Transcript Optimal Pairs Trading: A Stochastic Control Approach

Introduction to Algorithmic Trading Strategies
Lecture 4
Optimal Pairs Trading by Stochastic Control
Haksun Li
[email protected]
www.numericalmethod.com
Outline
Problem formulation
Ito’s lemma
Dynamic programming
Hamilton-Jacobi-Bellman equation
Riccati equation
Integrating factor






2
Reference
Optimal Pairs Trading: A Stochastic Control
Approach. Mudchanatongsuk, S., Primbs, J.A., Wong,
W. Dept. of Manage. Sci. & Eng., Stanford Univ.,
Stanford, CA.

3
Basket Creation vs. Trading
In lecture 3, we discussed a few ways to construct a
mean-reverting basket.
In this and the next lectures, we discuss how to trade a
mean-reverting asset, if such exists.


4
Stochastic Control
We model the difference between the log-returns of
two assets as an Ornstein-Uhlenbeck process.
We compute the optimal position to take as a function
of the deviation from the equilibrium.
This is done by solving the corresponding the
Hamilton-Jacobi-Bellman equation.



5
Formulation
Assume a risk free asset 𝑀𝑡 , which satisfies


𝑑𝑀𝑡 = 𝑟𝑀𝑡 𝑑𝑡
Assume two assets,𝐴𝑡 and 𝐵𝑡 .
Assume 𝐵𝑡 follows a geometric Brownian motion.



𝑑𝐵𝑡 = 𝜇𝐵𝑡 𝑑𝑡 + 𝜎𝐵𝑡 𝑑𝑧𝑡
𝑥𝑡 is the spread between the two assets.


6
𝑥𝑡 = log 𝐴𝑡 − log 𝐵𝑡
Ornstein-Uhlenbeck Process
We assume the spread, the basket that we want to
trade, follows a mean-reverting process.


𝑑𝑥𝑡 = 𝑘 𝜃 − 𝑥𝑡 𝑑𝑡 + 𝜂𝑑𝜔𝑡
𝜃 is the long term equilibrium to which the spread
reverts.
𝑘 is the rate of reversion. It must be positive to ensure
stability around the equilibrium value.


7
Instantaneous Correlation
Let 𝜌 denote the instantaneous correlation coefficient
between 𝑧 and 𝜔.


8
𝐸 𝑑𝜔𝑡 𝑑𝑧𝑡 = 𝜌𝑑𝑡
Univariate Ito’s Lemma

Assume



𝑑𝑋𝑡 = 𝜇𝑡 𝑑𝑡 + 𝜎𝑡 𝑑𝐵𝑡
𝑓 𝑡, 𝑋𝑡 is twice differentiable of two real variables
We have

9
𝑑𝑓 𝑡, 𝑋𝑡 =
𝜕𝑓
𝜕𝑡
+ 𝜇𝑡
𝜕𝑓
𝜕𝑥
+
𝜎𝑡 2 𝜕 2 𝑓
2 𝜕𝑥 2
𝑑𝑡 + 𝜎𝑡
𝜕𝑓
𝜕𝑥
𝑑𝐵𝑡
Log example


For G.B.M., 𝑑𝑋𝑡 = 𝜇𝑋𝑡 𝑑𝑡 + 𝜎𝑋𝑡 𝑑𝑧𝑡 , 𝑑 log 𝑋𝑡 =?
𝑓 𝑥 = log 𝑥
𝜕𝑓

=0
𝜕𝑡
𝜕𝑓
1

=
𝜕𝑥
𝑥
𝜕2 𝑓
1

=− 2
𝜕𝑥 2
𝑥

𝑑 log 𝑋𝑡 =

𝜎2
−
2
= 𝜇
10
1
𝜇𝑋𝑡
𝑋𝑡
+
𝜎𝑋𝑡 2
2
𝑑𝑡 + 𝜎𝑑𝐵𝑡
1
− 2
𝑋𝑡
𝑑𝑡 +
1
𝜎𝑋𝑡
𝑋𝑡
𝑑𝐵𝑡
Multivariate Ito’s Lemma

Assume



𝑋𝑡 = 𝑋1𝑡 , 𝑋2𝑡 , ⋯ , 𝑋𝑛𝑡 is a vector Ito process
𝑓 𝑥1𝑡 , 𝑥2𝑡 , ⋯ , 𝑥𝑛𝑡 is twice differentiable
We have

𝑑𝑓 𝑋1𝑡 , 𝑋2𝑡 , ⋯ , 𝑋𝑛𝑡

=
𝜕
𝑛
𝑖=1 𝜕𝑥
1
 +
2
𝑛
𝑖=1
11
𝑖
𝑓 𝑋1𝑡 , 𝑋2𝑡 , ⋯ , 𝑋𝑛𝑡 𝑑𝑋𝑖 𝑡
𝜕2
𝑛
𝑗=1 𝜕𝑥 𝜕𝑥 𝑓
𝑖
𝑗
𝑋1𝑡 , 𝑋2𝑡 , ⋯ , 𝑋𝑛𝑡 𝑑 𝑋𝑖 , 𝑋𝑗 𝑡
Multivariate Example


log 𝐴𝑡 = 𝑥𝑡 + log 𝐵𝑡
𝐴𝑡 = exp 𝑥𝑡 + log 𝐵𝑡
𝜕𝐴𝑡

𝜕𝑥𝑡
𝜕𝐴𝑡

𝜕𝐵𝑡
= exp 𝑥𝑡 + log 𝐵𝑡 = 𝐴𝑡
= exp 𝑥𝑡 +
𝜕 2 𝐴𝑡

𝜕𝑥𝑡 2
=
𝜕𝐴𝑡
𝜕𝑥𝑡
𝜕 2 𝐴𝑡

𝜕𝐵𝑡 2
=
𝜕 𝐴𝑡
𝜕𝐵𝑡 𝐵𝑡
𝜕 2 𝐴𝑡

𝜕𝐵𝑡 𝜕𝑥𝑡
12
1
log 𝐵𝑡
𝐵𝑡
=
=
= 𝐴𝑡
=0
𝜕 𝜕𝐴𝑡
𝜕𝐵𝑡 𝜕𝑥𝑡
=
𝜕𝐴𝑡
𝜕𝐵𝑡
=
𝐴𝑡
𝐵𝑡
𝐴𝑡
𝐵𝑡
What is the Dynamic of Asset At?

𝜕𝐴𝑡 =
𝜕 2 𝐴𝑡
𝜕𝐵𝑡 𝜕𝑥


=
𝜕𝐴𝑡
𝑑𝑥𝑡
𝜕𝑥
𝜕𝐴𝑡
+
𝑑𝐵𝑡
𝜕𝐵𝑡
1 𝜕 2 𝐴𝑡
+
2 𝜕𝑥 2
2
+
𝑑𝑥𝑡 𝑑𝐵𝑡
𝐴𝑡
𝐴𝑡 𝑑𝑥𝑡 + 𝑑𝐵𝑡
𝐵𝑡
+
1
𝐴
2 𝑡
𝑑𝑥𝑡
= 𝐴𝑡 𝑘 𝜃 − 𝑥𝑡 𝑑𝑡 + 𝜂𝑑𝜔𝑡 +
1
𝐴𝑡 𝜂2 𝑑𝑡
2
13
𝑑𝑥𝑡
𝐴𝑡
+ 𝜌𝜂𝜎𝐵𝑡 𝑑𝑡
𝐵𝑡
2
+
𝐴𝑡
𝐵𝑡
𝐴𝑡
𝐵𝑡
𝑑𝑥𝑡 𝑑𝐵𝑡
𝜇𝐵𝑡 𝑑𝑡 + 𝜎𝐵𝑡 𝑑𝑧𝑡 +
Dynamic of Asset At

𝜕𝐴𝑡 = 𝐴𝑡 𝑘 𝜃 − 𝑥𝑡 𝑑𝑡 + 𝜂𝑑𝜔𝑡 + 𝐴𝑡 𝜇𝑑𝑡 + 𝜎𝑑𝑧𝑡 +
1
𝐴𝑡 𝜂2 𝑑𝑡 + 𝐴𝑡 𝜌𝜂𝜎𝑑𝑡
2


= 𝐴𝑡 𝑘 𝜃 − 𝑥𝑡 + 𝜇 +
𝐴𝑡 𝜎𝑑𝑧𝑡
= 𝐴𝑡
14
1 2
𝜂
2
𝜇 + 𝑘 𝜃 − 𝑥𝑡 +
+ 𝜌𝜂𝜎 𝑑𝑡 + 𝐴𝑡 𝜂𝑑𝜔𝑡 +
1 2
𝜂
2
+ 𝜌𝜂𝜎 𝑑𝑡 + 𝜎𝑑𝑧𝑡 + 𝜂𝑑𝜔𝑡
Notations



𝑉𝑡 : the value of a self-financing pairs trading portfolio
ℎ𝑡 :the portfolio weight for stock A
ℎ𝑡 = −ℎ𝑡 :the portfolio weight for stock B
15
Self-Financing Portfolio Dynamic

𝑑𝑉𝑡
𝑉𝑡

= ℎ𝑡
16
= ℎ𝑡
𝑑𝐴𝑡
𝐴𝑡
+ ℎ𝑡
𝑑𝐵𝑡
𝐵𝑡
𝜇+𝑘 𝜃−
𝑑𝑀𝑡
𝑀𝑡
1 2
𝑥𝑡 + 𝜂
2
+
+ 𝜌𝜂𝜎 𝑑𝑡 + 𝜎𝑑𝑧𝑡 +
Power Utility

Investor preference:



17
𝑈 𝑥 = 𝑥𝛾
𝑥≥0
0<𝛾<1
Problem Formulation

max 𝐸 𝑉𝑇 𝛾 , s.t.,

𝑉 0 = 𝑣0 , x 0 = 𝑥0
ℎ𝑡
𝑑𝑥𝑡 = 𝑘 𝜃 − 𝑥𝑡 𝑑𝑡 + 𝜂𝑑𝜔𝑡



𝑑𝑉𝑡 = ℎ𝑡 𝑑𝑥𝑡 = ℎ𝑡 𝑘 𝜃 − 𝑥𝑡 𝑑𝑡 + ℎ𝑡 𝜂𝑑𝜔𝑡
Note that we simplify GBM to BM of 𝑉𝑡 , and remove
some constants.
18
Dynamic Programming

Consider a stage problem to minimize (or maximize) the
accumulated costs over a system path.
Cost = 𝑐3 +
s11
2
2
𝑡=0 𝑐𝑡
S21
5
3
s0
3
1
4
s12
s3
S22
5
5
3
s12
19
t=0
s23
time
t=1
t=2
Dynamic Programming Formulation

State change: 𝑥𝑘+1 = 𝑓𝑘 𝑥𝑘 , 𝑢𝑘 , 𝜔𝑘






𝑘: time
𝑥𝑘 : state
𝑢𝑘 : control decision selected at time 𝑘
𝜔𝑘 : a random noise
Cost: 𝑔𝑁 𝑥𝑁 + 𝑁−1
𝑘=0 𝑔𝑘 𝑥𝑘 , 𝑢𝑘 , 𝜔𝑘
Objective: minimize (maximize) the expected cost.

20
We need to take expectation to account for the noise, 𝜔𝑘 .
Principle of Optimality

Let 𝜋 ∗ = 𝜇0 ∗ , 𝜇1 ∗ , ⋯ , 𝜇𝑁−1 ∗ be an optimal policy for
the basic problem, and assume that when using 𝜋 ∗ , a
give state 𝑥𝑖 occurs at time 𝑖 with positive probability.
Consider the sub-problem whereby we are at 𝑥𝑖 at time
𝑖 and wish to minimize the “cost-to-go” from time 𝑖 to
time 𝑁.


𝐸 𝑔𝑁 𝑥𝑁 +
𝑁−1
𝑘=0 𝑔𝑘
𝑥𝑘 , 𝑢𝑘 , 𝜔𝑘
Then the truncated policy 𝜇𝑖 ∗ , 𝜇𝑖+1 ∗ , ⋯ , 𝜇𝑁−1 ∗ is
optimal for this sub-problem.
21
Dynamic Programming Algorithm

For every initial state 𝑥0 , the optimal cost 𝐽∗ 𝑥𝑘 of the
basic problem is equal to 𝐽0 𝑥0 , given by the last step
of the following algorithm, which proceeds backward
in time from period 𝑁 − 1 to period 0:

 𝐽𝑘
22
𝐽𝑁 𝑥𝑁 = 𝑔𝑁 𝑥𝑁
𝑥𝑘 = min 𝐸 𝑔𝑘 𝑥𝑘 , 𝑢𝑘 , 𝜔𝑘 + 𝐽𝑘+1 𝑓𝑘 𝑥𝑘 , 𝑢𝑘 , 𝜔𝑘
𝑢𝑘
Value function

Terminal condition:



𝐺 𝑇, 𝑉, 𝑥 = 𝑉 𝛾
DP equation:

𝐺 𝑡, 𝑉𝑡 , 𝑥𝑡 = m𝑎𝑥 𝐸 𝐺 𝑡 + 𝑑𝑡, 𝑉𝑡+𝑑𝑡 , 𝑥𝑡+𝑑𝑡

𝐺 𝑡, 𝑉𝑡 , 𝑥𝑡 = m𝑎𝑥 𝐸 𝐺 𝑡, 𝑉𝑡 , 𝑥𝑡 + Δ𝐺
ℎ𝑡
ℎ𝑡
By Ito’s lemma:

1
2
23
1
2
Δ𝐺 = 𝐺𝑡 𝑑𝑡 + 𝐺𝑉 𝑑𝑉 + 𝐺𝑥 𝑑𝑥 + 𝐺𝑉𝑉 𝑑𝑉
𝐺𝑥𝑥 𝑑𝑥
2
+ 𝐺𝑉𝑥 𝑑𝑉 𝑑𝑥
2
+
Hamilton-Jacobi-Bellman Equation

Cancel 𝐺 𝑡, 𝑉𝑡 , 𝑥𝑡 on both LHS and RHS.
Divide by time discretization, Δ𝑡.
Take limit as Δ𝑡 → 0, hence Ito.

0 = m𝑎𝑥 𝐸 Δ𝐺

m𝑎𝑥 𝐸 𝐺𝑡 𝑑𝑡 + 𝐺𝑉 𝑑𝑉 + 𝐺𝑥 𝑑𝑥 + 2 𝐺𝑉𝑉 𝑑𝑉



ℎ𝑡
1
ℎ𝑡
0
The optimal portfolio position is ℎ𝑡 ∗ .
24
2
1
+ 2 𝐺𝑥𝑥 𝑑𝑥
2
+ 𝐺𝑉𝑥 𝑑𝑉 𝑑𝑥
=
HJB for Our Portfolio Value

1
m𝑎𝑥 𝐸 𝐺𝑡 𝑑𝑡 + 𝐺𝑉 𝑑𝑉 + 𝐺𝑥 𝑑𝑥 + 2 𝐺𝑉𝑉 𝑑𝑉
0
2
ℎ𝑡
1
+ 2 𝐺𝑥𝑥 𝑑𝑥
2
+ 𝐺𝑉𝑥 𝑑𝑉 𝑑𝑥
𝐺𝑡 𝑑𝑡 + 𝐺𝑉 ℎ𝑡 𝑘 𝜃 − 𝑥𝑡 𝑑𝑡 + ℎ𝑡 𝜂𝑑𝜔𝑡 + 𝐺𝑥 𝑑𝑥 +

m𝑎𝑥 𝐸
ℎ𝑡
1
𝐺
2 𝑉𝑉
ℎ𝑡 𝑘 𝜃 − 𝑥𝑡 𝑑𝑡 + ℎ𝑡 𝜂𝑑𝜔𝑡
2
1
+ 2 𝐺𝑥𝑥 𝑑𝑥
2
+
=0
𝐺𝑉𝑥 ℎ𝑡 𝑘 𝜃 − 𝑥𝑡 𝑑𝑡 + ℎ𝑡 𝜂𝑑𝜔𝑡 𝑑𝑥
𝐺𝑡 𝑑𝑡 + 𝐺𝑉 ℎ𝑡 𝑘 𝜃 − 𝑥𝑡 𝑑𝑡 + ℎ𝑡 𝜂𝑑𝜔𝑡 +
𝐺𝑥 𝑘 𝜃 − 𝑥𝑡 𝑑𝑡 + 𝜂𝑑𝜔𝑡 +

m𝑎𝑥 𝐸
ℎ𝑡
1
𝐺 ℎ𝑘
2 𝑉𝑉 𝑡
1
𝐺 𝑘
2 𝑥𝑥
𝜃 − 𝑥𝑡 𝑑𝑡 + ℎ𝑡 𝜂𝑑𝜔𝑡
2
2
+
𝜃 − 𝑥𝑡 𝑑𝑡 + 𝜂𝑑𝜔𝑡
+
𝐺𝑉𝑥 ℎ𝑡 𝑘 𝜃 − 𝑥𝑡 𝑑𝑡 + ℎ𝑡 𝜂𝑑𝜔𝑡 × 𝑘 𝜃 − 𝑥𝑡 𝑑𝑡 + 𝜂𝑑𝜔𝑡
25
=0
=
Taking Expectation



All 𝜂𝑑𝜔𝑡 disapper because of the expectation operator.
Only the deterministic 𝑑𝑡 terms remain.
Divide LHR and RHS by 𝑑𝑡.
𝐺𝑡 + 𝐺𝑉 ℎ𝑡 𝑘 𝜃 − 𝑥𝑡

m𝑎𝑥 𝐺𝑥 𝑘 𝜃 − 𝑥𝑡
ℎ𝑡
26
+
1
𝐺
2 𝑉𝑉
ℎ𝑡 𝜂
+𝐺𝑉𝑥 ℎ𝑡 𝜂2
+
2
+
1
𝐺𝑥𝑥 𝜂2
2
=0
Dynamic Programming Solution


Solve for the cost-to-go function, 𝐺𝑡 .
Assume that the optimal policy is ℎ𝑡 ∗ .
27
First Order Condition

Differentiate w.r.t. ℎ𝑡 .

𝐺𝑉 𝑘 𝜃 − 𝑥𝑡


∗
ℎ𝑡 = −
+ ℎ𝑡 ∗ 𝐺𝑉𝑉 𝜂2 + 𝐺𝑉𝑥 𝜂2 = 0
𝐺𝑉 𝑘 𝜃−𝑥𝑡 +𝐺𝑉𝑥 𝜂2
𝐺𝑉𝑉 𝜂 2
In order to determine the optimal position, ℎ𝑡 ∗ , we
need to solve for 𝐺 to get 𝐺𝑉 , 𝐺𝑉𝑥 , and 𝐺𝑉𝑉 .
28
The Partial Differential Equation (1)
𝐺𝑡 −
𝐺𝑉
𝐺𝑉 𝑘 𝜃−𝑥𝑡 +𝐺𝑉𝑥 𝜂 2
𝐺𝑉𝑉
𝜂2
𝐺𝑥 𝑘 𝜃 − 𝑥𝑡

𝑘 𝜃 − 𝑥𝑡
+
𝐺𝑉 𝑘 𝜃−𝑥𝑡 +𝐺𝑉𝑥 𝜂2
1
2
𝐺 𝜂
2 𝑉𝑉
𝐺𝑉𝑉 𝜂 2
1
𝐺𝑥𝑥 𝜂2 −
2
2
𝐺
𝑘
𝜃−𝑥
+𝐺
𝜂
𝑉
𝑡
𝑉𝑥
𝐺𝑉𝑥 𝜂2
𝐺 𝜂2
𝑉𝑉
29
+
2
+
=0
The Partial Differential Equation (2)
𝐺𝑡 −
𝐺𝑉 𝑘 𝜃 − 𝑥𝑡
𝐺𝑉 𝑘 𝜃−𝑥𝑡 +𝐺𝑉𝑥 𝜂 2
𝐺𝑉𝑉
𝐺𝑥 𝑘 𝜃 − 𝑥𝑡
𝜂2
+
2
1 𝐺𝑉 𝑘 𝜃−𝑥𝑡 +𝐺𝑉𝑥 𝜂 2
+
2
𝐺𝑉𝑉 𝜂2
1
𝐺𝑥𝑥 𝜂2 −
2
𝐺𝑉 𝑘 𝜃−𝑥𝑡 +𝐺𝑉𝑥 𝜂 2
𝐺𝑉𝑥
𝐺

𝑉𝑉
30
+
=0
Dis-equilibrium

Let 𝑏 = 𝑘 𝜃 − 𝑥𝑡 . Rewrite:
2
𝐺𝑉 𝑏+𝐺𝑉𝑥 𝜂 2
1 𝐺𝑉 𝑏+𝐺𝑉𝑥 𝜂 2
1
2
 𝐺𝑡 − 𝐺𝑉 𝑏
+
𝐺
𝑏
+
+
𝐺
𝜂
𝑥
𝐺𝑉𝑉 𝜂2
2
𝐺𝑉𝑉 𝜂 2
2 𝑥𝑥
𝐺𝑉 𝑏+𝐺𝑉𝑥 𝜂 2
𝐺𝑉𝑥
=0
𝐺𝑉𝑉
 Multiply by 𝐺𝑉𝑉 𝜂 2 .
 𝐺𝑡 𝐺𝑉𝑉 𝜂 2 − 𝐺𝑉 𝑏 𝐺𝑉 𝑏 + 𝐺𝑉𝑥 𝜂 2 + 𝐺𝑥 𝑏𝐺𝑉𝑉 𝜂 2 +
1
1
2
2
𝐺𝑉 𝑏 + 𝐺𝑉𝑥 𝜂
+ 𝐺𝑥𝑥 𝐺𝑉𝑉 𝜂4 − 𝐺𝑉𝑥 𝜂2 𝐺𝑉 𝑏 +
2
2
31
−
Simplification

Note that

−𝐺𝑉 𝑏 𝐺𝑉 𝑏 + 𝐺𝑉𝑥 𝜂2 +
1
2
𝐺𝑉𝑥 𝜂2 𝐺𝑉 𝑏 + 𝐺𝑉𝑥 𝜂2 = −


𝐺𝑉 𝑏 + 𝐺𝑉𝑥 𝜂2
1
2
2
−
𝐺𝑉 𝑏 + 𝐺𝑉𝑥 𝜂2
2
The PDE becomes
𝐺𝑡 𝐺𝑉𝑉
32
𝜂2
+ 𝐺𝑥 𝑏𝐺𝑉𝑉
𝜂2
1
+ 𝐺𝑥𝑥 𝐺𝑉𝑉 𝜂4
2
1
−
2
𝐺𝑉 𝑏 +
The Partial Differential Equation (3)

2
𝐺𝑡 𝐺𝑉𝑉 𝜂 + 𝐺𝑥 𝑏𝐺𝑉𝑉 𝜂
33
2
1
+ 𝐺𝑥𝑥 𝐺𝑉𝑉 𝜂4
2
1
−
2
𝐺𝑉 𝑏 +
Ansatz for G









𝐺 𝑡, 𝑉, 𝑥 = 𝑓 𝑡, 𝑥 𝑉 𝛾
𝐺 𝑇, 𝑉, 𝑥 = 𝑉 𝛾
𝑓 𝑇, 𝑥 = 1
𝐺𝑡 = 𝑉 𝛾 𝑓𝑡
𝐺𝑉 = 𝛾𝑉 𝛾−1 𝑓
𝐺𝑉𝑉 = 𝛾 𝛾 − 1 𝑉 𝛾−2 𝑓
𝐺𝑥 = 𝑉 𝛾 𝑓𝑥
𝐺𝑉𝑥 = 𝛾𝑉 𝛾−1 𝑓𝑥
𝐺𝑥𝑥 = 𝑉 𝛾 𝑓𝑥𝑥
34
Another PDE (1)

𝑉 𝛾 𝑓𝑡 𝛾 𝛾 − 1 𝑉 𝛾−2 𝑓𝜂2 + 𝑉 𝛾 𝑓𝑥 𝑏𝛾 𝛾 − 1 𝑉 𝛾−2 𝑓𝜂2 +
1 𝛾
1
𝛾−2
4
𝑉 𝑓𝑥𝑥 𝛾 𝛾 − 1 𝑉 𝑓𝜂 − 𝛾𝑉 𝛾−1 𝑓𝑏 +
2
35
2
Ansatz for 𝑓
1 2
𝜂 𝑓𝑓𝑥𝑥
2
𝛾
−
2 𝛾−1
2
𝑏
𝑓
𝜂

𝑓𝑓𝑡 + 𝑏𝑓𝑓𝑥 +

𝑓 𝑡, 𝑥 = 𝑔 𝑡 exp 𝑥𝛽 𝑡 + 𝑥 𝛼 𝑡 = 𝑔 exp 𝑥𝛽 + 𝑥 2 𝛼
𝑓𝑡 = 𝑔𝑡 exp 𝑥𝛽 + 𝑥 2 𝛼 + 𝑔 exp 𝑥𝛽 + 𝑥 2 𝛼 𝑥𝛽𝑡 +

36
+ 𝜂𝑓𝑥
2
=0
Boundary Conditions




𝑓
𝑔
𝛼
𝛽
37
𝑇, 𝑥 = 𝑔 𝑇 exp 𝑥𝛽 𝑇 + 𝑥 2 𝛼 𝑇
𝑇 =1
𝑇 =0
𝑇 =0
=1
Yet Another PDE (1)

𝑔 exp 𝑥𝛽 + 𝑥 2 𝛼 𝑔𝑡 exp 𝑥𝛽 + 𝑥 2 𝛼 + 𝑔 exp 𝑥𝛽 + 𝑥 2 𝛼 𝑥𝛽𝑡 + 𝑥 2 𝛼𝑡
1
𝑏𝑔 exp 𝑥𝛽 + 𝑥 2 𝛼 𝑔 exp 𝑥𝛽 + 𝑥 2 𝛼 𝛽 + 2𝑥𝛼 + 2 𝜂2 𝑔 exp 𝑥𝛽 +
38
+
Yet Another PDE (2)


𝜆=
𝛾
2 𝛾−1
2
𝑔𝑡 + 𝑔 𝑥𝛽𝑡 + 𝑥 𝛼𝑡 + 𝑏𝑔 𝛽 + 2𝑥𝛼
39
1 2
+ 𝜂 𝑔
2
𝛽+
Expansion in 𝑥

1
𝑔𝑡 + 𝑔𝑥𝛽𝑡 + 𝑔𝑥 2 𝛼𝑡 + 𝑏𝑔𝛽 + 2𝑥𝛼𝑏𝑔 + 2 𝜂2 𝑔 𝛽 2 + 4𝑥 2 𝛼 2 + 4𝑥𝛼𝛽 + 𝜂2 𝑔𝛼 −
𝜆𝑔

𝑏2
𝜂2
+ 𝜂2 𝛽 2 + 4𝜂2 𝑥 2 𝛼 2 + 2𝑏𝛽 + 4𝑏𝑥𝛼 + 4𝜂2 𝑥𝛼𝛽
=0
1
𝑔𝑡 + 𝑔𝑥𝛽𝑡 + 𝑔𝑥 2 𝛼𝑡 + 𝑘 𝜃 − 𝑥 𝑔𝛽 + 2𝑥𝛼𝑘 𝜃 − 𝑥 𝑔 + 2 𝜂2 𝑔 𝛽 2 + 4𝑥 2 𝛼 2 +
40
Grouping in 𝑥
1
𝜆𝑔
𝑔𝑡 + 𝑘𝑔𝛽𝜃 + 2 𝜂2 𝑔𝛽 2 + 𝜂2 𝑔𝛼 − 𝜂2 𝑘 2 𝜃 2 − 𝜆𝑔𝜂2 𝛽 2 − 2𝜆𝑔𝑘𝛽𝜃 + 𝑔𝛽𝑡 −

41
The Three PDE’s (1)

𝑔𝛼𝑡 − 2𝛼𝑘𝑔 +
2𝜂2 𝑔𝛼 2
0

𝑔𝛽𝑡 − 𝑘𝑔𝛽 + 2𝛼𝑘𝑔𝜃 +
𝜆𝑔 2
− 2𝑘
𝜂
− 4𝜆𝑔𝜂2 𝛼 2 + 4𝜆𝑔𝑘𝛼 =
2𝜂2 𝑔𝛼𝛽
4𝜆𝑔𝑘𝜃𝛼 − 4𝜆𝑔𝜂 2 𝛼𝛽 = 0

𝑔𝑡 + 𝑘𝑔𝛽𝜃 +
2𝜆𝑔𝑘𝛽𝜃 = 0
42
1 2
𝜂 𝑔𝛽 2
2
2
+ 𝜂 𝑔𝛼 −
+
𝜆𝑔 2
2 2𝑘 𝜃
𝜂
𝜆𝑔 2 2
𝑘 𝜃
𝜂2
+ 2𝜆𝑔𝑘𝛽 −
− 𝜆𝑔𝜂2 𝛽 2 −
PDE in 𝛼

𝛼𝑡 +
2𝜂2

𝛼𝑡 =
𝜆 2
𝑘
2
𝜂
43
−
4𝜆𝜂 2
𝛼2
+ 4𝜆𝑘 − 2𝑘 𝛼 −
𝜆 2
𝑘
𝜂2
+ 2𝑘 1 − 2𝜆 𝛼 + 2𝜂2 2𝜆 − 1 𝛼 2
=0
PDE in 𝛽, 𝛼

𝛽𝑡 − 𝑘𝛽 + 2𝜂2 𝛼𝛽 + 2𝜆𝑘𝛽 − 4𝜆𝜂2 𝛼𝛽 − 4𝜆𝑘𝜃𝛼 +
𝜆
2 2 𝑘 2 𝜃 + 2𝛼𝑘𝜃 = 0
𝜂

𝛽𝑡 = 𝑘 − 2𝜂2 𝛼 − 2𝜆𝑘 + 4𝜆𝜂2 𝛼 𝛽 + 4𝜆𝑘𝜃𝛼 −
44
PDE in 𝛽, 𝛼, 𝑔

𝑔𝑡 + 𝑘𝑔𝛽𝜃 +
2𝜆𝑔𝑘𝛽𝜃 = 0

𝑔𝑡 = −𝑘𝑔𝛽𝜃 −
2𝜆𝑔𝑘𝛽𝜃

1 2
𝜂 𝑔𝛽 2
2
1 2
𝜂 𝑔𝛽 2
2
𝑔𝑡 = 𝑔 −𝑘𝛽𝜃 −
45
+
1 2 2
𝜂 𝛽
2
𝜂2 𝑔𝛼
−
−
𝜂2 𝑔𝛼
2
−𝜂 𝛼
𝜆𝑔 2 2
𝑘 𝜃
𝜂2
− 𝜆𝑔𝜂2 𝛽 2 −
𝜆𝑔 2 2
+ 2𝑘 𝜃
𝜂
𝜆 2 2
+ 2𝑘 𝜃
𝜂
+ 𝜆𝑔𝜂2 𝛽 2 +
+ 𝜆𝜂 2 𝛽2 +
Riccati Equation

A Riccati equation is any ordinary differential equation
that is quadratic in the unknown function.
𝜆 2
𝑘
𝜂2
+ 2𝑘 1 − 2𝜆 𝛼 + 2𝜂2 2𝜆 − 1 𝛼 2

𝛼𝑡 =

𝛼𝑡 = 𝐴0 + 𝐴1 𝛼 + 𝐴2 𝛼 2
46
Solving a Riccati Equation by Integration

Suppose a particular solution, 𝛼1 , can be found.

𝛼 = 𝛼1 + is the general solution, subject to some
𝑧
boundary condition.
1
47
Particular Solution


Either 𝛼1 or 𝛼2 is a particular solution to the ODE.
This can be verified by mere substitution.
𝛼1,2 =
48
−𝐴1 ± 𝐴1 2 −4𝐴2 𝐴0
2𝐴2
𝑧 Substitution

Suppose 𝛼 =

1
𝑧

=

=

=
1
𝛼1 + .
𝑧
1 2
= 𝐴0 + 𝐴1 𝛼1 + + 𝐴2 𝛼1 +
𝑧
1
1
𝛼1
2
𝐴0 + 𝐴1 𝛼1 + 𝐴1 + 𝐴2 𝛼1 + 𝐴2 2 + 2𝐴2
𝑧
𝑧
𝑧
1
1
𝛼1
2
𝐴0 + 𝐴1 𝛼1 + 𝐴1 + 𝐴2 𝛼1 + 𝐴2 2 + 2𝐴2
𝑧
𝑧
𝑧
𝐴1 +2𝛼1 𝐴2
𝐴2
2
𝐴0 + 𝐴1 𝛼1 + 𝐴2 𝛼1 +
+ 2
𝑧
𝑧
1
𝑧
goes to 0 by the definition of 𝛼1
49
Solving 𝑧
1
𝑧

=
1
 − 2𝑧
𝑧

𝐴2
𝑧2
+
𝐴2
𝑧2
1st order linear ODE


𝐴1 +2𝛼1 𝐴2
+
𝑧
𝐴1 +2𝛼1 𝐴2
=
𝑧
𝑧 + 𝐴1 + 2𝛼1 𝐴2 𝑧 = −𝐴2
𝑧 𝑡 =
50
−𝐴2
𝐴1 +2𝛼1 𝐴2
+ 𝐶exp − 𝐴1 + 2𝛼1 𝐴2 𝑡
Solving for 𝛼
1

𝛼 = 𝛼1 +

boundary condition:


exp −
𝐴1 +2𝛼1 𝐴2 𝑡
𝛼 𝑇 =0
𝛼1 +
1
−𝐴2
+𝐶
𝐴1 +2𝛼1 𝐴2

𝐶exp −

𝐶 = exp
51
−𝐴2
+𝐶
𝐴1 +2𝛼1 𝐴2
exp −
𝐴1 +2𝛼1 𝐴2 𝑇
=0
1
𝐴2
𝐴1 + 2𝛼1 𝐴2 𝑇 = − +
𝛼1
𝐴1 +2𝛼1 𝐴2
𝐴2
1
𝐴1 + 2𝛼1 𝐴2 𝑇
−
𝐴1 +2𝛼1 𝐴2
𝛼1
𝛼 Solution (1)

𝛼 = 𝛼1 +

= 𝛼1 +
1
−𝐴2
+𝐶
𝐴1 +2𝛼1 𝐴2
exp

= 𝛼1 +
= 𝛼1 +
52
𝐴1 +2𝛼1 𝐴2 𝑡
1
−𝐴2
+
𝐴1 +2𝛼1 𝐴2

exp −
𝐴1 +2𝛼1 𝐴2 𝑇
𝐴2
1
−
𝐴1 +2𝛼1 𝐴2 𝛼1
exp
− 𝐴1 +2𝛼1 𝐴2 𝑡
1
−𝐴2
+
𝐴1 +2𝛼1 𝐴2
exp
−𝛼1 𝐴2 +exp
𝐴1 +2𝛼1 𝐴2 𝑇−𝑡
𝐴2
1
−
𝐴1 +2𝛼1 𝐴2 𝛼1
𝛼1 𝐴1 +2𝛼1 𝐴2
𝐴1 +2𝛼1 𝐴2 𝑇−𝑡 𝛼1 𝐴2 −𝐴1 −2𝛼1 𝐴2
𝛼 Solution (2)

𝛼 = 𝛼1 +

= 𝛼1 1 −

= 𝛼1 1 −
53
−𝛼1 𝐴2 +exp
𝛼1 𝐴1 +2𝛼1 𝐴2
𝐴1 +2𝛼1 𝐴2 𝑇−𝑡
𝐴1 +2𝛼1 𝐴2
𝐴2 +exp 𝐴1 +2𝛼1 𝐴2 𝑇−𝑡
𝐴
1 +1
1+ 𝛼 𝐴
1 2
−𝐴1 −𝛼1 𝐴2
𝐴1
+𝐴2
𝛼1
𝐴1
+2𝛼1
𝐴2
exp
𝐴1 +2𝛼1 𝐴2 𝑇−𝑡
𝛼 Solution (3)

𝛼 𝑡 =
54
𝑘
2𝜂 2
1− 1−𝛾 +
2 1−𝛾
2
1− 1−𝛾
1+ 1−
exp
2𝑘
1−𝛾
𝑇−𝑡
Solving 𝛽
𝜆 2
𝑘 𝜃
𝜂2

𝛽𝑡 = 𝑘 − 2𝜂2 𝛼 − 2𝜆𝑘 + 4𝜆𝜂2 𝛼 𝛽 + 4𝜆𝑘𝜃𝛼 − 2

Let 𝜏 = 𝑇 − 𝑡

𝛽 𝜏 =𝛽 𝑇−𝑡

𝛽𝜏 𝜏 = −𝛽𝑡 𝜏

−𝛽𝜏 𝜏 = 𝛽𝑡 𝜏 = 𝑘 − 2𝜂2 𝛼 𝜏 − 2𝜆𝑘 + 4𝜆𝜂2 𝛼 𝜏 𝛽 𝜏 +
4𝜆𝑘𝜃𝛼 𝜏 − 2

𝜆 2
𝑘 𝜃
𝜂2
− 2𝛼 𝜏 𝑘𝜃
𝛽𝜏 𝜏 = −𝑘 + 2𝜂2 𝛼 + 2𝜆𝑘 − 4𝜆𝜂2 𝛼 𝛽 + −4𝜆𝑘𝜃𝛼 + 2
55
− 2𝛼𝑘𝜃
𝜆 2
𝑘 𝜃
𝜂2
+
First Order Non-Constant Coefficients

𝛽𝜏 = 𝐵1 𝛽 + 𝐵2

𝐵1 𝜏 = 2𝜆 − 1 𝑘 + 2𝜂 2 𝛼 1 − 2𝜆

𝐵2 𝜏 = 2𝛼𝑘𝜃 1 − 2𝜆 + 2
56
𝜆 2
𝑘 𝜃
𝜂2
Integrating Factor (1)


𝛽𝜏 − 𝐵1 𝛽 = 𝐵2
We try to find an integrating factor 𝜇 = 𝜇 𝜏 s.t.


𝑑𝜏
𝜇𝛽 = 𝜇
𝑑𝛽
𝑑𝜏
+𝛽
𝑑𝜇
𝑑𝜏
= 𝜇𝐵2
Divide LHS and RHS by 𝜇𝛽.


𝑑
1 𝑑𝛽
𝛽 𝑑𝜏
+
1 𝑑𝜇
𝜇 𝑑𝜏
=
𝐵2
𝛽
By comparison,

57
−𝐵1 =
1 𝑑𝜇
𝜇 𝑑𝜏
Integrating Factor (2)

𝜇 = exp

𝜇𝛽 =


𝑑𝜇
𝜇
−𝐵1 𝑑𝜏 =

𝛽=
𝛽=
58
= log 𝜇 + 𝐶
−𝐵1 𝑑𝜏
𝜇𝐵2 𝑑𝜏 + 𝐶
𝜇𝐵2 𝑑𝜏+𝐶
𝜇
exp
−𝐵1 𝑑𝑢 𝐵2 𝑑𝜏+𝐶
exp
−𝐵1 𝑑𝜏
𝛽 Solution
𝜏
exp
0
𝜏
−𝐵1 𝑢
0
𝜏
exp 0 −𝐵1

𝛽=

𝛽 𝜏 =
𝜏
exp 0 𝐵1 𝑢 𝑑𝑢
59
𝑑𝑢 𝐵2 𝑠 𝑑𝑠
𝑢 𝑑𝑢
𝜏
0
exp
+𝐶
𝑠
−𝐵1
0
𝑢 𝑑𝑢 𝐵2 𝑠 𝑑𝑠 + 𝐶
𝐵1 , 𝐵2
𝜏
𝐵
0 1

 𝐼2
60
=
𝜏
𝑠 𝑑𝑠 = 0 2𝜆 − 1 𝑘 + 2𝜂2 1 −
𝜏
𝑠
exp 0 −𝐵1 𝑢 𝑑𝑢 𝐵2 𝑠 𝑑𝑠
0
2𝜆 𝛼 𝑠 𝑑𝑠
𝛽 Solution

𝛽 𝑡 =
61
𝑘𝜃
𝜂2
1+ 1−𝛾
exp
2𝑘
1−𝛾
2
1− 1−𝛾
1+ 1−
𝑇−𝑡
exp
−1
2𝑘
1−𝛾
𝑇−𝑡
Solving 𝑔

𝑔𝑡 = 𝑔 −𝑘𝛽𝜃 −
62
1 2 2
𝜂 𝛽
2
− 𝜂2 𝛼
𝜆 2 2
+ 2𝑘 𝜃
𝜂
+ 𝜆𝜂 2 𝛽2 +
Computing the Optimal Position



ℎ 𝑡
=−
=−

=

=

=

=
63
∗
=−
𝐺𝑉 𝑘 𝜃−𝑥𝑡 +𝐺𝑉𝑥 𝜂2
𝐺𝑉𝑉 𝜂2
𝛾𝑉 𝛾−1 𝑓 𝑘 𝜃−𝑥𝑡 +𝛾𝑉 𝛾−1 𝑓𝑥 𝜂2
𝛾 𝛾−1 𝑉 𝛾−2 𝑓𝜂2
𝑉𝑓 𝑘 𝜃−𝑥𝑡 +𝑉𝑓𝑥 𝜂2
𝛾−1 𝑓𝜂2
𝑓 𝑘 𝜃−𝑥𝑡 +𝑓𝑥 𝜂2
𝑉
−
𝛾−1 𝜂2
𝑓
𝑉
𝑓𝑥 2
−
𝑘 𝜃 − 𝑥𝑡 + 𝜂
𝛾−1 𝜂2
𝑓
𝑉
2 𝛽 + 2𝛼𝑥
𝑘
𝜃
−
𝑥
+
𝜂
𝑡
1−𝛾 𝜂2
𝑉
𝑘
− 2 𝑥𝑡 − 𝜃 + 2𝛼𝑥 + 𝛽
1−𝛾
𝜂
The Optimal Position

ℎ 𝑡

ℎ 𝑡
64
𝑉𝑡
=
1−𝛾
𝑘
∗
~− 2
𝜂
∗
𝑘
− 2
𝜂
𝑥𝑡 − 𝜃
𝑥𝑡 − 𝜃 + 2𝛼 𝑡 𝑥𝑡 + 𝛽 𝑡
P&L for Simulated Data

The portfolio increases from $1000 to $4625 in one year.
65
Parameter Estimation



Can be done using Maximum Likelihood.
Evaluation of parameter sensitivity can be done by
Monte Carlo simulation.
In real trading, it is better to be conservative about the
parameters.



Better underestimate the mean-reverting speed
Better overestimate the noise
To account for parameter regime changes, we can use:


66
a hidden Markov chain model
moving calibration window