00) = 1. Now follow the outline in the Lawler in orer to fin, y m = y m) := r m 00). 1.5 Homework 5. Due Friay, May 6, 016 Look at from lecture note exercises: 1.14, 1.15 Han in lecture note exercises: 1.16, 1.18 Han in from Lawler 5.4, 5.7a, 5.8a, I have change the inexing a bit since Lawler s choices are a bit confusing.

2 1 Math 85 Homework Problem List for S Homework 6. Due Friay, May 13, 016 Look at from lecture note exercises: 1.19, 1.0, 1.1, 1.4 Look at from Lawler 5.13 Han in lecture note exercises: 1., 1.3 Han in from Lawler 5.7b, 5.9*, 5.14 *Correction to 5.9 The conition, Pf x) = g x) for x S \ A, shoul rea Pf x) f x) = g x) for x S \ A. 1.7 Homework 7. Due Friay, May 0, 016 Look at from lecture note exercises: 1.8, 1.30 Han in lecture note exercises: 1.5, 1.6, 1.7, Homework 8. Due Friay, May 7, 016 Look at from lecture note exercises: 1.31, 1.3, 1.33, 1.41 Han in lecture note exercises: 1.34, 1.35, 1.36, Homework 9. Due Friay, June 3, 016 Look at from lecture note exercises: 1.37, 1.38, 1.4 Han in lecture note exercises: 1.39, 1.40 Exercise 1.1 Optional). Let W 0 := X Ω) W. Finish the proof of Proposition?? using the following outline; 1. Use the fact that σ Y ) σ X) to show for each s S there exists B s W 0 W such that {Y = s} = {X B s }.. Show B s B s = for all s, s S with s s. 3. Show X Ω) = W 0 := s S B s. Now fix a point s S an then efine, f : W S by setting f w) = s when w W \ W 0 an f w) = s when w B s W Verify that Y = f X). Exercise 1.. Suppose that X an Y are two integrable ranom variables such that E X Y = Y an E Y X = X. Fin EX an EY. Exercise 1.3. Let {X i } i=1 be i.i.. ranom variables with E X i < for all i an let S m := X X m for m = 1,,.... Show E S m S n = m n S n for all m n. Hint: observe by symmetry that there is a function h : R R such that E X i S n ) = h S n ) inepenent of i. Exercise 1.4. Let τ : Ω N be a function. Verify the following are equivalent; 1. τ is a stopping time.. {τ n} F X n for all n N {τ > n} F X n for all n N 0. Also show that if τ is a stopping time then {τ = } F X. Exercise 1.5. If τ an σ are two stopping times shows, σ τ = min {σ, τ}, σ τ = max {σ, τ}, an σ + τ are all stopping times. Exercise 1.6 Hitting time after a stopping time). Let σ be any stopping time. Show are both stopping times. τ 1 = inf {n σ : X n B} an τ = inf {n > σ : X n B}. Exercise 1.7 Invariant istributions an expecte return times). Suppose that {X n } n=0 is a Markov chain on a finite state space S etermine by one step transition probabilities, p x, y) = P X n+1 = y X n = x). For x S, let R x := inf {n > 0 : X n = x} be the first passage time 3 to x. We will assume here that E y R x < for all x, y S. Use the first step analysis to show, E x R y = z:z y p x, z) E z R y ) Now further assume that π : S 0, 1 is an invariant istributions for p, that is x S π x) = 1 an x S π x) p x, y) = π y) for all y S, i.e. πp = π. By multiplying Eq. 1.1) by π x) an summing on x S, show, π y) E y R y = 1 for all y S = π y) = 1 E y R y > 0. 1.) Apply Theorem?? using X 1, S n) = X i, S n) for 1 i n. 3 R x is the first return time to x when the chain starts at x. Page: job: 85hmsolns macro: svmonob.cls ate/time: 5-May-016/11:45

3 1.9 Homework 9. Due Friay, June 3, Exercise 1.8 Uniqueness of solutions to n orer recurrence relations). Let a, b, c be real numbers with a 0 c, α, β Z {± } with α < β, an g : Z α, β) R be a given function. Show that there is exactly one function u : α, β Z R with prescribe values on two two consecutive points in α, β Z which satisfies the secon orer recurrence relation: au x + 1) + bu x) + cu x 1) = f x) for all x Z α, β). 1.3) are for α < x < β. Show; if u an w both satisfy Eq. 1.3) an u = w on two consecutive points in α, β) Z, then u x) = w x) for all x α, β Z. Exercise 1.9 General homogeneous solutions). Let a, b, c be real numbers with a 0 c, α, β Z {± } with α < β, an suppose {u x) : x α, β Z} solves the secon orer homogeneous recurrence relation au x + 1) + bu x) + cu x 1) = 0 for all x Z α, β), 1.4) i.e. Eq. 1.3) with f x) 0. Show: 1. for any λ C, aλ x+1 + bλ x + cλ x 1 = λ x 1 p λ) 1.5) where p λ) = aλ + bλ + c is the characteristic polynomial associate to Eq. 1.3). Let λ ± = b± b 4ac a be the roots of p λ) an suppose for the moment that b 4ac 0. From Eq. 1.3) it follows that for any choice of A ± R, the function, w x) := A + λ x + + A λ x, 1.6) solves Eq. 1.3) for all x Z.. Show there is a unique choice of constants, A ± R, such that the function u x) is given by u x) := A + λ x + + A λ x for all α x β. 3. Now suppose that b = 4ac an λ 0 := b/ a) is the ouble root of p λ). Show for any choice of A 0 an A 1 in R that w x) := A 0 + A 1 x) λ x 0 1.7) solves Eq. 1.3) for all x Z. Hint: Differentiate Eq. 1.5) with respect to λ an then set λ = λ Again show that any function u solving Eq. 1.3) is of the form u x) = A 0 + A 1 x) λ x 0 for α x β for some unique choice of constants A 0, A 1 R. Exercise Let w x) := P x XHa,b = b ) := P X Ha,b = b X 0 = x ). 1. Use first step analysis to show for a < x < b that w x) = 1 w x + 1) + w x 1)) 1.8) provie we efine w a) = 0 an w b) = 1.. Use the results of Exercises 1.8 an 1.9 to show 3. Let P x XHa,b = b ) = w x) = 1 x a). 1.9) b a { min {n : Xn = b} if {X H b := n } hits b otherwise be the first time {X n } hits b. Explain why, { X Ha,b = b } {H b < } an use this along with Eq. 1.9) to conclue that P x H b < ) = 1 for all x < b. By symmetry this result hols true for all x Z.) Exercise The goal of this exercise is to give a secon proof of the fact that P x H b < ) = 1. Here is the outline: 1. Let w x) := P x H b < ). Again use first step analysis to show that w x) satisfies Eq. 1.8) for all x with w b) = 1.. Use Exercises 1.8 an 1.9 to show that there is a constant, c, such that w x) = c x b) + 1 for all x Z. 3. Explain why c must be zero to again show that P x H b < ) = 1 for all x Z. Exercise 1.1. Let H = H a,b an u x) := E x H := E H X 0 = x. 1. Use first step analysis to show for a < x < b that u x) = 1 u x + 1) + u x 1)) ) with the convention that u a) = 0 = u b).. Show that u x) = A 0 + A 1 x x 1.11) solves Eq. 1.10) for any choice of constants A 0 an A Choose A 0 an A 1 so that u x) satisfies the bounary conitions, u a) = 0 = u b). Use this to conclue that E x H a,b = ab + b + a) x x = a b x) + bx x. 1.1) Page: 3 job: 85hmsolns macro: svmonob.cls ate/time: 5-May-016/11:45

4 4 1 Math 85 Homework Problem List for S016 Exercise Let {X n } n=0 be the fair ranom walk on Z as in Exercise 1.10) starting at 0 an let N A N := E k=0 1 Xk =0 enote the expecte number of visits to 0. Using Sterling s formula an integral approximations for N n=1 1 n to argue that A N c N for some constant c > 0. Exercise Construct an example of a martingale, {M n } n=0 such that E M n as n. In particular, {M n } n=1 will be a martingale which is not of the form M n = E Fn X for some X L 1 P ). Hint: try taking M n = n k=0 Z k for a juicious choice of {Z k } k=0 which you shoul take to be inepenent, mean zero, an having E Z n growing rather rapily. Exercise Show that M n := n 1 0, n for n N 0 as efine in Example?? is a martingale. Exercise Suppose that {Z n } n=0 are inepenent ranom variables such that σ := EZn < an EZ n = 0 for all n 1. As in Example??, let S n := n k=0 Z k be the martingale relative to the filtration, Fn Z := σ Z 0,..., Z n ). Show M n := S n σ n for n N 0 is a martingale. Hint, make use of the basic properties of conitional expectations. Exercise 1.17 Quaratic Variation). Suppose {M n } n=0 is a square integrable martingale. Show; 1. E Mn+1 Mn F n = E M n+1 M n ) F n. Conclue from this that the Doob ecomposition of M n is of the form, where A n := 1 k n M n = N n + A n E k M) F k 1 = 1 k n E M k M k 1 ) F k 1. In particular we see from this that M n is a sub-martingale.. Show that N n above may be expresse as, N 0 = M 0 an for n 1, N n = M 0 + n M k 1 k M + k=1 n k=1 ) k M) E k M) F k 1. From Example?? an Proposition?? below one may see irectly that the expression above is a martingale. 3. If we further assume that M k M k 1 is inepenent of F k 1 for all k = 1,,..., explain why, A n = E M k M k 1 ). 1 k n Exercise Suppose that {X n } n=0 is a Markov chain on S with one step transition probabilities, P, an F n := σ X 0,..., X n ). Let f : S R be a boune for simplicity) function an set g x) = Pf) x) f x). Show M n := f X n ) g X j ) for n N 0 is a martingale. 0 j<n Exercise Suppose Harriet has 7 ollars. Her plan is to make one ollar bets on fair coin tosses until her wealth reaches either 0 or 50, an then to go home. What is the expecte amount of money that Harriet will have when she goes home? What is the probability that she will have 50 when she goes home? Exercise 1.0. Consier a contract that at time N will be worth either 100 or 0. Let S n be its price of the contract at time 0 n N. If S n is a martingale, an S 0 = 47, then what is the probability that the contract will be worth 100 at time N? Exercise 1.1. Pero plans to buy the contract in the previous problem at time 0 an sell it the first time T at which the price goes above 55 or below 15. What is the expecte value of S T? You may assume that the value, S n, of the contract is boune there is only a finite amount of money in the worl up to time N. Also note, by assumption, T N. Exercise 1.. For a < 0 < b with a, b Z, let τ = σ a σ b. Explain why {Sn} τ n=0 is a boune martingale use this to show P τ = ) = 0. Hint: make use of the fact that S n S n 1 = Z n = 1 for all n an hence the only way lim n Sn τ can exist is if it stops moving! Exercise 1.3. Show P σ b < σ a ) = a b + a 1.13) an use this to conclue P σ b < ) = 1, i.e. every b N is almost surely visite by S n. Hint: As in Exercise 1. notice that {Sn} τ n=0 is a boune martingale where τ := σ a σ b. Now compute E S τ = E Sτ τ in two ifferent ways. Exercise 1.4. Let τ := σ a σ b. In this problem you are aske to show E τ = a b with the ai of the following outline. Page: 4 job: 85hmsolns macro: svmonob.cls ate/time: 5-May-016/11:45

5 1.9 Homework 9. Due Friay, June 3, Use Exercise 1.17 above to conclue N n := S n n is a martingale.. Now show 0 = EN 0 = EN τ n = ES τ n E τ n. 1.14) 3. Now use DCT an MCT along with Exercise 1.3 to compute the limit as n in Eq. 1.14) to fin E σ a σ b = E τ = b a. 1.15) 4. By consiering the limit, a in Eq. 1.15), show E σ b =. Exercise 1.5 Some Discrete Distributions). Let p 0, 1 an λ > 0. In the two parts below, the istribution of N will be escribe. In each case fin the generating function see Proposition??) an use this to verify the state values for EN an Var N). 1. Geometricp) : P N = k) = p 1 p) k 1 for k N. P N = k) is the probability that the k th trial is the first time of success out a sequence of inepenent trials with probability of success being p.) You shoul fin EN = 1/p an Var N) = 1 p p.. Poissonλ) : P N = k) = λk k! e λ for all k N 0. You shoul fin EN = λ = Var N). Exercise 1.6. Let T = E λ) be as in Definition??. Show; 1. ET = λ 1 an. Var T ) = λ. Exercise 1.7 Discrete time M.C. jump-hol escription). Let S be a countable or finite set Ω, P, {X n } n=0 ) be a Markov chain with transition kernel, P := {p x, y)} x,y S an let ν x) := P X 0 = x) for all x S. For simplicity let us assume there are no absorbing states, 4 i.e. p x, x) < 1 for all x S) an then efine Q x,y = q x, y) where { px,y) q x, y) := 1 px,x) if x y 0 if x = y. Let j k enote the time of the k th jump of the chain {X n } n=0 so that j 1 := inf {n > 0 : X n X 0 } an j k+1 := inf {n > j k : X n X jk } with the convention that j 0 = 0. Further let σ k := j k+1 j k enote the time spent between the k th an k + 1) th jump of the chain {X n } n=0, see Figure??. Show; 4 A state x is absorbing if p x, x) = 1 since in this case there is no chance for the chain to leave x once it hits x. 1. For {x k } n k=0 S with x k x k 1 for k = 1,..., n an m 0,..., m n 1 N, show P n k=0 {X jk = x k } n 1 k=0 {σ k = m k } ) n 1 = ν x 0 ) p x k, x k ) mk 1 1 p x k, x k )) q x k, x k+1 ). 1.16) k=0. Summing the previous formula on m 0,..., m n 1 N, conclue n 1 P n k=0 {X jk = x k }) = ν x 0 ) q x k, x k+1 ), i.e. this shows {Y k := X jk } k=0 is a Markov chain with transition kernel, Q x,y = q x, y). 3. Conclue, relative to the conitional probability measure, P n k=0 {X j k = x k }), that {σ k } n 1 k=0 are inepenent geometric σ k = Geo 1 p xk, x k )) for 0 k n 1. Exercise 1.8. Show; if {a n } n=1 0, 1), then 1 a n ) = 0 n=1 k=0 a n =. n=1 The implication, =, hols even if a n = 1 is allowe. Exercise 1.9. Suppose that S = {1,,..., n} an A is a matrix such that A i,j 0 for i j an n j=1 A i,j = 0 for all i. Show P t = e ta := n=0 t n n! An 1.17) is a time homogeneous Markov kernel. Hints: 1. To show P t i, j) 0 for all t 0 an i, j S, write P t = e tλ e tλi+a) where λ > 0 is chosen so that λi + A has only non-negative entries.. To show j S P t i, j) = 1, compute t P t1 where 1 is the column vector of all 1 s. Exercise For 0 s < t <, show by inuction on n that t s)n a n s, t) := w 1... w n =. n! s w 1 w w n t Page: 5 job: 85hmsolns macro: svmonob.cls ate/time: 5-May-016/11:45

6 6 1 Math 85 Homework Problem List for S016 Exercise If n N an g : n R boune non-negative) function, then E g W 1,..., W n ) = g w 1, w,..., w n ) λ n e λwn w 1... w n. 1.18) n Exercise 1.3. Show N t = Poi λt) for all t > 0. Exercise Show for 0 s < t < that N t N s = Poi λ t s)) an {N s, N t N s } are inepenent. Hint: for m, n N 0 show using the few exercises above that λs λs)m λt s) λ t s))n P 0 N s = m, N t = m + n) = e e. 1.19) m! n! This result easily generalizes to show {N t } t 0 has inepenent increments an using this one easily verifies that {N t } t 0 is a Markov process with transition kernel given as in Eq.??). Exercise 1.34 Expecte return times). Let A be a infinitesimal generator of a continuous time Markov chain {X t } t 0 on S with 0 < a x = A x,x K < for all x S. Further let R y := inf {t S 0 : X t = y} be the first return time 5 to y on or after the first jump an m x,y := E x R y. Use the first jump analysis to show, m xy = 1 + Q x,z m zy = 1 + a x a x z y z / {x,y} A xz a x m zy, 1.0) where Q x,z := 1 x y A x,y /a x. You may fin it useful to observe; if H y := inf {t 0 : X t = y} is the first hitting time of y, then E x H y = E x R y if x y an E y H y = 0. Exercise Suppose that X an Y are inepenent normal ranom variables. Show; 1. Z = X, Y ) is a normal ranom vector, an. W = X + Y is a normal ranom variable. 3. If N is a stanar normal ranom variable an X is any normal ranom variable, show X = σn + µ where µ = EX an σ = Var X). 5 We require t S 0 so that if the chain starts at y it must first leave y before we count being at y again a return to y. Exercise 1.36 Brownian Motion). Let {B t } t 0 be a Brownian motion as in Definition??. 1. Explain why {B t } t 0 is a time homogeneous Markov process with transition operator, P t f) x) = p t x, y) f y) y 1.1) where R p t x, y) = 1 πt e 1 t y x. 1.) In more etail use Proposition?? an Eq. 1.4) below to argue, E f B t ) F s = P t s f) B s ). 1.3). Show by irect computation that P t P s = P t+s for all s, t > 0. Hint: probably the easiest way to o this is to make use of Exercise 1.35 along with the ientity, P t f) x) = E f x + ) tz, 1.4) where Z = N 0, 1). 3. Show by irect computation that p t x, y) of Eq. 1.) satisfies the heat equation, t p t x, y) = 1 x p t x, y) = 1 y p t x, y) for t > Suppose that f : R R is a twice continuously ifferentiable function with compact support. Show where t P tf = AP t f = P t Af for all t > 0, Af x) = 1 f x). Note: you may use uner the above assumptions) without proof the fact that it permissible to interchange the t an x erivatives with the integral in Eq. 1.1). Exercise 1.37 h transforms of B t ). Let {B t } t R+ be a imensional Brownian motion. Show the following processes are martingales; 1. M t = u B t ) where u : R R is a Harmonic function, u = 0, such that sup t T E u B t ) + u B t ) < for all T <.. M t = λ B t for all λ R. Page: 6 job: 85hmsolns macro: svmonob.cls ate/time: 5-May-016/11:45

7 1.9 Homework 9. Due Friay, June 3, M t = e λyt cos λx t ) an M t = e λyt sin λx t ) where B t = X t, Y t ) is a two imensional Brownian motion an λ R. 4. M t = B t t. 5. M t = a B t ) b B t ) a b) t for all a, b R. 6. M t := e λ Bt λ t/ for any λ R. Exercise 1.38 Compensators). Let {B t } t R+ be a imensional Brownian motion. Fin the compensator, A t, for each of the following square integrable martingales. 1. M t = u B t ) where u : R R is a harmonic function, u = 0, such that sup E u B t ) + u B t ) < for all T <. t T Verify the hypothesis of Corollary?? to show A t = t. M t = λ B t for all λ R. 3. M t = B t t. 4. M t := e λ Bt λ t/ for any λ R. 0 u B τ ) τ. Exercise 1.39 The Laplace transform of τ b ). Let b > 0 an τ b := inf {t > 0 : B t = b}. Using with λ > 0) the martingale, M t := e λbt 1 λt, show using the optional sampling Theorem?? that E 0 e µτ b = e b µ for all µ ) Note: for 0 t τ b that e λbt e λb which is useful when applying DCT. Exercise 1.40 Quaratic Variation). Let Π m := { 0 = t m 0 < t m 1 < < t m n m = T } be a sequence of partitions such that mesh Π m ) 0 as m. Further let Show Q m := n m i=1 i B) := n m i=1 lim E Q m T ) = 0. m B t m i B t m i 1 ). 1.6) Also show that mesh Π m ) < = E Q m T ) < m=1 m=1 = lim m Q m = T a.s. These results are often abbreviate by the writing, B t = t. Hints: it is useful to observe; 1) Q m T = n m i=1 i B) i t an ) using i B = i tn 0, 1) one easily shows there is a constant, c < such that E i B) i t = c i t). 1.7) Exercise Let q x) be a polynomial 6 in x, Z = N 0, 1), an u t, x) := E q x + ) tz 1.8) 1 = e 1 t y x) q y) y 1.9) πt Show u satisfies the heat equation, R t u t, x) = 1 u t, x) for all t > 0 an x R, x with u 0, x) = q x). Hints: Make use of Lemma?? along with the fact which is easily prove here) that u t, x) = E t t q x + tz). You will also have to use the corresponing fact for the x erivatives as well. Exercise 1.4. Let q x) be a polynomial in x, Z = N 0, 1), an = Show ) ) n E q Z) = e / q 0) := q) 0) = n! n! n n q) 0) n=0 n=0 x. 6 Actually, q x) can be any twice continuously ifferentiable function which along with its erivatives grow slower than e εx for any ε > 0. Page: 7 job: 85hmsolns macro: svmonob.cls ate/time: 5-May-016/11:45

8 where the above sum is actually a finite sum since n q 0 if n > eg q. Hint: let u t) := E q tz ). From your proof of Exercise 1.41 you shoul be able to see that u t) = 1 E q) tz ). This latter equation may be iterate in orer to fin u n) t) for all n 0. With this information in han you shoul be able to finish the proof with the ai of Taylor s theorem. Exercise Suppose X 1 an X are two inepenent Gaussian ranom ) variables with X i = N 0, σ i for i = 1,. Show X1 + X is Gaussian an X 1 + X = N 0, σ 1 + σ ). Hint: use Remark??.)