➀ ➁ ➂ ➃ ➄ ➅ ➆ ➇ ➈ ➉ Lecture on Stochastic Processes (by Lijun Bo) 2

Stochastic Processes stoprocess@yahoo.com.cn 111111

➀ ➁ ➂ ➃ ➄ ➅ ➆ ➇ ➈ ➉ Lecture on Stochastic Processes (by Lijun Bo) 2

: Stochastic Processes? (Ω, F, P), I t I, X t (Ω, F, P), X = {X t, t I}, X t (ω) : I Ω R, I, : I = [0, ), I = {0} N, I = [0, ) R 3

: State Space (t, ω), X t (ω) : Sample Path ω Ω, t X t (ω) X 4

1. : Simple Branching Process 5

Z n n ( ) Z n,j n j (i.i.d. ) Z n+1 = Z n,1 + Z n,2 + + Z n,zn. Z = {Z n ; n {0} N} 6

2. Poisson : Poisson Process N = {N t ; t 0} Poisson, (1) N 0 = 0, (2) t > 0, r.v. N t λt Poisson, P(N t = n) = (λt)n e λt, n {0} N. n! (3) t > s > 0, r.v.s N t N s N t s, (N u ; u s). Poisson M/M/1 M/G/1 7

3. Poisson : Compound Poisson Risk Process X = {X t ; t 0} Poisson, X t = x + ct N t i=1 U i, ( 0 U i = 0) i=1 x c N = {N t ; t 0} -Poisson ([0, t] ) (U i ; i 1) - (i.i.d. ) N. 8

Poisson 9

4. : Brownian Motion (Wiener Process) W = {W t ; t 0} ( ), (1) W 0 = 0, (2) t > 0, r.v. W t N(0, t), (3) t > s > 0, r.v.s W t W s W t s, (W u ; u s) Bachelier (1900), Einstein (1905) and Wiener (1923,1924). (µ σ2 X = {e 2 )t+σw t ; t 0} X = { W t ; t 0}. 10

11

5. : Stochastic Differential Equation (SDE) (Itô) dx t = b(x t )dt + σ(x t )dw t, X 0 = x. ( ) X = {X t ; t 0} σ2 (µ X = {e 2 )t+σw t ; t 0} SDE: dx t X t = µdt + σdw t, X 0 = x > 0. Black and Scholes SDE Black-Scholes 12

6. : Martingale X = {X t ; t 0}, E[X t X u, u s] = X s, t > s. W Poisson Ñ = {N t λt; t 0} Itô (Non-arbitrary Pricing Theory) 13

7. Lévy : Lévy Process Lévy W, Poisson N Poisson Lévy Lévy 14

α- (α-stable process) 15

: Finite-Dimensional Distribution Function? X = {X t ; t I} (Ω, F, P) t 1, t 2,, t n I, F n (t 1,..., t n ; x 1,..., x n ) = P(X t1 x 1,..., X tn x n ), x i S (i = 1, 2,..., n), X X, F = {F n (t 1,..., t n ; x 1,..., x n ); x i S, t i I, i = 1,..., n, n N}. 16

1. ( : symmetry). (i 1,..., i n ) (1,..., n) (permutation), F n (t i1,..., t in ; x i1,..., x in ) = F n (t 1,..., t n ; x 1,..., x n ). 2. ( : Consistent). m < n, F m (t 1,..., t m ; x 1,..., x m ) = F n (t 1,..., t m, t m+1,..., t n ; x 1,..., x m, +,..., + ). 17

F? X,, (F n, n N), X F n? 18

Answer: YES! Daniell (1918), Kolmogorov (1933). ( Brownian Motion). 19

: Characteristic Function X = {X t ; t I} (Ω, F, P), t 1,..., t n I, [ ( )] n ϕ(t 1,..., t n ; u 1,..., u n ) = E exp i u k X tk, i = 1, u i R, X Fourier k=1 20

1. W 2 F 2 (t 1, t 2 ; x 1, x 2 ) n ϕ(t 1,..., t n ; u 1,..., u n ) 21

Answer: Let 0 t 1 < t 2 <. Then F 2 (t 1, t 2 ; x 1, x 2 ) = P(W t1 x 1, W t2 x 2 ) = P(W t1 x 1, W t2 W t1 + W t1 x 2 ) = = = = P(W t1 x1 x1 P(W t1 x 1, W t2 W t1 + W t1 x 2, W t1 dy) P(W t1 x 1, W t2 W t1 + W t1 x 2 W t1 dy) dy) P(W t2 W t1 x 2 y W t1 dy)ϕ t1 (y)dy ϕ t2 t 1 (x 2 y)ϕ t1 (y)dy. 22

Let 0 t 1 < < t n < and Y 1 := W t1, Y 2 := W t2 W t1,..., Y n := W tn W tn 1. Then they are independent, and W t2 = Y 1 + Y 2,..., W tn = n Y i. i=1 Consequently, ϕ(t 1,..., t n ; u 1,..., u n ) = E [exp (i((u 1 + u n )Y 1 + (u 2 + u n )Y 2 + + u n Y n ))] = E [exp (i(u 1 + u n )Y 1 )] E [exp (iu n Y n )] = e 1 2 (u 1+ u n ) 2 t 1 e 1 2 u2 n (t n t n 1 ). 23

X = {X t ; t I} (Ω, F, P) 1. : Mean Function m X (t) = E[X t ], t I. 2. : Mean Square Deviation Function Φ X (t) = E[ X t 2 ], t I. 3. : Variance Function D X (t) = Var[X t ] = E[ X t m X (t) 2 ] = Φ X (t) m 2 X(t), t I. 24

4. : Correlation Function R X (s, t) = E[X s X t ], s, t I. 5. : Covariance Function C X (s, t) = Cov(X s, X t ) = E[(X s m X (s))(x t m X (t))] = R X (s, t) m X (s)m X (t), s, t I. 25

1. W 2. Poisson N 3. Poisson X 26

1. Answer: Note that W t N(0, t). Then for all t > 0, m W (t) = E[W t ] = 0 Φ W (t) = E[ W t 2 ] = t D W (t) = Φ W (t) m 2 W (t) = t R W (s, t) = E[W s W t ] = E[W s (W t W s )] + E[ W s 2 ] = E[W s ]E[W t W s ] + s = s, if s < t = E[W t ]E[W s W t ] + t = t, if t < s = s t, for all s, t 0 C W (s, t) = R W (s, t) m W (s)m W (t) = s t, for all s, t 0. 27

2. Poisson Answer: Note that N t Poisson(λt). Then for all t > 0, m N (t) = E[N t ] = λt Φ N (t) = E[ N t 2 ] = λt(1 + λt) D N (t) = Φ N (t) m 2 N(t) = λt R N (s, t) = E[N s N t ] = E[N s (N t N s )] + E[ N s 2 ] = E[N s ]E[N t s ] + λs(1 + λs) = λs(1 + t), if s < t = E[N t ]E[N s t ] + λt(1 + λt) = λt(1 + s), if t < s = λs t(1 + s t), for all s, t 0 C N (s, t) = R N (s, t) m N (s)m N (t) = λs t(1 + s t) λ 2 st, for all s, t 0. 28

3. Poisson Answer: Note that Then [ Nt ] E U i i=1 X t = u + ct = E = = [ E [ Nt i=1 N t i=1 U i, U i N t ]] = {U i, i 1} i.i.d.. E n=0 [ Nt i=1 [ n ] E U i N t = n P(N t = n) n=0 i=1 [ n ] E U i P(N t = n) = E [U 1 ] n=0 i=1 = E [U 1 ] m N (t) = λte [U 1 ], U i N t = n ] P(N t = n) np(n t = n) n=0 29

and E N t i=1 2 U i = = E n=0 n i,j=1 [ n E n=0 + n=0 i=1 U i U j P(N t = n) U 2 i ] P(N t = n) n E[U i ]E[U j ]P(N t = n) i j = λte[u 2 1 ] + E 2 [U 1 ] P 2 np(n t = n) n=0 = λte[u 2 1 ] + (Φ N (t) m N (t))e 2 [U 1 ]. 30

Let M t = N t i=1 U i. Then R M (s, t) = E[M s M t ] = E[M s (M t M s )] + E[ M s 2 ] = E[M s ]E[M t s ] + E[ M s 2 ], if s < t = E[M t ]E[M s t ] + E[ M t 2 ], if t < s. 31

: 2-DIM STOCHASTIC PROCESS X = {X t ; t I} Y = {Y t ; t I} (Ω, F, P), X = (X, Y ) (Ω, F, P) X : Joint Finite-Dimensional Distribution Function t 1,..., t m, t 1,..., t n I x 1,..., x m S X, y 1,..., y n S Y, F m,n (t 1,..., t m, x 1,..., x m ; t 1,..., t n, y 1,..., y n ) = P(X t1 x 1,..., X tm x m, Y t 1 y 1,..., Y t n y n ) X : 2-DIM STOCHASTIC PROCESS 32

(X, Y ) : Independence of Stoch. Processes F X m F Y n X Y, (X, Y ) F m,n : F m,n (t 1,..., t m, x 1,..., x m ; t 1,..., t n, y 1,..., y n ) = F X m (t 1,..., t m, x 1,..., x m )F Y n (t 1,..., t n, y 1,..., y n ), X Y. : 2-Dim Brownian Motion W 1 = {W 1 t ; t 0} W 2 = {W 2 t ; t 0}, (W 1, W 2 ). 33

(X, Y ) : (Cross) Correlation Function R XY (s, t) = E(X s Y t ), s, t I. : (Cross) Covariance Function C XY (s, t) = E[(X s m X (s))(y t m Y (t))] = R XY (s, t) m X (s)m Y (t), s, t I. Problem: R W 1 W 2(s, t) =? C W 1 W 2(s, t) =? (X, Y ) 34

N 1 = {N 1 (t); t 0} N 2 = {N 2 (t); t 0} λ 1 > 0 λ 2 > 0 Poisson N = N 1 + N 2 = {N 1 (t) + N 2 (t); t 0} Poisson λ 1 + λ 2 35

Answer: (1) N 0 = N 1 (0) + N 2 (0) = 0. 36

(2) For each t > 0 and n {0} N, P(N t = n) = P(N 1 (t) + N 2 (t) = n) = = = = n m=0 n m=0 n m=0 n m=0 P(N 2 (t) = n m, N 1 (t) = m) P(N 2 (t) = n m)p(n 1 (t) = m), (λ 1 t) n m e λ 1t (n m)! = 1 n! e (λ 1+λ 2 )t n m=0 = 1 n! e (λ 1+λ 2 )t (λ 1 + λ 2 ) n t n (λ 2 t) m e λ 2t m! P(N 1 (t) + N 2 (t) = n, N 1 (t) = m) n! (n m)!m! (λ 1t) n m (λ 2 t) m 37

(3) t > s > 0, N t N s = (N 1 (t) N 1 (s)) + (N 2 (t) N 2 (s)) = N 1 (t s) + N 2 (t s) = N t s. N 1 (t) N 1 (s) (N 1 (u); u s) N 2 (t) N 2 (s) (N 2 (u); u s), N 1 N 2, N t N s (N 1 (u) + N 2 (u); u s) 38

: Complex-Valued Stoch. Process? X = {X t ; t I} Y = {Y t ; t I} (Ω, F, P) Z t = X t + iy t, i = 1, Z = {Z t ; t I} (Ω, F, P) 39

m Z (t) = E[Z t ] = E[X t ] + ie[y t ] = m X (t) + im Y (t). Φ Z (t) = E[ Z t 2 ] = E[ X t 2 ] + E[ Y t 2 ] = Φ X (t) + Φ Y (t). 40

D Z (t) = E[ Z t m Z (t) 2 ] = E[ (X t m X (t)) + i(y t m Y (t)) 2 ] = D X (t) + D Y (t). R Z (s, t) = E[Z s Z t ] = E[(X s iy s )(X t + iy t )] = R X (s, t) + R Y (s, t) + i(r XY (s, t) R Y X (s, t)). 41

C Z (s, t) = E [ ] (Z s m Z (s))(z t m Z (t)) = E{[(X s m X (s)) i(y s m Y (s))] [(X t m X (t)) + i(y t m Y (t))] } = C X (s, t) + C Y (s, t) + i(c XY (s, t) C Y X (s, t)). Prove: Φ Z (t) = R Z (t, t), C Z (s, t) = R Z (s, t) m Z (s)m Z (t). 42

Z 1 = {Z 1 (t); t I} Z 2 = {Z 2 (t); t I}, (Z 1, Z 2 ) R Z1 Z 2 (s, t) = E [ ] Z 1 (s)z 2 (t). C Z1 Z 2 (s, t) = E [ ] (Z 1 (s) m Z1 (s))(z 2 (t) m Z2 (t)). 43

( P50: 2.6.1) : Z t = n X k e i(ω 0t+Φ k ), i = 1, t R, k=1 ω 0 > 0, n N, X 1,..., X n, Φ 1,..., Φ n, E[X k ] = 0, D(X k ) = σ k > 0, Φ k U[0, 2π] (k = 1, 2,..., n). Z = {Z t ; t R} m Z (t) R Z (s, t) (s, t R) ( P50: 2.6.1) 44

Answer: [ n ] m Z (t) = E X k e i(ω 0t+Φ k ) = k=1 n E[X k ]E k=1 [ e i(ω 0t+Φ k ) ] = 0, E[X k ] = 0. 45

Answer: [ n ] n R Z (s, t) = E X k e i(ω 0s+Φ k ) X l e i(ω 0t+Φ l ) k=1 [ n = E X k e i(ω 0s+Φ k ) k=1 = e iω0(t s) E n k,l=1 [ n = e iω0(t s) E k=1 [ n = e iω0(t s) E l=1 ] n X l e i(ω 0t+Φ l ) l=1 X k X l e i(φ k Φ l ) X 2 k ] ] + e iω0(t s) E Xk 2 k=1 k=1 = e iω 0(t s) n X k X l e i(φ k Φ l ) k l n σk. 2 46

: Second-Moment Process X = {X t ; t I} (Ω, F, P) ( ), Φ X (t), X Φ X (t) = E[ X t 2 ] <, t I, Prove: (Hint: Cauchy-Schwarz ) 1. X m X (t) (E[ X t ] < )? 2. X R X (s, t) (E[ X s X t ] < )? 47

(1) : R X (s, t) = R X (t, s), s, t I. (2) : n 1, t 1,..., t n I λ 1,..., λ n, n k=1 n R X (t k, t l ) λ k λ l 0. l=1 48

( ): Gaussian Process X = {X t ; t I} (Ω, F, P), n 1 t 1,..., t n I, (X t1,..., X tn ) n, X ( ) : n (X t1,..., X tn ) n (X t1,..., X tn ) n : f(x) = 1 (2π) n/2 C 1/2 e 1 2 (x b) T C 1 (x b), x = (x 1,..., x n ) T, b = (E[X t1 ],..., E[X tn ]) T C n n (X t1,..., X tn ) 49

W ( Wiener ) Proof: : X = (W t1, W t2,..., W tn ), Y = (W t1, W t2 W t1,..., W tn W tn 1 ). X = YA, 50

A = 1 1 1 0 1 1 0 0 1 0 0 1 n n Y n N(0, C), 0 = (0, 0,..., 0) T, 51

C = t 1 0 0 0 0 t 2 t 1 0 0 0 0 0 t n t n 1 n n X N(0, A T CA). n N(0, A T CA) 52

X = {X t ; t 0} (Ω, F, P), t 1 < t 2 t 3 < t 4, [ ] E (X t2 X t1 )(X t4 X t3 ) = 0, X 53

X = {X t ; t [a, b]} X a = 0, : (1) R X (s, t) = Φ X (s t), (2) C X (s, t) = D X (s t) + m X (s t) 2 m X (s)m X (t), (3) t Φ X (t). 54

Proof: (1) a s t b, R X (s, t) = E[X s X t ] = E[X s (X t X s + X s )] = E[X s (X t X s )] + E[ X s 2 ] = E[(X s X a )(X t X s )] + Φ X (s), (X a = 0) = Φ X (s t). (2) C X (s, t) = R X (s, t) m X (s)m X (t). 55

Proof: (3) a s t b, 0 E[ X s X t 2 ] = E [ ] (X s X t )(X s X t ) = Φ X (s) + Φ X (t) R X (s, t) R X (t, s) = Φ X (s) + Φ X (t) Φ X (s) Φ X (s) = Φ X (t) Φ X (s). 56

X = {X t ; t 0} (Ω, F, P), n 3 t 1 < t 2 < < t n, X t2 X t1, X t3 X t2,..., X tn X tn 1, X 57

, 58

Proof: : Y 1 = X t1, Y 2 = X t2 X t1,..., Y n = X tn t n 1. (Y 1,..., Y n ) : ϕ(t 1,..., t n ; u 1,..., u n ) = E [e i(u 1X t1 + +u n X )] t n ] = E [e i(u 1+ +u n )Y 1 +i(u 2 + +u n )Y 2 + +iu n Y n = ϕ Y1 (u 1 + + u n )ϕ Y2 (u 2 + + u n ) ϕ Yn (u n ) = ϕ Xt1 (u 1 + + u n )ϕ Xt2 X t1 (u 2 + + u n ) ϕ Xt n X t n 1 (u n ). 59

Poisson : N = {N t ; t 0}, t > 0, N t t N (1) N 0 = 0, (2) N t {0} N, t > 0 ( S = {0} N), (3) 0 s < t, N t N s ( ), (4) 0 s < t, N t N s (s, t] 60

61

Poisson N = {N t ; t 0} λ > 0 Poisson, : (1) t, Poisson (t, t + t] λ t + o( t), P(N t+ t N t = 1) = λ t + o( t). (2) t, Poisson (t, t + t] o( t), P(N t+ t N t 2) = o( t). Prove: P(N t+ t N t = 0) = 1 λ t + o( t). 62

Proof: Poisson, P(N t+ t N t = 1) = P(N t = 1) = λ te λ t = λ t (1 + (λ t) + (λ t)2 2! + (λ t)3 3! ) + = λ t + (λ t) 2 + (λ t)3 2! = λ t + o( t). + 63

P(N t+ t N t 2) = P(N t 2) = P(N t = 2) + P(N t = 3) + = (λ t)2 2! = o( t). e λ t + (λ t)3 3! e λ t + 64

N = {N t ; t 0} : (1),(2), λ > 0 Poisson Proof: : t > 0, N t λt Poisson 65

p 0 (t) = P(N t = 0). (1) (2), p 0 (t + t) = P(N t+ t = 0)., t 0, p 0 (t) = e λt. p 0 (t + t) = P(N t+ t N t + N t = 0) = P(N t+ t N t + N t = 0, N t = 0) = P(N t+ t N t = 0, N t = 0) = P(N t+ t N t = 0)p 0 (t) = (1 λ t + o( t))p 0 (t). p 0 (t + t) p 0 (t) t = λp 0 (t) + o( t). t p 0(t) = λp 0 (t), p 0 (0) = 1. 66

p k (t + t) = P(N t+ t = k), k 1. p k (t + t) = P(N t+ t = k) = P(N t = k, N t+ t N t = 0) +P(N t = k 1, N t+ t N t = 1) k + P(N t = k i, N t+ t N t = i) i=2 = p k (t)p 0 ( t) + p k 1 (t)p 1 ( t) + o( ). t 0, p k (t + t) p k (t) t = λp k (t) + λp k 1 (t) + o( t). t p k(t) + λp k (t) = λp k 1 (t), p 1 (0) = 1, p 0 (t) = e λt, k = 2, 3,.... 67

p k (t) = (λt)k k! e λt, k = 1, 2,.... 68

Poisson N = {N t ; t 0} λ Poisson τ 0 = 0, τ 1 = min{t > 0; N t = 1}, τ 2 = min{t > τ 1 ; N t = 2},, τ n = min{t > τ n 1 ; N t = n}, {τ 0, τ 1,..., τ n,... } Poisson 69

T 1 = τ 1 τ 0, T 2 = τ 2 τ 1,..., T n = τ n τ n 1,... {T 1, T 2,..., T n,... } Poisson : {T 1, T 2,..., T n,... } λ 70

Proof: Poisson T 1,..., T n... (1) T 1 : t 0, F 1 (t) = P(T 1 t) = 1 P(T 1 > t) = 1 P(τ 1 > t) = 1 P(N t = 0) = 1 e λt. (2) T n (n 2): t 0 s 1,..., s n 1 0, F n (t) = P(T n t) = 1 P(T n > t) = 1 P(T n > t T 1 = s 1,..., T n 1 = s n 1 ) = 1 P(N t+s1 + +s n 1 N s1 + +s n 1 = 0) = 1 P(N t = 0) = 1 e λt. 71

, Poisson τ n Γ(n, λ) Γ(α, β) β α Γ(α) f(t) = tα 1 e βt, t 0, 0, t < 0. Proof: n λ Γ(n, λ) 72

Prove: For 0 s < t, P(τ 1 s N t = 1) = s t. 73

Tau_1 s t 74

: PAGE 59: 2.7.10 N = {N t ; 0} λ > 0 Poisson, (τ 1,..., τ n ) N t = n (0, t) n U 1,..., U n U (1) < < U (n), : (τ 1,..., τ n ) N t = n n! t, 0 u f (n) (u 1,..., u n ) = n 1 < < u n < t, 0, otherwise. : PAGE 59: 2.7.10 75

Proof: h 1,..., h n > 0 0 u 1 < τ 1 < u 1 +h 1 < u 2 < τ 2 < u 2 +h 2 < < u n < τ n < u n +h n t, 76

P( n k=1(u k < τ k u k + h k ) N t = n) = P( n k=1 (N u k +h k N uk = 1), N t = n) P(N t = n) = P(N h 1 = 1,, N hn = 1, N t = n) P(N t = n) = P(N h 1 = 1,, N hn = 1, N t n k=1 N h k = 0) P(N t = n) = P(N h 1 = 1,, N hn = 1, N t n k=1 h = 0) k P(N t = n) n k=1 = P(N h k = 1)P(N t n k=1 h = 0) k P(N t = n) = n! t n h 1 h n. 77