1. Random WALK, BROWNIAN MOTION and SDEs
Continuation…

2. Stochastic Differential Equations
ODE:
deterministic
solution is a function
SDE:
probabilistic
solution is a stochastic process (has different realizations or sample paths)

3. Stochastic Differential Equations
Suppose we have an SDE 𝑑𝑥=𝑓 𝑡,𝑥 𝑑𝑡+𝑔 𝑡,𝑥 𝑑 𝐵 𝑡 or simply, 𝑑𝑥=𝑓𝑑𝑡+𝑔𝑑 𝐵 𝑡 This DE is written in differential form (not in derivative form like ODEs) because Brownian motion is continuous but NOT differentiable.

4. Stochastic Differential Equations
Suppose we have an SDE 𝑑𝑥=𝑓 𝑡,𝑥 𝑑𝑡+𝑔 𝑡,𝑥 𝑑 𝐵 𝑡 Remark: 𝒅 𝑩 𝒕 is called white noise.

5. Stochastic Differential Equations
Suppose we have an SDE
𝑑𝑥=𝑓 𝑡,𝑥 𝑑𝑡+𝑔 𝑡,𝑥 𝑑 𝐵 𝑡
Remark: This also means 𝒅𝒙 ~ 𝑵 𝒇𝒅𝒕, 𝒈 𝟐 𝒅𝒕
𝑓𝑑𝑡 is the mean change
𝑔𝑑 𝐵 𝑡 is the error
𝑔 2 𝑑𝑡 is the variance of the error

6. Stochastic Differential Equations
Suppose we have an SDE 𝑑𝑥=𝑓 𝑡,𝑥 𝑑𝑡+𝑔 𝑡,𝑥 𝑑 𝐵 𝑡 This SDE is equivalent to the integral equation 𝑥 𝑡 =𝑥 𝑡 𝑓 𝑠,𝑥 𝑑𝑠 + 0 𝑡 𝑔 𝑠,𝑥 𝑑 𝐵 𝑠

7. Stochastic Differential Equations
Suppose we have an SDE 𝑑𝑥=𝑓 𝑡,𝑥 𝑑𝑡+𝑔 𝑡,𝑥 𝑑 𝐵 𝑡 This SDE is equivalent to the integral equation 𝑥 𝑡 =𝑥 𝑡 𝑓 𝑠,𝑥 𝑑𝑠 + 𝟎 𝒕 𝒈 𝒔,𝒙 𝒅 𝑩 𝒔 The integral in red font is called an Ito integral.

8. Stochastic Differential Equations
Let 𝑎= 𝑡 0 < 𝑡 1 <…< 𝑡 𝑛−1 < 𝑡 𝑛 =𝑏 be a grid of points on the interval [𝑎,𝑏]. The Riemann integral is defined as a limit 𝑎 𝑏 𝑓 𝑡 𝑑𝑡 = lim Δ 𝑡 𝑖 →0 𝑖=1 𝑛 𝑓( 𝑡 𝑖 ) Δ 𝑡 𝑖 where Δ 𝑡 𝑖 = 𝑡 𝑖 − 𝑡 𝑖−1 and 𝑡 𝑖−1 ≤ 𝑡 𝑖 ≤ 𝑡 𝑖

9. Stochastic Differential Equations
Let 𝑎= 𝑡 0 < 𝑡 1 <…< 𝑡 𝑛−1 < 𝑡 𝑛 =𝑏 be a grid of points on the interval [𝑎,𝑏]. The Ito integral is defined as a limit 𝑎 𝑏 𝑔 𝑡 𝑑 𝐵 𝑡 = lim Δ 𝑡 𝑖 →0 𝑖=1 𝑛 𝑔( 𝑡 𝑖−1 ) Δ 𝐵 𝑡 𝑖 where Δ 𝐵 𝑡 𝑖 = 𝐵 𝑡 𝑖 − 𝐵 𝑡 𝑖−1 .

10. Kiyosi Ito

11. Stochastic Differential Equations
Example: 𝑑𝑥=𝑟𝑑𝑡+𝜎𝑑 𝐵 𝑡 where 𝑟,𝜎 are constants 𝑥 0 =𝑥 0 If 𝜎=0, we have the ODE 𝑑𝑥 𝑑𝑡 =𝑟 with solution 𝑥 𝑡 = 𝑥 0 +𝑟𝑡.

12. Stochastic Differential Equations
Example: 𝑑𝑥=𝑟𝑑𝑡+𝜎𝑑 𝐵 𝑡 where 𝑟,𝜎 are constants 𝑥 0 =𝑥 0 If 𝜎 can be any real number, we integrate both sides 0 𝑡 𝑑𝑥 = 0 𝑡 𝑟𝑑𝑠 + 0 𝑡 𝜎𝑑 𝐵 𝑠 .

13. Stochastic Differential Equations
Example: 0 𝑡 𝑑𝑥 = 0 𝑡 𝑟𝑑𝑠 + 0 𝑡 𝜎𝑑 𝐵 𝑠 𝑥 𝑡 −𝑥 0 =𝑟𝑡+𝜎 𝐵 𝑡 The solution is the stochastic process: 𝑥 𝑡 = 𝑥 0 +𝑟𝑡+𝜎 𝐵 𝑡

14. Stochastic Differential Equations
Example: The term 𝑟𝑡 is the drift The term 𝜎 𝐵 𝑡 is the diffusion of Brownian motion

15. Stochastic Differential Equations
Solving SDEs analytically: we can use the Ito formula (Ito’s lemma). This is an analogue to the chain rule in conventional calculus. If 𝑥=ℎ(𝑡,𝑦) then 𝑑𝑥= 𝜕ℎ 𝜕𝑡 𝑑𝑡+ 𝜕ℎ 𝜕𝑦 𝑑𝑦+ 1 2 𝜕 2 ℎ 𝜕 𝑦 2 𝑑𝑦𝑑𝑦

16. Stochastic Differential Equations
We also use the following identities: 𝑑𝑡 𝑑𝑡=0 𝑑𝑡 𝑑 𝐵 𝑡 =𝑑 𝐵 𝑡 𝑑𝑡=0 𝑑 𝐵 𝑡 𝑑 𝐵 𝑡 =𝑑𝑡
× 𝑑𝑡 𝑑 𝐵 𝑡 𝑑𝑡 0 0 𝑑 𝐵 𝑡 0 𝑑𝑡

17. Stochastic Differential Equations
Remark: Recall Taylor Series expansion of ℎ(𝑥): ℎ 𝑥 =ℎ 𝑥 + ℎ ′ 𝑥 𝑥− 𝑥 ℎ′′ 𝑥 𝑥− 𝑥 2 +…

18. Stochastic Differential Equations
Remark: ℎ 𝑥 −ℎ 𝑥 = ℎ ′ 𝑥 𝑥− 𝑥 ℎ′′ 𝑥 𝑥− 𝑥 2 +… Replace: ∆𝑥=𝑥− 𝑥 ∆ℎ(𝑥)=ℎ 𝑥 −ℎ 𝑥

19. Stochastic Differential Equations
Remark: ∆ℎ(𝑥)= ℎ ′ 𝑥 ∆𝑥+ 1 2 ℎ′′ 𝑥 ∆𝑥 2 +… Replace the difference by differential, 𝑑ℎ(𝑥)= ℎ ′ 𝑥 𝑑𝑥+ 1 2 ℎ′′ 𝑥 𝑑𝑥 2 +…

20. Stochastic Differential Equations
Remark: In conventional calculus, we compute for differential by 𝑑ℎ(𝑥)= ℎ ′ 𝑥 𝑑𝑥 where higher terms are negligible for small changes in 𝑥.

21. Stochastic Differential Equations
Remark: In stochastic calculus, we need to keep up to second-order terms: 𝑑ℎ(𝑥)= ℎ ′ 𝑥 𝑑𝑥+ 1 2 ℎ′′ 𝑥 𝑑𝑥 2 This is exact, i.e., third-order and higher-order terms are zero.

22. Stochastic Differential Equations
Derivation of the Ito formula: Taylor series expansion of ℎ(𝑡,𝑦) (in two- variables): ℎ 𝑡,𝑦 =ℎ 𝑡 , 𝑦 + ℎ 𝑡 𝑡 , 𝑦 𝑡− 𝑡 + ℎ 𝑦 𝑡 , 𝑦 𝑦− 𝑦 ℎ 𝑡𝑡 𝑡 , 𝑦 𝑡− 𝑡 2 + 2ℎ 𝑡𝑦 𝑡 , 𝑦 𝑡− 𝑡 𝑦− 𝑦 + ℎ 𝑦𝑦 𝑡 , 𝑦 𝑦− 𝑦 2 ℎ 𝑡𝑡 𝑡 , 𝑦 𝑡− 𝑡 2 + 2ℎ 𝑡𝑦 𝑡 , 𝑦 𝑡− 𝑡 𝑦− 𝑦 + ℎ 𝑦𝑦 𝑡 , 𝑦 𝑦− 𝑦 2 +…

23. Stochastic Differential Equations
Derivation of the Ito formula: Taylor series expansion of ℎ(𝑡,𝑦) (in two- variables): 𝑑ℎ 𝑡,𝑦 = ℎ 𝑡 𝑡 , 𝑦 𝑑𝑡+ ℎ 𝑦 𝑡 , 𝑦 𝑑𝑦 ℎ 𝑡𝑡 𝑡 , 𝑦 𝑑𝑡 2 + 2ℎ 𝑡𝑦 𝑡 , 𝑦 𝑑𝑡𝑑𝑦+ ℎ 𝑦𝑦 𝑡 , 𝑦 𝑑𝑦 2 +…

24. Stochastic Differential Equations
Derivation of the Ito formula: Taylor series expansion of ℎ(𝑡,𝑦) (in two- variables): 𝑑ℎ 𝑡,𝑦 = ℎ 𝑡 𝑡 , 𝑦 𝑑𝑡+ ℎ 𝑦 𝑡 , 𝑦 𝑑𝑦 ℎ 𝑡𝑦 𝑡 , 𝑦 𝑑𝑡𝑑𝑦+ ℎ 𝑦𝑦 𝑡 , 𝑦 𝑑𝑦 2

25. Stochastic Differential Equations
Derivation of the Ito formula: Note that 𝑦 is a function of 𝑡 and 𝐵 𝑡 , and 𝑑𝑦 contains 𝑑𝑡 and 𝑑 𝐵 𝑡 . 2ℎ 𝑡𝑦 𝑡 , 𝑦 𝑑𝑡𝑑𝑦=0

26. Stochastic Differential Equations
Derivation of the Ito formula: Taylor series expansion of ℎ(𝑡,𝑦) (in two- variables): 𝑑ℎ 𝑡,𝑦 = ℎ 𝑡 𝑡 , 𝑦 𝑑𝑡+ ℎ 𝑦 𝑡 , 𝑦 𝑑𝑦+ 1 2 ℎ 𝑦𝑦 𝑡 , 𝑦 𝑑𝑦 2 Hence, if 𝑥=ℎ(𝑡,𝑦) then 𝑑𝑥= 𝜕ℎ 𝜕𝑡 𝑑𝑡+ 𝜕ℎ 𝜕𝑦 𝑑𝑦+ 1 2 𝜕 2 ℎ 𝜕 𝑦 2 𝑑𝑦𝑑𝑦

27. Stochastic Differential Equations
Why 𝒅 𝑩 𝒕 𝒅 𝑩 𝒕 =𝒅𝒕? 0= 𝑡 0 < 𝑡 1 <…< 𝑡 𝑛−1 < 𝑡 𝑛 =𝑡 0 𝑡 𝑑 𝐵 𝑠 2 = lim Δ 𝑡 𝑖 →0 𝑖=1 𝑛 Δ 𝐵 𝑡 𝑖 2 lim Δ 𝑡 𝑖 →0 𝑖=1 𝑛 Δ 𝐵 𝑡 𝑖 2 = lim 𝑛→∞ 𝑖=1 𝑛 Δ 𝐵 𝑡 𝑖 2

28. Stochastic Differential Equations
Why 𝒅 𝑩 𝒕 𝒅 𝑩 𝒕 =𝒅𝒕? WLoG, suppose Δ 𝑡 𝑛 =Δ 𝑡 𝑛−1 =…=Δ 𝑡 1 . It means 𝑡=𝑛Δ 𝑡 1 . 𝑡 𝑛Δ 𝑡 1 𝑖=1 𝑛 Δ 𝐵 𝑡 𝑖 2 where Δ 𝑡 1 →0, 𝑛→∞. 𝑡 1 𝑛 𝑖=1 𝑛 Δ 𝐵 𝑡 𝑖 Δ 𝑡 1 2

29. Stochastic Differential Equations
Why 𝒅 𝑩 𝒕 𝒅 𝑩 𝒕 =𝒅𝒕?
Recall: Δ 𝐵 𝑡 𝑖 Δ 𝑡 𝑖 ~𝑁(0,1)
Remark: Chi-squared distribution with k degrees of freedom ( 𝜒 2 (𝑘)) is the distribution of a sum of the squares of k independent standard normal random variables
Let 𝑌= 𝑖=1 𝑛 Δ 𝐵 𝑡 𝑖 Δ 𝑡 ~ 𝜒 2 (𝑛)
or 𝑌 𝑖 = Δ 𝐵 𝑡 𝑖 Δ 𝑡 ~ 𝜒 2 (1), 𝑖=1,2,…,𝑛

30. Stochastic Differential Equations
Why 𝒅 𝑩 𝒕 𝒅 𝑩 𝒕 =𝒅𝒕? 𝑡 1 𝑛 𝑖=1 𝑛 Δ 𝐵 𝑡 𝑖 Δ 𝑡 1 2 =𝑡 𝑌 1 + 𝑌 2 +…+ 𝑌 𝑛 𝑛 Remark: By the law of large numbers (𝑛→∞), the sample mean of n i.i.d. chi-squared random variables of degree k is k. Since 𝑌 𝑖 ~ 𝜒 2 (1) then 𝑌 1 + 𝑌 2 +…+ 𝑌 𝑛 𝑛 =1.

31. Stochastic Differential Equations
Why 𝒅 𝑩 𝒕 𝒅 𝑩 𝒕 =𝒅𝒕? 𝑡 1 𝑛 𝑖=1 𝑛 Δ 𝐵 𝑡 𝑖 Δ 𝑡 1 2 =𝑡 𝑌 1 + 𝑌 2 +…+ 𝑌 𝑛 𝑛 =𝑡 Therefore 0 𝑡 𝑑 𝐵 𝑠 2 = lim Δ 𝑡 𝑖 →0 𝑖=1 𝑛 Δ 𝐵 𝑡 𝑖 2 =𝑡. Which means 𝑑 𝐵 𝑠 2 =𝑑𝑡.

32. Stochastic Differential Equations
Why 𝒅𝒕𝒅𝒕=𝟎? WLoG, suppose Δ 𝑡 𝑛 =Δ 𝑡 𝑛−1 =…=Δ 𝑡 1 . It means 𝑡=𝑛Δ 𝑡 𝑡 𝑑𝑠 2 = lim Δ 𝑡 𝑖 →0 𝑖=1 𝑛 Δ 𝑡 𝑖 2 = lim Δ 𝑡 1 →0 𝑖=1 𝑛 Δ 𝑡 1 2 = lim Δ 𝑡 1 →0 𝑛 Δ 𝑡 1 2 Since t is constant, lim Δ 𝑡 1 →0 𝑛 Δ 𝑡 1 2 = lim Δ 𝑡 1 →0 𝑡Δ 𝑡 1 =𝑡 lim Δ 𝑡 1 →0 Δ 𝑡 1 =0

33. Stochastic Differential Equations
Why 𝒅 𝑩 𝒕 𝒅𝒕=𝟎? WLoG, suppose Δ 𝑡 𝑛 =Δ 𝑡 𝑛−1 =…=Δ 𝑡 1 . It means 𝑡=𝑛Δ 𝑡 𝑡 𝑑𝐵 𝑠 𝑑𝑠 = lim Δ 𝑡 𝑖 →0 𝑖=1 𝑛 Δ 𝐵 𝑡 𝑖 Δ 𝑡 𝑖 = lim Δ 𝑡 1 →0 𝑖=1 𝑛 Δ 𝐵 𝑡 1 Δ 𝑡 1 = lim Δ 𝑡 1 →0 𝑛Δ 𝐵 𝑡 1 Δ 𝑡 1 Since t is constant and Δ 𝐵 𝑡 1 = Δ 𝑡 1 𝑁(0,1), lim Δ 𝑡 1 →0 𝑛Δ 𝐵 𝑡 1 Δ 𝑡 1 = 𝑡 lim Δ 𝑡 1 →0 Δ 𝐵 𝑡 1 =0

34. Stochastic Differential Equations
We can guess a solution and use Ito’s Lemma to verify that the solution satisfies the SDE If 𝑥=ℎ(𝑡,𝑦) then 𝑑𝑥= 𝜕ℎ 𝜕𝑡 𝑑𝑡+ 𝜕ℎ 𝜕𝑦 𝑑𝑦+ 1 2 𝜕 2 ℎ 𝜕 𝑦 2 𝑑𝑦𝑑𝑦

35. Stochastic Differential Equations
Example 1: Solve 𝑑𝑥=𝑑𝑡+2 𝐵 𝑡 𝑑 𝐵 𝑡 The solution is 𝑥(𝑡)= 𝐵 𝑡 2 .

36. Stochastic Differential Equations
Example 1: Solve 𝑑𝑥=𝑑𝑡+2 𝐵 𝑡 𝑑 𝐵 𝑡 Let our guess be 𝑥=ℎ 𝑡,𝑦 = 𝐵 𝑡 2 = 𝑦 2 where 𝑦= 𝐵 𝑡 . Ito formula: 𝑑𝑥= 𝜕ℎ 𝜕𝑡 𝑑𝑡+ 𝜕ℎ 𝜕𝑦 𝑑𝑦+ 1 2 𝜕 2 ℎ 𝜕 𝑦 2 𝑑𝑦𝑑𝑦 𝑑𝑥=0𝑑𝑡+2𝑦𝑑𝑦 𝑑𝑦𝑑𝑦

37. Stochastic Differential Equations
Example 1: 𝑑𝑥=0𝑑𝑡+2𝑦𝑑𝑦 𝑑𝑦𝑑𝑦 𝑑𝑥=2 𝐵 𝑡 𝑑 𝐵 𝑡 +𝑑 𝐵 𝑡 𝑑 𝐵 𝑡 𝒅𝒙=𝟐 𝑩 𝒕 𝒅 𝑩 𝒕 +𝒅𝒕 Hence, the solution is 𝑥(𝑡)= 𝐵 𝑡 2 .

38. Stochastic Differential Equations
Example 2: Solve 𝑑𝑥=𝑟𝑥𝑑𝑡+𝜎𝑥𝑑 𝐵 𝑡 where 𝑟,𝜎 constant Initial condition: 𝑥 0 = 𝑥 0 The solution is the geometric Brownian motion 𝑥 𝑡 = 𝑥 0 exp 𝑟− 1 2 𝜎 2 𝑡+𝜎 𝐵 𝑡 . Note: geometric Brownian motion is the underlying model for the Black–Scholes equations that are used to price financial derivatives

39. Stochastic Differential Equations
Example 2: Solve 𝑑𝑥=𝑟𝑥𝑑𝑡+𝜎𝑥𝑑 𝐵 𝑡 Let our guess be 𝑥=ℎ 𝑡,𝑦 = 𝑥 0 𝑒 𝑦 where 𝑦= 𝑟− 1 2 𝜎 2 𝑡+𝜎 𝐵 𝑡 . Ito formula: 𝑑𝑥= 𝜕ℎ 𝜕𝑡 𝑑𝑡+ 𝜕ℎ 𝜕𝑦 𝑑𝑦+ 1 2 𝜕 2 ℎ 𝜕 𝑦 2 𝑑𝑦𝑑𝑦 𝑑𝑥=0𝑑𝑡+ 𝑥 0 𝑒 𝑦 𝑑𝑦+ 1 2 𝑥 0 𝑒 𝑦 𝑑𝑦𝑑𝑦

40. Stochastic Differential Equations
Example 2: 𝑑𝑥=0𝑑𝑡+ 𝑥 0 𝑒 𝑦 𝑑𝑦+ 1 2 𝑥 0 𝑒 𝑦 𝑑𝑦𝑑𝑦 Note: 𝑑𝑦= 𝑟− 1 2 𝜎 2 𝑑𝑡+𝜎 𝑑𝐵 𝑡 , which means 𝑑𝑦𝑑𝑦= 𝜎 2 𝑑𝑡 𝑑𝑥= 𝑥 0 𝑒 𝑦 𝑟− 1 2 𝜎 2 𝑑𝑡+𝜎 𝑑𝐵 𝑡 𝑥 0 𝑒 𝑦 𝜎 2 𝑑𝑡

41. Stochastic Differential Equations
Example 2: 𝑑𝑥= 𝑥 0 𝑒 𝑦 𝑟𝑑𝑡− 𝑥 0 𝑒 𝑦 1 2 𝜎 2 𝑑𝑡+ 𝑥 0 𝑒 𝑦 𝜎 𝑑𝐵 𝑡 𝑥 0 𝑒 𝑦 𝜎 2 𝑑𝑡 𝑑𝑥= 𝑥 0 𝑒 𝑦 𝑟𝑑𝑡+ 𝑥 0 𝑒 𝑦 𝜎 𝑑𝐵 𝑡 Since 𝑥= 𝑥 0 𝑒 𝑦 , 𝒅𝒙=𝒓𝒙𝒅𝒕+𝝈𝒙 𝒅𝑩 𝒕 Hence, the solution is 𝑥 𝑡 = 𝑥 0 exp 𝑟− 1 2 𝜎 2 𝑡+𝜎 𝐵 𝑡 .

42. Stochastic Differential Equations
Example 3: Solve 𝑑𝑥= 𝐵 𝑡 𝑑𝑡+𝑡𝑑 𝐵 𝑡 Initial condition: 𝑥 0 =𝑐 The solution is 𝑥 𝑡 =𝑡 𝐵 𝑡 +𝑐.

43. Stochastic Differential Equations
Example 4: Solve 𝑑𝑥=(1− 𝐵 𝑡 2 ) 𝑒 −2𝑥 𝑑𝑡+2 𝐵 𝑡 𝑒 −𝑥 𝑑 𝐵 𝑡 Initial condition: 𝑥 0 =0 The solution is 𝑥 𝑡 = ln (1+ 𝐵 𝑡 2 ) .

44. Stochastic Differential Equations
Example 5: Solve 𝑑𝑥= 𝐵 𝑡 𝑑𝑡+ 3 9 𝑥 2 𝑑 𝐵 𝑡 Initial condition: 𝑥 0 =0 The solution is 𝑥 𝑡 = 1 3 𝐵 𝑡 3 .

45. Stochastic Differential Equations
Example 6: Solve 𝑑𝑥= −1 2 𝑥𝑑𝑡+ 1− 𝑥 2 𝑑 𝐵 𝑡 Initial condition: 𝑥 0 =0 The solution is 𝑥 𝑡 = sin 𝐵 𝑡 .