Project 6:
Chaos in a financial model
Simon Wu, Ziyun Zeng, Michael Li, Jiaye Zhang, Haochen Li, Yixin
Ma
Agenda
• Introduction of the model
• Stability of the balance points of the model
• Graphs
• Time delay model
Introduction of the financial model
• 𝑥 =𝑧+ 𝑦−𝑎 𝑥
• 𝑦 = 1 − 𝑏𝑦 − 𝑥 2
• 𝑧 = −𝑥 − 𝑐𝑧
The economic meaning of variables and parameters:
• x: interest rate
• y: the investment demand
• z: price index
𝑥 =𝑧+ 𝑦−𝑎 𝑥
• 𝑥 = 𝑓1 𝑦 − 𝑎 𝑥 + 𝑓2 𝑧
• 𝑓1 , 𝑓2 are constants and a is the amount saving.
What does this equation mean in reality?
• change of interest rate is influenced by 2 parts, i.e. price index
and contradictions from the investment market.
2
𝑦 = 1 − 𝑏𝑦 − 𝑥
• 𝑦 = 𝑓3 𝑟 − 𝑏𝑦 − 𝛼𝑥 2
• 𝑓3 , b, 𝛼 are constants, r is the natural growth rate
• What does this equation mean in reality?
• Change of investment demand is affected by parts, i.e.
investment demand and interest rate.
𝑧 = −𝑥 − 𝑐𝑧
• 𝑧 = −𝑓4 𝑥 − 𝑐𝑧
• 𝑓4 is constant and c is the elasticity of demand of commercial
market
• What does this equation mean in reality?
• Change of price index is affected by 2 parts, i.e. the interest rate
and current price index.
𝑥 =𝑧+ 𝑦−𝑎 𝑥
𝑦 = 1 − 𝑏𝑦 − 𝑥 2
𝑧 = −𝑥 − 𝑐𝑧
Balance point: 𝑥 = 𝑦 = 𝑧 = 0
①𝑃=
1
0, , 0
𝑏
if 𝑐 − 𝑏 − 𝑎𝑏𝑐 ≤ 0 (∵ 𝑐𝑥 2 = 𝑐 − 𝑏 − 𝑎𝑏𝑐)
1
𝑐−𝑏−𝑎𝑏𝑐 1+𝑎𝑐
1
,
,∓
𝑐
𝑐
𝑐
② 𝑃 = 0, 𝑏 , 0 , Q± = ±
𝑐−𝑏−𝑎𝑏𝑐
𝑐
if 𝑐 − 𝑏 − 𝑎𝑏𝑐 ≥ 0
Stability theorem (Jacobian-->eigenvalue)
If all eigenvalues of J have strictly negative real part then the solution is stable.
If all eigenvalues of J have strictly positive real part then the solution is unstable.
If the eigenvalues of J have strictly negative real part and strictly positive real
part then the solution is a saddle.
1
① Set 𝑋 = 𝑥, 𝑌 = 𝑦 − 𝑏 , 𝑍 = 𝑧 ⟹
1
−𝑎
𝑏
𝑋=
𝑋 + 𝑍 + 𝑋𝑌
and P = X0 , Y0 , 𝑍0 = 0,0,0
𝑌 = −𝑏𝑌 − 𝑋 2
𝑍 = −𝑋 − 𝑐𝑍
𝑋=
② Set 𝑋 = 𝑥 −
𝑐−𝑏−𝑎𝑏𝑐
1+𝑎𝑐
,
𝑌
=
𝑦
−
,𝑍 = 𝑧 +
𝑐
𝑐
𝑐−𝑏−𝑎𝑏𝑐
⟹
𝑐
𝑋
+ 𝑍 + 𝑋𝑌 +
𝑐
𝑌 = −𝑏𝑌 − 𝑋 2 − 2
𝑐−𝑏−𝑎𝑏𝑐
𝑌
𝑐
𝑐−𝑏−𝑎𝑏𝑐 and Q
𝑋
𝑐
±
= (0,0,0)
𝑍 = −𝑋 − 𝑐𝑍
By noticing 𝑋 ≈ 𝐽 𝑋0 𝑋 − 𝑋0 ,
linearize this set of non-linear equations: 𝐽 P =
① Characteristic equation: det 𝜆𝐼 − 𝐽 = 𝜆 + 𝑏
1
① Eigenvalues: 𝜆1 = −𝑏, 𝜆2 =
− 𝑐+𝑎−𝑏 −
1 2
𝑐+𝑎−𝑏
2
1
−𝑎
𝑏
0
1
0
−𝑏 0
−1
0 −𝑐
1
𝑐
𝜆2 + 𝑐 + 𝑎 − 𝑏 𝜆 + 1 + 𝑎𝑐 − 𝑏 = 0
𝑐
−4 𝑎𝑐−𝑏+1
1
, 𝜆3
− 𝑐+𝑎−𝑏 +
1 2
𝑐+𝑎−𝑏
2
𝑐
−4 𝑎𝑐−𝑏+1
satisfing 𝜆2 + 𝑐 + 𝑎 −
1
Only study on balance point 𝑃 = 0, 𝑏 , 0
Case 1
𝑐
1
𝑐 − 𝑏 − 𝑎𝑏𝑐 < 0 ⟹ 1 + 𝑎𝑐 − 𝑏 > 0 𝑎𝑛𝑑 𝑐 + 𝑎 − 𝑏 > 0 ⟹ 𝜆1 < 0, 𝜆2 < 0, 𝜆3 < 0 (only consider real
part of eigenvalues)
• Balance point is stable convergence.
Case 2
1
𝑐 − 𝑏 − 𝑎𝑏𝑐 < 0 𝑎𝑛𝑑 𝑐 + 𝑎 − < 0 ⟹ 𝜆1 < 0, 𝜆2 > 0, 𝜆3 > 0
𝑏
• Balance point is saddle.
Case 3
𝑐 − 𝑏 − 𝑎𝑏𝑐 > 0
𝜆1 , 𝜆2 < 0, 𝜆3 > 0 ⟹balance point P is a saddle point.
x-t diagram（initial value
(x=1,y=3,z=8)
a=0.3
a=6
a=2
a=9
y-t diagram （initial value x=1,y=3,z=8)
a=0.3
a=6
a=2
a=9
z-t diagram （initial value x=1,y=3,z=8)
a=0.3
a=6
a=2
a=9
x-t diagram a=2
initial value (x=1, y=3, z=8)
initial value (x=1, y=0, z=8)
initial value (x=8, y=3, z=8)
initial value (x=1, y=3, z=2)
a=1
a=5
a=9
initial value x=1,y=3,z=8
time delay
• What is time delay?
• One is the time lag between the time economic decisions are made and the time the
decisions bear fruit.
•
• simplified finance model
• By adding time-delayed feedbacks to system, we can get the new system, where s1,
s2, and s3 are time delays and K1, K2, and K3 indicate the strengths of the feedbacks.
• Hopf bifurcation of the model with time delay
• we only discuss the time delay on Y (the investment demand)
• Lemma1 When c−b−abc≤0, i.e.,1+ac−bc >0, system (1) has a unique equilibrium
• P0 (0,1/b,0)
• We calculated X=sqrt(1-ab-b/c) or X=-sqrt(1-ab-b/c)
• So if c−b−abc≤0, i.e., 1-ab-b/c≤0. The system has a single equilibrium iff 1-ab-b/c=0.
• So X=0. Substitute and get P0 (0,1/b,0)
• When c−b−abc > 0, we have 2 equilibrium solutions. X=sqrt(1-ab-b/c) or X=-sqrt(1ab-b/c)
• Substitute and get P+ and P• So we have three equilibria in total.
•
• First Let x=x, y=y−1/b, z=z
• linearization of system (2) at the equilibrium (0, 0, 0) can be written as:
• We know that the equilibrium point will not be stable when both exponentials in the
solution converge to 0, which implies that both eigenvalues need to be smaller than 0.
• So we need to prove when 1+ac−bc >0 and c+a−1/b >0, λ1 λ2 are real and smaller
than 0.
• We can use lemma4 to prove.
•
• Lemma 4 The two roots λ1,2 = 1/2(−P ±sqrt(P −4Q)) of (4) have always negative real
parts when τ ≥ 0 with the condition (H1)
• Let P = c+a – 1/b , Q = 1 + ac − bc ; it’s easy to prove P2−4Q>0. So λ1,2 are real.
•
•
Code and Graph
Evolution of x and z
• Interpretation of Time History Plots
1. Stable Oscillations:
stable cyclic pattern, common
energy is conserved
2. Damped Oscillations:
loses energy over time, moving towards equilibrium
3. Unstable Oscillations:
The system is unstable and may exhibit chaotic behavior.
Purpose of the model
• By analyzing these behaviors
• fine-tune the parameters to achieve desired system dynamics
• such as ensuring stable oscillations for practical applications in
engineering or scientific systems.
Thanks a lot!