= p(t)(1 λδt + o(δt)) (from axioms) Now p(0) = 1, so c = 0 giving p(t) = e λt as required. 5 For a non-homogeneous process we have

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1 . (a (i I: P(exactly event occurs in [t, t + δt = λδt + o(δt, [o(δt/δt 0 as δt 0]. II: P( or more events occur in [t, t + δt = o(δt. III: Occurrence of events after time t is indeendent of occurrence of events before t. Let X(t be the number of realizations by time t and let (t = P(X(t = 0 seen (t + δt = P(0 realizations in [0, t and 0 realizations in [t, t + δt (t + δt (t δt d(t dt = (t( λδt + o(δt (from axioms = (tλ + o(δt δt = (tλ log((t = λt + c (b (iii Now (0 =, so c = 0 giving (t = e λt as required. 5 For a non-homogeneous rocess we have So, μ(t = (t = e μ(t where μ(t = t 0 t 0 λ(u du + sin(u du = [u cos(u] t 0 = t cos(t +, method seen giving (t = ex(cos(t t. Let X(t, t be the number of questions answered in in [t, t, then method seen E(X(t, t + δt = δt + o(δt. + t Let D(t = number questions answered by t in deterministic model. Then D(t + δt = D(t + δt + o(δt + t D(t + δt D(t = δt + t + o(δt δt dd = dt + t D(t = log( + t + c and since D(0 = 0, we have c = 0 D(t = log( + t 6 MS/MSSolutions Page of 9

2 . (a (i Let Π(s be the gf for X, and let μ = E(X and μ n = E(Z n, so art seen μ = Π ( μ n = Π n( where Π n(s is the gf of Z n = Y Y Zn with Y i being the number of offsring of individual i in generation n. From standard gf results, we have Π n (s = Π n [Π(s] Π n(s = Π n [Π(s] Π (s Π n( = Π n [Π(] Π ( = Π n (Π ( so μ = μ n μ = μ n μ =... = μ n. as E(X = we have E(Z n = n =. 5 Let σ = var(x and let σ n = var(z n. Π n(s = Π n [Π(s] Π (s Π n(s = Π n [Π(s] Π (s + Π n [Π(s] Π (s ( Now Π( =, Π ( = μ, Π ( = σ μ + μ. Also, since σ n = Π n( + μ n μ n, we have From (, Π n( = σ n μ n + μ n and Π n ( = σ n μ n + μ n. Π n( = Π n (Π ( + Π n (Π ( σ n μ n + μ n = (σ n μ n + μ n μ + μ n (σ μ + μ σ n = μ σ n + μ n σ Leading to σ n = μ n σ ( + μ + μ μ n So, as μ = we have var(z n = σ n = nσ. 7 MS/MSSolutions Page of 9

3 (b (i We have Giving, Π(s = P(X = is i = α + ( αs. i=0 Π (s = ( αs So μ = Π ( = ( α. Let P(ultimate extinction = θ, then. μ θ = ultimate extinction certain.. μ > θ < ultimate extinction not certain. μ > when ( α > α <. So, when α < ultimate extinction is not certain. θ = smallest ositive solution of θ = Π(θ, and θ = : We know that θ = is a solution: θ = α + ( αθ θ θ + = 0 θ θ + = 0 θ θ + = (θ (θ + θ roots of θ + θ are ± 5, therefore method seen Probability of extinction = 5 MS/MSSolutions Page of 9

4 . (a (i Show that P(ever reach 0 starts at i = P(ever reach i starts at i P(ever reach i starts at i. P(ever reach 0 starts at = A A... A }{{} (satial homogeneity i times = (A i condition on first ste: A = P(visit origin start from = qp(visit origin start from 0 + P(visit origin start from (iii A A + q = 0 = q + A Solving A A + q = 0 gives A = or A = Now look for solutions in [0, ]: also, take ositive solutions: + q ± < 0, + q + + q + q 9 A = (Noting that when = /, A =. { + + q if > / if / unseen 5 MS/MSSolutions Page 5 of 9

5 (b (i Let X t be the value of the share on day t. Let Y t be the change in value of the share on day t. Then, X t = X 0 + Y Y t. X 0 = 00 Y i = 0. with robability with robability 0. 0 with robability 0. Giving E(Y i = = = 0.. var(y i = E(Yi E (Y i E(Yi = (0. 0. = = = 0.09 var(y i = = = E(X t = t = μ t For large t var(x t = 0.07t = σ t 5 X t X 0 = t Y i N(0.0t, 0.07t i= X t X 0 0.t 0.07t N(0, ( 0 ( P(X 65 > 0 = Φ ( 0 μ 65 = Φ σ 65 MS/MSSolutions Page 6 of 9

6 . (a (i A Markov chain is irreducible if it has only one communicating class, i.e. there is a ath of non-zero robability from state i to state j and back again for all i, j in the samle sace. A Markov chain is aeriodic if all states have eriod, i.e. if where (n ij gcd{n : (n ij > 0} =. is the robability of going from i to j in n stes. (b (i State sace = {0,,, } (Number of balls in the first urn P = transition diagram: seen 0 (iii Irreducible, finite state sace there is a unique stationary distribution. (iv Need aeriodicity for this distribution to also be limiting. Here the Markov chain is eriodic with eriod, so the stationary distribution is not limiting. (v Find stationary distribution, π from, π = πp, i=0 π i = and π 0 = π π 0 + π = π + π = π = π π = π 0 π = π 0 π = π 0 π 0 + π + π + π = π 0 + π 0 + π 0 + π 0 = π 0 = 8 Giving, π = ( 8, 8, 8, 8 Mean recurrence time to state 0, μ 0 is given by: μ 0 = π 0 = 8 MS/MSSolutions Page 7 of 9 5

7 5. (a (i Define seen = d dt P (t t=0 the transition rate matrix with elements q ij, and let P (t have element ij (t, i, j in the samle sace. The forward differential equations are given by: d P (t = P (t dt d dt ij(t = k ik (tq jk i, j. π i =, π = 0 (or π = P (tπ, t. i (b (i Let state 0 good mood, and state bad mood. 0 (δt = P(0 in [t, t + δt = α δt + o(δt 0 (δt = P( 0 in [t, t + δt = β δt + o(δt We have, by definition, { + δt qii + o(δt i = j ij (δt = δt q ij + o(δt i j small δt So, as the rows of sum to zero, we have ( α α = β β assume true for n = k, let n = k +. unseen = k+ = k = (( k from assumtion = ( k ( α + αβ α αβ αβ β αβ + β ( α α = ( α β β β k+ = ( k ( = ( k = ( α β true for n = k+ if true for n = k, as true for n = ( = ( α β 0, result follows by induction. MS/MSSolutions Page 8 of 9

8 ex(t = I + = I + n= t n n! (n (t( n n=0 n! = I + {ex(t( } = I + α + β + ex(t( (iii Backward differential equations: d dt P (t = d { I + dt α + β + = ex(t( { P (t = I + α + β + } ex(t( ex(t( = + α + β + ex(t( ( ( = + + ex(t( α + β = ex(t( } i.e. P (t satisfies the backward differential equations: d dt P (t = P (t. (iv stationary distribution satisfies ( α α (π 0 π β β = (0 0 π 0 + π =. απ 0 + βπ = 0 and π 0 + π = απ 0 + β( π 0 = 0 π 0 = β α + β π = β α + β = α α + β giving π = ( β α + β, α α + β MS/MSSolutions Page 9 of 9