Statistical learning and optimal control:

A framework for biological learning and motor control

Lecture 3: Introduction to optimal control

Reza Shadmehr

Johns Hopkins School of Medicine

*w *w *w

y y y

x x xEstimation: Given observations x and y, estimate the hidden state w. Your estimates have no bearing on your observations.Example: classical conditioning. The actions of the learner has no effect on the stimuli.

Control: figure out the u that you need to give so that your observations y behave as you want them to.Example: operant conditioning, where the learner’s actions affect whether it gets rewarded or not.

r r

(0)x

(1)y

(0)u

(1)x ( )px

( 1)pu

( )py

The linear quadratic tracking problem

( )

( ) ( )

( 1) ( ) ( )

k

k k

k k k

B

A C

r

y x

x x u

We are trying to track a reference trajectory r(k)

We observe y(k), which is related to x(k)

We generate command u(k), which causes a change in x(k)

We wish to find the control sequence u(0), u(1), …, u(p-1) such that we minimize the cost function,

Given the constraint that:

1

( 1) ( 1) ( 1) ( 1) ( 1) ( ) ( ) ( )

0

p Tk k k k k k T k k

k

J T L

y r y r u u

( 1) ( ) ( )

( 1) ( 1)

k k k

k k

A C

B

x x u

y x

qx1

qx1

mx1

nx1

control costtracking cost

(1) (0) (0)

(2) (1) (1) 2 (0) (0) (1)

(3) (2) (2) 3 (0) 2 (0) (1) (2)

1( ) (0) 1 ( )

0

kk k k j j

j

A C

A C A AC C

A C A A C AC C

A A C

x x u

x x u x u u

x x u x u u u

x x u

(1) (0)

(2) (2)

( ) ( 1)

h h

p p

x u

x ux u

x u

( 1) ( ) ( )k k kA C x x uSuppose we have a linear dynamical system:We have the history of inputs u(k), where k=0…p-1. We want to write the history of state x(k).

mx1 nx1mxnmxm

22 2

3 2 3 2

0 0 0 0 0 00 00 0 0 0

0 0 0 0 0

0

0 0p

I CA CA I AC CA C

E F FA A I A C AC C

A A A I A C A C AC CCA

(p.m)x(p.n)

(p.m)x(p.m)

(0)h hE F x x u

(p.n)x1(p.m)x1

(p.m)xm

(0)0

h h

T Th h h h h h

T Th h h h h h

h h

G

J T L

J G T G L

E F

y x

y r y r u u

x r x r u u

x x u

control costtracking cost

(1) (1)(1) (1)

(2) (2)

( ) ( )( ) ( )

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0

h hp p

p p

T L

T LT L

T L

r y

r y

r y

0 0 0

0 0 0

0 0 0

0 0

B

BG

B

constraint

Total cost

Constraint minimization with Lagrange multipliers: Example 1

Suppose we want to find the point (xs,ys) along the line y=mx+b that is closest to the point (xo,yo).

x

y

1/ 22 200 0

0

2 2

0 0

,

distance

,

0g x y

xxx x y y

yy

J x y x x y y

y mx b

cost

constraint

y mx b

x

y0y mx b

The points along each line are of equal cost

0 0,x y

We want to find point (xs,ys) that belongs to the line, and among the points that belong to the line, gives us the smallest cost.

,s sx y

x

y , 0g x y

1

dgmdx

dg

dy

x

y

,J x y c

0

0

2

2

dJx xdx

dJ y ydy

x

y

The point where the line meets the cost contour is where the vector normal to the constraint and the vector normal to the cost are in the same direction.

Vector normal to the constraint

Vector normal to the cost

The point that we are looking for satisfies the condition:

dJ dg

dx dxdJ dg

dy dy

Lagrange multiplier

0

0

2

2

dJx xdx

dJ y ydy

1

dgmdx

dg

dy

0

0

2

2 1

0

x x m

y y

y mx b

This is 2 equations with 3 unknowns.

Here is our 3rd equation.

120 0

0

1

2

s

s s

s

x m my x mb

y mx b

y y

Suppose the milkmaid wants to get to the cow so that she travels the shortest distance possible, given the constraint that she first washes her milk pan in the river. So we want to find the shortest route that includes a line from the milkmaid to the river edge, and a line from the river edge to the cow. Find the point P that minimizes the following:

x

y

milkmaid

cow

Example from: Steuard Jensen

,

, 0

m c

m c

x xx xf x y

y yy y

g x y

An ellipse can be defined as the set of points P for which the total distance from one focus to P and then to the other focus is constant.

If we keep points M and C as the foci of this ellipse, then as soon as we have an ellipse that touches the river edge, we have found the point P that is our solution.

Note that at point P, the normal vector to g(x) and the normal vector to f(x) are in the same direction.

cost

constraint

, ,

, ,

df x y dg x y

dx dxdf x y dg x y

dy dy

Constraint minimization with Lagrange multipliers: Example 2

Constraint minimization with Lagrange multipliers

A scalar constraint

In order to minimize the scalar function:

Subject to scalar constraint:

We form an augmented cost:

J x

0g x

aJ J g x x x

Note that when we find the x that satisfies the constraint g(x), g(x) will be zero and so we have not changed our cost function.

0adJ dJ dg

d d d

x x xTo minimize the augmented cost, we have:

So, to find the x that minimizes the cost subject to the constraint, we find the (x, lambda) that satisfies:

This should look familiar from last two examples.

0adJg

d x

0

0

a

a

dJ

ddJ

d

x

Constraint minimization with Lagrange multipliers

A “vector” constraint

In order to minimize the scalar function:

Subject to constraint:

We form an augmented cost with two Lagrange multipliers:

J x

1

2

0

0

g

g

xg x

x

1

2

TaJ J

x x λ g x

λ

To minimize the augmented cost, we have:

So, to find the x that minimizes the cost subject to the constraint, we find the (x, lambda1, lambda2) that satisfies:

We have as many multipliers as we have constraints.

1

2

0

0

0

a

a

a

dJ

ddJ

d

dJ

d

x

11

22

0

0

0

a

a

a

dJ dJ d

d d ddJ

gd

dJg

d

gλ

x x x

x

x

(0)

,

,

T Th h h h h h h h

h h h h

J G T G L

g E F

x u x r x r u u

x u x x u 0

(0),T T T

a h h h h h h h h h hJ G T G L E F x u x r x r u u λ x x u

Cost: a scalar

constraint: a vector

2 2

2

,

T Tah h

h

T Tah

h

ah h

dJG TG G T

d

dJL F

d

dJg

d

x r λ 0x

u λ 0u

x u 0λ

Eq. 1

Eq. 2

Solve for lambda in Eq. 1, and then plug it into Eq. 2.

(0)

(0)

(0)

1 (0)

2 2

2 2

T Th h h h

T T T T Th h h

T T T T T Th h h

T T T T T Th h

G T G G T GE GF

L F F G T GE GF

L F G TGF F G T GE

L F G TGF F G T GE

λ r x r x u

u λ r x u

u u r x

u r x

T TdA A A

d x x x

x

2

k

2

kb

x

m 0xu

0

2

2

kx

kx x

bx

u

Force in the bottom spring

Force in the viscous element

Force in the top spring

Force in the motor command

0

0

0

2 2

2

2

k kmx x bx u x x

kkx x bx u

xk x bx u

Let us construct a simple model of the eye’s dynamics and produce a saccade using optimal control

k b

mu

If we re-define x so that we measure it from xo/2, then the equivalent system is shown on right, where the equilibrium point of the spring is at x=0.

x

mx kx bx u

1 2

1 1

2 2

1

0 1 0 00

10

1 0c c

mx kx bx u

k bx x x u

m m m

x x x x

x xk b

x x um m m

A C

y

x x u

x

System dynamics in continuous form

Our observation

Goal: find the motor commands that move the mass (the eye) to a certain location by a certain time while minimizing a cost the depends on endpoint accuracy and motor commands.

First step: re-formulate the system dynamics from continuous to discrete time.

Second step: solve the optimum control problem.

Relating discrete and continuous representation of a linear system(approximate solution)

1

c cc

c c

k k kd d

d k k kd d

t A t C tG

t B t D t

A CG

B D

x x u

y x u

x x u

y x u

Simple (but approximate method) is to use Euler’s approximation:

c c

c c

d c

d c

d c

d c

t t tt

tt t t t t

A t C t t t

I A t t C t t

A I A t

C C t

B B

D D

x xx

x x x

x u x

x u

Continuous system

Discrete system

Solution of continuous LTI state equations (scalar condition)

x t ax tSuppose that our state is a scalar variable and the state update equation is of the form:

0 expx t x atThe solution will have the exponential form:

0 0

0

0

0

exp exp exp

exp exp

exp exp

exp 0 exp

exp 0 exp exp

exp 0 exp

x t ax t cu t

at x t at ax t at cu t

dat x t at cu t

dt

at x t at cu t dt

a x x at cu t dt

x a x a at cu t dt

x a x a t cu t dt

Suppose that our state is a scalar variable and the state update depends on an external input u(t):

Matrix exponential

t A tx xSuppose that our state is a vector variable:

0 expt Atx xWe can imagine that the solution will have a “matrix exponential” form:

For any square matrix A, the matrix exponential exp(A) is a square matrix function. We can compute it using Taylor series expansion.

In Matlab, exp(A) is computed as expm(A).

In Mathematica, use MatrixExp[A].

2 2

20 0 0

22

0

02 !

exp2 !

exp!

n n

nt t t

nn

nn

n

df d f t d f tf t f t

dt dt dt n

t tAt I At A A

n

tAt A

n

Some properties of the matrix exponential

1 2 1 2

exp(0 )

exp exp exp

exp exp exp

t I

A t t At At

dAt A At At A

dt

Using Taylor series expansion, one can show the following properties of the matrix exponential:

Other properties of the matrix exponential:

exp( )exp( )

exp exp exp only if 0

A A I

A B A B AB BA

Solution of continuous LTI state equations (vector condition)

0 0

0

0

0

exp exp exp

exp exp

exp exp

exp 0 exp

exp 0 exp exp

exp 0 exp

t A t C t

At t At A t At C t

dAt t At C t

dt

At t At C t dt

A At C t dt

A A At C t dt

A A t C t dt

x x u

x x u

x u

x u

x x u

x x u

x x u

Solution of discrete LTI state equations

1

1 0 0

2 1 1 0 0 12

10 1

0

k k k

kk mk k m

m

A C

A C

A C A AC C

A A C

x x u

x x u

x x u x u u

x x u

Relating discrete and continuous representation of a linear system

0

1 1

11

0

0

1

exp 0 exp exp

1

exp 1 0 exp 1 exp

exp 1 0 exp 1 exp

exp 1 exp

exp exp 0 exp exp

t

c c c

k k

kkc c c c

k

c c c c

k

c c ck

c c c c c

t A t A t A C d

t k

A k A k A C d

A k A k A C d

A k A C d

A A k A k A C d

x x u

x x

x x u

x u

u

x u

0

1

1

11

1

exp 1 exp

exp exp 1

exp exp 1

exp exp

k

k

c c ck

kkc c ck

kk k

c c c ck

k kc c c c

A k A C d

A A k C d

A A A k C

A A I A C

u

x u

x u

x u

Assume that u(t) is constant between the two sampling intervals.

Discrete and continuous representation of a linear system

(noise free scenario)

1

c cc

c c

k k kd d

d k k kd d

t A t C tG

t B t D t

A CG

B D

x x u

y x u

x x u

y x u

Continuous system

Discrete system

1

sampling interval

exp

exp

d c

d c c c

d c

d c

A A

C A A I C

B B

D D

11 exp exp 1

c c

T

kk kc c ck

t A t C t

E t Q t

A A k C d

x x u

u u

x x u

We note that for small , the term inside the exponential is near zero over the range k to (k+1). Therefore, we can approximate the matrix exponential with an identity matrix.

11

( )

1 1( ) ( )

1

1

expkk k

c ck

k kd

Tk kk k Tc ck k

k T Tc ck

k T Tc ck

Tc c

A C d

A

E E C d C d

E C C d

C E C d

C QC

x x u

x w

w w u u

u u

u u

State noise in continuous domain

Equivalent state noise in discrete domain

State noise in the continuous system

y c

T

t B t t

E t R t

x v

v v

Suppose that we imagine that we average the sample y(t) over the discrete interval to get our discrete sample:

Measurement noise in continuous domain

Equivalent state noise in discrete domain

( 1) ( 1)( )

( 1)( )

1 1y

1

k kk

c

k k

kk

c

k

t dt B t t dt

B t dt

y x v

x v

Noise in discrete domain is:

( 1)( )

( 1)( ) ( )

2

2

1

1

1

kk

k

kk k T

k

t dt

E t dtd

R

R

v v

v v v v

Measurement noise in the continuous system

Discrete and continuous representation of a linear system with noise

1

Tc c u u u

Tc y y y

k k k k k k Td d u u u

k k k k k Td y y y

t A t C t t E t Q t

t B t t E t R t

A C E Q

RB E

x x u ε ε ε

y x ε ε ε

x x u

y x

Continuous system

Equivalent discrete system

1

sampling interval

exp

exp

d c

d c c c

d c

A A

C A A I C

B B

1 2

1 1

2 2

( 1) ( )

1

3 . / 0.65 . . / 0.004 . /

0 1 0 00

10

1 0

0.001sec

exp

exp

c c

k k

c

c c c

k N m rad b N m s rad m kg m rad

x x x x

x xk b

x x um m m

A C

y

A C

A A

C A A I C

x x u

x

x x u

(0)0

h h

T Th h h h h h

T Th h h h h h

h h

G

J T L

J G T G L

E F

y x

y r y r u u

x r x r u u

x x u

Continuous time model of the eye

Discrete time model of the eye

Optimal control problem

0 0.02 0.04 0.06 0.08

-2

0

2

4

0 0.02 0.04 0.06 0.080

2

4

6

0 0.02 0.04 0.06 0.080

0.025

0.05

0.075

0.1

0.125

0.15

0.175

0 0.02 0.04 0.06 0.080

500000

1106

1.5106

2106

0 0.02 0.04 0.06 0.080

0.025

0.05

0.075

0.1

0.125

0.15

0.175

(1)(1)

(2)

( )( )

0 0 0 1 0 0 0

0 1 0 00 0 0

0 0 00 0 00 0 10 0

hp

p

T

TT L

T

r

r

r

Make a 30 deg (~0.5 rad) saccade in 0.5 seconds

Time (sec)

r T

u x x

0 0.02 0.04 0.06 0.080

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 0.02 0.04 0.06 0.08

0

2

4

6

8

10

0 0.02 0.04 0.06 0.08

-2

0

2

4

6

8

Po

siti

on

(ra

d)

Vel

oci

ty (

rad

/s)

Mo

tor

com

man

d (

N.m

)

5 deg

10 deg

15 deg

20 deg

Eye muscle activity for a 10 deg saccade.

Time (sec)

1 1 2

1

2

1 2

1 . / 0.15 . . / 0.1 . / 0.15 . /

: position of the right hand

: position of the left hand

: cursor position (the observable variable)

k N m rad b N m s rad m kg m rad m kg m rad

x

x

y

y x x

Resolving redundancies

Suppose that we have a cursor that its position depends on the sum of positions of left and right joysticks. Suppose that the left joystick is heavier than the right joystick. We want to move the cursor to some location. How much should we move each joystick?

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5

-1

0

1

2y

1x

2x

1u

2ur

( 1) ( )

( 1) ( 1)

(0)

(0)

1 (0)

k k

k k

h h

h h

T T Ta h h h h h h h h

T T T T T Th h

A C

B

G

E F

J T L E F

L F G TGF F G T GE

x x u

y x

y x

x x u

y r y r u u λ x x u

u r x

Summary: Optimal control of a linear system with quadratic cost

Issues with the control policy:

• What if the system gets perturbed during the control policy? With the current approach, there is no compensation for the perturbation.

• In reality, both the state update equation and the measurement equation are subject to noise. How do we take that into account?

• To resolve this, we need a way to figure out what command to produce, given that we find ourselves at some state x at some time k. Once we figure this out, we will consider the situation where we cannot measure x directly, but have noise to deal with. Our best estimate will be through the Kalman filter. This will link estimation with control.

1( ) ( ) ( ) ( 1) ( 1) ( 1)

0

pk T k k k T k k

k

J L T

u u y y

(0)

(0) (1) ( 1)

(1) (2) ( ) (1) (2) ( )

, , ,

, , , , , ,

p

p pB B B

x

u u u

y y y x x x

Starting at state

Sequence of actions

Observations

Cost to minimize

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

( ) ( 1) ( 1) ( ) ( 1) ( 1)

( 1) ( ) ( 1) ( 1) ( ) ( 1)

( 1) ( ) ( 1)

( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( 1) (

2

p p T p p p T T p p

p T p

Tp p p p p p

p T T p p p T T p p

p T T p p

p p T p p p T p p

J T B T B

W B T B

J A C W A C

A W A C W C

C W A

J T L J

y y x x

x u x u

x x u u

u x

y y u u

)

( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( )

( 1)( 1) ( 1) ( ) ( 1) ( ) ( 1)

( 1)

1( 1) ( 1) ( ) ( ) ( 1)

1( 1) ( 1) ( ) ( )

( 1) ( 1) ( 1)

2 2 2 0

p

p T T p p p T p p p

pp p T p p T p p

p

p p T p T p p

p p T p T p

p p p

B T B L J

dJL C W C C W A

d

L C W C C W A

G L C W C C W A

G

x x u u

u u xu

u x

u x

Cost at the last time point

Cost-to-go at the next to the last time point

Note that at the last time step, cost is a quadratic function of state

( ) ( 1) ( ) ( 1) ( 1) ( ) ( 1) ( 1) ( ) ( 1)

( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( )

( 1) ( 1) ( ) ( 1)

( 1) ( 1) ( ) ( 1) ( 1) ( ) (

2

2

p p T T p p p T T p p p T p p

p p T T p p p T p p p

p T T p T p p

p T p T p p p T T p p

J A W A C W C W BA

J B T B L J

B T B A W A

L C W C C W A

x x u u u x

x x u u

x x

u u u x

1)

( 1) ( 1) ( ) ( 1)

( 1) ( 1) ( 1) ( ) ( 1) ( 1) ( 1) ( ) ( 1)2

p T T p T p p

p T p T p T p p p p T T p p

B T B A W A

G L C W C G C W A

x x

x x u x

We will now show that if we choose the optimal u at step p-1, then cost to go is once again a quadratic function of state x.

Can be simplified to:( ) ( 1)T p pA W CG

( 1) ( 1) ( 1) ( ) ( ) ( 1) ( 1)

( 1) ( 1) ( ) ( 1) ( 1)

( 1) ( 1) ( 1)

p p T T p T p T p p p

p T T p T p p p

p T p p

J B T B A W A A W CG

B T B A W A CG

W

x x

x x

x x

Can be simplified to: ( 1) ( ) ( 1)2 p T T p pA W CG x

We just showed that for the last time step, the cost to go is a quadratic function of x:

( ) ( ) ( ) ( )p p T p pJ Wx x

The optimal u to at time point p-1 minimizes cost to go J(p-1):

1( 1) ( 1) ( ) ( )

( 1) ( 1) ( 1)

p p T p T p

p p p

G L C W C C W A

G

u x

If at time point p-1 we indeed carry out this optimal policy u, then the cost to go at time p-1 also becomes a linear function of x:

( 1) ( 1) ( 1) ( 1)

( 1) ( 1) ( ) ( 1)

p p T p p

p T p T p p

J W

W B T B A W A CG

x x

If we now repeat the process and find the optimal u for time point p-2, it will be:

1( 2) ( 2) ( 1) ( 1)

( 2) ( 2) ( 2)

p p T p T p

p p p

G L C W C C W A

G

u x

And if we apply the optimal u at time points p-2 and p-1, then the cost to go at time point p-2 will be a quadratic function of x:

( 2) ( 2) ( 2) ( 2)

( 2) ( 2) ( 1) ( 2)

p p T p p

p T p T p p

J W

W B T B A W A CG

x x

So in general, if for time points t+1, …, p we calculated the optimal policy for u, then the above gives us a recipe to compute the optima policy for time point t.

( 1) ( ) ( )

( 1) ( 1)

1(0) ( 1) ( 1) ( 1) ( ) ( ) ( )

0

k k k

k k

pk T k k k T k k

k

A C

B

J T L

x x u

y x

y y u u

Summary of the linear quadratic tracking problem

( ) ( ) ( ) ( )

( ) ( )

1( 1) ( 1) ( ) ( )

( 1) ( 1) ( 1)

( 1) ( 1) ( 1) ( 1)

( 1) ( 1) ( ) ( 1)

p p T p p

p T p

p p T p T p

p p p

p p T p p

p T p T p p

J W

W B T B

G L C W C C W A

G

J W

W B T B A W A CG

x x

u x

x x

(0)x

(1)y

(0)u

(1)x ( )px

( 1)pu

( )py

Cost to go

1(0) (0) (1) (1)

(0) (0) (0)

(0) (0) (0) (0)

(0) (0) (1) (0)

T T

T

T T

G L C W C C W A

G

J W

W B T B A W A CG

u x

x x

The procedure is to compute the matrices W and G from the last time point to the first time point.

1 2

1 1

2 2

( 1) ( )

1

3 . / 0.45 . . / 0.3 . /

0 1 0 00

10

1 0

0.01sec

exp

exp

c c

k k

c

c c c

k N m rad b N m s rad m kg m rad

x x x x

x xk b

x x um m m

A C

y

A C

A A

C A A I C

x x u

x

x x u

Continuous time model of the elbow

Discrete time model of the elbow

Modeling of an elbow movement

Goal: Reach a target at 30 deg in 300 ms time and hold it there for 100 ms.

Unperturbed movement Arm held at start for 200ms Force pulse to the arm for 50ms

0 0.1 0.2 0.3 0.4sec

0

0.1

0.2

0.3

0.4

0.5

noitisoP

0.05 0.1 0.15 0.2 0.25 0.3 0.35sec

-75

-50

-25

0

25

50

75

rotoMdnammoc

0 0.1 0.2 0.3 0.4sec

0

0.1

0.2

0.3

0.4

0.5

noitisoP

0.05 0.1 0.15 0.2 0.25 0.3 0.35sec

-5

0

5

10

15

rotoMdnammoc

0 0.1 0.2 0.3 0.4sec

0

0.1

0.2

0.3

0.4

0.5

noitisoP0.05 0.1 0.15 0.2 0.25 0.3 0.35

sec

-30

-20

-10

0

10

rotoMdnammoc

0 0.1 0.2 0.3 0.4sec

1

1.25

1.5

1.75

2

2.25

2.5

L

L

0 0.1 0.2 0.3 0.4sec

0

500000

1 106

1.5 106

2 106

soPtsoc T

0 0.1 0.2 0.3 0.4sec

0

5000

10000

15000

20000

leVtsoc T

0 0.1 0.2 0.3 0.4 0.5 0.6sec

0

500000

1106

1.5106

2106

soPtsoc

Movement with a via point: we set the cost to be high at the time when we are supposed to be at the via points.

0 0.1 0.2 0.3 0.4 0.5 0.6sec

-10

0

10

20

30

rotoMdnammoc

0 0.1 0.2 0.3 0.4 0.5 0.6sec

0

0.2

0.4

0.6

0.8

noitisoP

0 0.1 0.2 0.3 0.4 0.5 0.6sec

0

200

400

600

800

soPniaG

T

G

( 1) ( ) ( )

( 1) ( 1)

1( ) ( ) ( ) ( 1) ( 1) ( 1)

0

0,

0,

k k kx x

k ky y

pk T k k k T k k

k

A C N Q

B N R

J L T

x x u ε ε

y x ε ε

u u y y

Stochastic optimal control

Biological processes have noise. For example, neurons fire stochastically in response to a constant input, and muscles produce a stochastic force in response to constant stimulation. Here we will see how to solve the optimal control problem with additive Gaussian noise.

Cost to minimize

Because there is noise, we are no longer able to observe x directly. Rather, the best we can do is to estimate it. As we saw before, for a linear system with additive noise the best estimate of state is through the Kalman filter. So our goal is to determine the best command u for the current estimate of x so that we can minimize the global cost function.

Approach: as before, at the last time point p the cost is a quadratic function of x. We will find the optimal motor command for time point p-1 so that it minimizes the expected cost to go. If we perform the optimal motor command at p-1, then we will see that the cost to go at p-1 is again a quadratic function of x.

Preliminaries: Expected value of a squared random variable. In the following example, we assume that x is the random variable.

2

22

2

2

var

var

var

var

var

T

TT

T

T

T

T

v x

x E x E x

E v E x

x E x

v

E E E

E v E tr

tr E

tr tr E E

tr E E

x x

x xx x x

xx

xx

x x x

x x x

1 2

2 2 21 2

21 1 2 1

22 1 2 2

21 2

2

1

Tn

Tn

n

T n

n n n

nT

ii

T T

r r r

r r r

r r r r r

r r r r r

r r r r r

tr r

tr

r

r r

rr

rr

r r rr

Scalar x

Vector x

var

T

T

v A

E v tr A E AE

x x

x x x

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

var var

p p T p p

p T T p p T py y

p T p

p p T p p T py y

p p p T p p py

p p T p p

J T

B T B T

W B T B

E J E W E T

tr W E W E tr T

tr W Q E W E

y y

x x ε ε

x x ε ε

x x x ε

x x

( )

( 1) ( 1) ( ) ( 1) ( 1) ( ) ( )

( 1) ( ) ( 1) ( 1) ( ) ( 1)

( 1) ( ) ( 1) ( ) ( )2

p

Tp p p p p p p

p T T p p p T T p p

p T T p p T p p

tr T Q

A C W A C tr W Q tr T Q

A W A C W C

A W C tr W Q tr T Q

x u x u

x x u u

x u

Cost at the last time point

( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( )

( 1) ( 1) ( ) ( 1) ( 1) ( 1) ( ) ( 1)

( 1) ( ) ( 1) ( ) ( ) ( 1)

( 1)( 1)

( 1)

2

2

p p T p p p T p p p

p T T p T p p p T p T p p

p T T p p p p T py y

pp T

p

J T L E J

B T B A W A L C W C

A W C tr W Q tr T Q T

dJL C W

d

y y u u

x x u u

x u ε ε

u

( ) ( 1) ( ) ( 1)

1( 1) ( 1) ( ) ( ) ( 1)

1( 1) ( 1) ( ) ( )

( 1) ( 1) ( 1)

2 0p p T p p

p p T p T p p

p p T p T p

p p p

C C W A

L C W C C W A

G L C W C C W A

G

u x

u x

u x

Cost-to-go at the next to the last time point

So we see that if our system has additive state or measurement noises, the optimal motor command remains the same as if the system had no noises at all. When we use the optimal policy at time point p-1, we see that, as before, the cost-to-go at p-1 is a quadratic function of x. The matrix W at p-1 remains the same as when the system had no noise.

The problem is that we do not have x. The best that we can do is to estimate x via the Kalman filter. We do this in the next slide.

1 21 1

1 2 2 2 2

2 2 2 3 2 32 2

1 2 2 3 2 32 2 2

1 21

1 2 2 2 2 2

1 1 1

ˆ

ˆ ˆ

ˆ ˆ ˆ

ˆ ˆ ˆ

ˆ ˆ

ˆ ˆ ˆ

ˆ

p pp p

p p p p p

p p p p p pp p

p p p p p pp p p

p pp

p p p p p p

p p p

G

A C

K B

A C AK B

A C AK B

G

u x

x x u

x x y x

x x u y x

x x

x x u y x

u x

On trial p-1, our best estimate of x is the prior.

We compute the prior for the current trial from the posterior of the last trial.

The posterior estimate.

Our short-hand way to note the prior estimate of x on trial p-1.

Although the noises in the system do not affect the gain G, the estimate of x is of course affected by the noises because the Kalman gain is influenced by them.

Kalman gain

( 1) ( ) ( )

( 1) ( 1)

1(0) ( ) ( ) ( ) ( 1) ( 1) ( 1)

0

0,

0,

k k kx x

k ky y

pk T k k k T k k

k

A C N Q

B N R

J L T

x x u ε ε

y x ε ε

u u y y

Summary of stochastic optimal control for a linear system with additive Gaussian noise and quadratic cost

( ) ( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( )

1( 1) ( 1) ( ) ( )

( 1) ( 1) ( 1)

( 1) ( 1) ( 1) ( 1) ( 1)

( 1) ( 1) ( ) ( 1)

( 1) ( )

ˆ

p p T p p p

p T p

p T p py y

p p T p T p

p p p

p p T p p p

p T p T p p

p p Ty

J W w

W B T B

w T tr T Q

G L C W C C W A

G

J W w

W B T B A W A CG

w tr W Q T

x x

ε ε

u x

x x

ε ( 1) ( )p py w ε

Cost to go at the start

Cost to go at the end

( ) ( 1) ( )

( ) ( ) ( )

1 1( ) ( )

0,

0,

ˆ ˆ ˆ

ˆvar

ˆ ˆ

ˆ ˆ

ˆ ˆ

n n nx x

n n ny y

n n n n n nn n

n n n n

n n T

T

T

A N Q

H N R

y H

P

tr P tr E

E tr

E

x x ε ε

y x ε ε

x x k x

x

x x x x

x x x x

x x x x

11 1( )

1( )

n n n nn T T

n n n nn

P H HP H R

P I H P

k

k

The duality of the Kalman filter and optimal control

In the estimation problem, we have a model of how we think the hidden states x are related to observations y. Given an observation y, we have a rule with which we can change our estimates.

Our objective is to minimize the trace of the variance of our estimate xhat. This variance is P. This trace is our scalar cost function, which is quadratic in terms of xhat. We minimize it by finding the optimal gain k.

If we use this optimal k, then we can compute the variance in the next time step. Our cost (i.e., variance) of course still remains quadratic in terms of xhat.

1 1( )

11 1 1 1

n n n n n nT n T

n n n n n n n nT T T

P AP A Q A I H P A Q

P A I P H HP H R H P A Q

k

( 1) ( ) ( )

( 1) ( 1)

1(0) ( ) ( ) ( ) ( 1) ( 1) ( 1)

0

( ) ( ) ( ) ( ) ( )

( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( 1)

0,

0,

k k kx x

k ky y

pk T k k k T k k

k

p p T p p p

p p T p p p T p p

A C N Q

B N R

J L T

J W w

J T L E J

x x u ε ε

y x ε ε

u u y y

x x

y y u u

( )

1 1 1

( 1) ( 1) ( 1) ( 1) ( 1)

ˆ

p

p p p

p p T p p p

G

J W w

u x

x x

The duality of the Kalman filter and optimal control, continued.

In the control problem, we have a model of how we think the hidden states x are related to commands u and observations y.

Our objective is to find the u that minimizes a scalar cost. To find this u, we run time backwards!

We start at the end time point and find the optimal u that minimizes the cost to go. When we find this u, we then move to the next time point and so on.

The cost to go is a quadratic function of hidden states. This is very similar to the Kalman filter, where the cost was a quadratic function of the hidden states as well.

1( 1) ( ) ( 1) ( ) ( ) ( 1)k T k k T k T k T kW A I W C L C W C C W A B T B

State noiseMeasurement noiseState uncertainty

11 1( ) n n n nn T TP H R HP H

k

1( ) ( ) ( 1) ( 1) ( )ˆk k T k T k kL C W C C W A u x

11 1 1 1n n n n n n n nT T TP A I P H R HP H H P A Q

So W is like an estimate of state uncertainty matrix, BTB is like state update noise Q, and L is like measurement noise R.

In optimal control, the motor commands are generated by applying a gain to the state. This gain is like the Kalman gain.

Duality of optimal control and Kalman filter, continued.

Motor cost

Kalman Filter

Optimal control

Weighting of state Tracking cost

A B

Noise characteristics of biological systems are not additive GaussianNoise in the motor output grows with the size of the motor command

The standard deviation of noise grows with mean force in an isometric task. Participants produced a given force with their thumb flexors. In one condition (labeled “voluntary”), the participants generated the force, whereas in another condition (labeled “NMES”) the experimenters stimulated their muscles artificially to produce force. To guide force production, the participants viewed a cursor that displayed thumb force, but the experimenters analyzed the data during a 4-s period in which this feedback had disappeared. A. Force produced by a typical participant. The period without visual feedback is marked by the horizontal bar in the 1st and 3rd columns (top right) and is expanded in the 2nd and 4th columns. B. When participants generated force, noise (measured as the standard deviation) increased linearly with force magnitude. Abbreviations: NMES, neuromuscular electrical stimulation; MVC, maximum voluntary contraction. From Jones et al. (2002) J Neurophysiol 88:1533.

Electrical stimulation of the muscle

Voluntary contraction of the muscle

Representing signal dependent noise

,N I 0 Vector of zero mean, variance 1 random variables

( 1) ( ) ( ) ( ) noise with standard deviation that linearly grows with

( 1) ( 1) ( ) noise with standard deviation that linearly grows with

k k k kx

k k ky

A B

H

x x u ε u

y x ε x

Zero mean Gaussian noise signal dependent motor noise

Zero mean Gaussian noise signal dependent sensory noise

( ) ( ) ( )1 11 1 1

( ) ( ) ( )( ) ( )2 22 2 2

1

1 2 2

( )( )

( )

0 0 0 0

motor noise 0 0 0 0

0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0

motor noise =

var motor noise va

k k k

k k kk k

kki i

i

ki

c u c u

c u c u

c

C C c

C

C

u

u

( )r k T T T Ti i ii

i i

C C C u uu

( )( 1) ( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( )

( 1) ( ) ( ) ( ) ( ) ( ) ( )

ˆ

( ) ( ) ( ) ( ) ( )

ˆ ˆ ˆ

0,1 0,1

, , ,

kk k k k kx i ii

kk k k ky i ii

k k k k k k k

i i

x x y y x

k T k k T k k

A B C

H D

A AK H B

N N

N Q N Q N Q

L T

x x u ε u

y x ε x

x x y x u

ε 0 ε 0 0

u u x x

Cost per step:

Control problem with signal dependent noise (Todorov 2005)

To find the motor commands that minimize the total cost, we start at the last time step p and work backwards. At time step p, the cost is a quadratic function of x. At time step p-1, we can find the optimal u that minimizes the cost to go. When we find this optimal u, the cost to go at p-1 will be a quadratic function of x plus a quadratic function of x-xhat. In general, by induction we can prove that as long as we apply the optimal u, the cost to go will have this quadratic form. This proof is due to E. Todorov, Neural Computation, 2005.

( ) ( ) ( ) ( )

( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( ) ( 1) ( 1) ( 1)

( ) ( ) ( ) ( ) ( ) ( ) ( 1) ( 1) ( 1)

( ) ( 1

ˆ, ,

ˆvar , ,

var

p p T p p

p p T p p T p p p p p p

Tp p p p p p p p p

p px i

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E J E T E tr T

Q C

x x

u u x x x x u

x x x x x u

x u

) ( 1)

( ) ( 1) ( 1) ( ) ( 1) ( 1)

( ) ( 1) ( ) ( 1)

( 1) (

(

1) ( 1) ( ) ( 1) ( )

( 1) ( )

) ( 1)

( ( 1 () ) 2

p T px

p T Tii

Tp p p p p p

p p T T p px i ii

p p T p T p p px

i ii

px

p T T p p p

C

E J A B T A B

tr T Q C T C

J T A

C C T

T A tr T Q

T B

C

CL B

u

x u x u

u u

x x

u u u

1) ( ) ( 1)

( 1)( ) ( 1) ( ) ( 1)

( 1)

1( 1) ( ) ( ) (

(

( ) 1

)

)

2 2 0

ˆ

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pT p p T p p

p

p T p Tp p p

px

x

B T A

dJL B T B B T A

d

L B T B B

C

T AC

x

u xu

u x

Cost at time step p (last time step)

Optimal u to minimize the cost-to-go at time step p-1

Cost-to-go at p-1

1( ) ( 1) ( 1) ( 1)

( ) ( ) ( )

( 1) ( 1) ( 1) ( ) ( 1) ( )

( 1) ( 1) ( ) ( 1)

( 1) ( 1) ( ) ( 1)

( 1) ( 1)

ˆ

ˆ ˆ

ˆ2

ˆ ˆ ˆ ˆ ˆ2

p T p p T p

p p p

p p T p T p p px

p T p T T p p

p T p T T p p

TT T T

p p T

G L B T B C B T A

G

J T A T A tr T Q

G B T A

G B T A

Z Z Z Z

J T

u x

x x

x x

x x

x x x x x x x x x x

x

( 1)

( 1)( 1)( 1)

( 1) ( ) ( ) ( 1) ( 1)

( 1) ( 1) ( ) ( 1) ( 1) ( 1) ( )

( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( 1) (

ˆ ˆ

px

pe pp

p T p p p p

W

Tp p T p p p p px

Ww

p p T p p p T p px e

A T A T BG

A T BG tr T Q

J W W w

e

x

x x x x

x x e e

1)p

J(p-1) is the cost-to-go at time step p-1, assuming that the optimal u is produced at p-1.Note that unlike the cost at time step p, this cost-to-go is quadratic in terms of x and the error in estimation of x. So now we need to show that if we continue to produce the optimal u at each time step, the cost-to-go remains in this form for all time steps.

( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( 1) ( 1)k k T k k k T k k kx eJ W W w x x e e

( ) ( ) ( ) ( ) ( ) ( ) ( 1) ( ) ( ) ( )ˆ, ,k k T k k T k k k k k kJ L T E J u u x x x x u

Conjecture: If at some time point k+1 the cost-to-go under an optimal control policy is quadratic in x and e, and provided that we produce a u that minimizes the cost-to-go at time step k, then the cost-to-go at time step k will also be quadratic.To prove this, our first step is to find the u that minimizes the cost-to-go at time step k, and then show the at the resulting optimal cost-to-go remains in the quadratic form above.

To compute the expected value term, we need to do some work on the term e.

( )

( 1) ( 1) ( 1)

( )( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( ) ( ) ( ) (

( ) ( ) ( )

) ( ) ( )

( )

( )

ˆ

ˆ ˆkk k ky i ii

k k k

kk k kx i ii

k k k k

k kk k k k k k k k kx i y ii ii i

k

k

A C

A AK H

A AK H C AK A D

B

K

H D

B

e x x

x ε u

x x

e ε ε x

u

u

x uε x

( )ky

( 1) ( ) ( ) ( ) ( ) ( )

( 1) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )ˆ

ˆ, ,

ˆvar , ,

k k k k k k

k k k k k k T T k k T Tx i i yi

k k k T T k T Ti i xi

E A AK H

Q C C AK Q K A

AK D D K A Q

e x x u e

e x x u u u

x x

(

( 1) ( 1) ( 1) ( ) ( ) ( ) ( ) ( ) ( 1) ( ) ( )

( 1) ( ) ( )

( 1) ( 1) ( 1) ( ) ( ) ( ) ( ) ( ) ( 1) ( ) ( )

( 1)

1)

ˆ, ,

ˆ, ,

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k kx x

k k T kx x

Tk T k k k k k k T k ke

i

k

i

ke

ke

E W A B W A B

tr W Q

E W A AK H W A AK H

tr W

C W C

Q

x x x x u x u x u

u u

e e x x u e e

( ) ( )ˆ

( ) ( )

( )

( 1)

( ) ( 1) ( ( ))

T ki e i

T k T T k ki

k k T Tx x y

k T k

e i

i

Ti

k kD K A W A

Q AK Q K

D

C W C

K

A

u u

x x

To compute the Expected value of J(k+1), we compute the Exp value of the two quadratic terms (the Exp value of the third term is zero as it is composed only of zero mean random variables).

( )kxC

( )keC

( )kD

( ) ( ) ( ) ( ) ( ) ( ) ( 1) ( ) ( ) ( )

( ) ( ) ( ) ( 1) ( ) ( ) ( 1) ( )

( )( ) ( ) ( 1) ( ) ( 1) ( )

( )

( ) ( )) ((

ˆ, ,

2

2 2 0

k k T k k T k k k k k k

k T k k T k k k T T k kx e x x

kk k T k

k k T kx e x

k T k kx e x xk

k L C C

J L T E J

L C C B W B B W A

dJL C C B W B

B W

B W Ad

u u x x x x u

u u u x

u xu

u

) 1 ( )11 ( ) ˆT k kxB B W A

x

Terms that do not depend on u

( )kG

( ) ( ) ( ) ( 1) ( ) ( ) ( ) ( 1) ( )

( ) ( ) ( 1) ( ) ( )

( 1) ( 1) ( ) ( )ˆ

( ) ( ) ( 1) ( ) (

( ) ( 1 1

)

) (( )

ˆ ˆ ˆ2k k T k T T k k k T k T T k kx x

k T k T k k k Tx

k k k k T Tx x e x x y

Tk T k k k ke

k T k T kx

k Tx

J G B W A G B W A

T A W A D

tr W Q W Q Q AK Q K A

A AK H W A AK H

T A W A A W

x x x x

x x

e e

x

( 1) ( 1) ( ) (

( 1)

)ˆ

( )

( ) ( )

( ) ( ) ( ) ( (

)

( ) ( ) ( 1) ( )

(

( ) ( )

( ) ) ))

k k k

TT k

k T Tx x e x x y

k

k T k

k T k k T k

k

k k k kx e

ke

k

k

kx

A W BG A AK H W

tr W Q W Q Q AK Q K A

A AK H

W w

BG D

W

x

e e

x x e e

So we just showed that if at some time point k+1 the cost-to-go under an optimal control policy is quadratic in x and e, and provided that we produce a u that minimizes the cost-to-go at time step k, then the cost-to-go at time step k will also be quadratic. Since we had earlier shown that at time step p-1 the cost is quadratic in x and e, we now have the solution to our problem.

( )( 1) ( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( )

( 1) ( ) ( ) ( ) ( ) ( ) ( )

ˆ

( ) ( ) ( ) ( ) ( )

ˆ ˆ ˆ

0,1 0,1

, , ,

kk k k k kx i ii

kk k k ky i ii

k k k k k k k

i i

x x y y x

k T k k T k k

A B C

H D

A AK H B

N N

N Q N Q N Q

L T

x x u ε u

y x ε x

x x y x u

ε 0 ε 0 0

u u x x

( ) ( ) ( )

1( ) ( 1) ( 1) ( 1) ( 1)

( ) ( ) ( 1) ( 1) ( ) ( ) ( 1) ( )

( ) ( 1) ( ) ( ) ( 1) ( )

( ) ( 1)

ˆk k k

k T k T k T k T ki x i i e i x xi i

k k T k T k k T k T T k kx x x i e ii

Tk T k k k k ke x e

k kx x

G

G L C W C C W C B W B B W A

W T A W A A W BG D K A W AK D

W A W BG A AK H W A AK H

w tr W Q W

u x

( 1) ( ) ( )ˆ

( ) ( ) ( ) ( )0 0

k k k T Te x x y

p p p px e

Q Q AK Q K A

W T W w

Cost per step

Summary: Control problem with signal dependent noise (Todorov 2005)

For the last time step

Computing a cost for the motor commands

(1) (0) (0)

(2) (1) (1) 2 (0) (0) (1)

(3) (2) (2) 3 (0) 2 (0) (1) (2)

1( ) (0) 1 ( )

0

1( ) 1 ( ) 1

0

var var

kk k k j j

j

k Tk k j j k j

j

A C

A C A AC C

A C A A C AC C

A A C

A C A C

x x u

x x u x u u

x x u x u u u

x x u

x u

( ) ( ) ( )

( ) 1 1

var j j T j

Tn k n k n

I

L A C A C

u u u

Because there is noise in the motor commands, it will produce variance in our state. The above equation shows that the variance at the end of the movement (at time k) is mostly influenced by the motor commands late in the movement, and less by commands that were produced early in the movement.

To see this, note that A is a matrix that when raised to a power, will become “smaller”. The larger the raised power, the smaller the resulting matrix will become. In the sum, we have a contribution from each motor command j. When j is zero (the very first command), A is raised to a very high power. The noise in this command will have little influence on the endpoint variance. When j is larger (commands near end of the movement), A is raised to a small power. The noise in these commands will have a great deal of influence on the endpoint variance. Therefore, we have a natural cost function for the motor commands: