Probability: Review

Pieter AbbeelUC Berkeley EECS

Many slides adapted from Thrun, Burgard and Fox, Probabilistic Robotics

Often state of robot and state of its environment are unknown and only noisy sensors available Probability provides a framework to fuse

sensory information Result: probability distribution over possible

states of robot and environment Dynamics is often stochastic, hence can’t

optimize for a particular outcome, but only optimize to obtain a good distribution over outcomes Probability provides a framework to reason in

this setting Result: ability to find good control policies for

stochastic dynamics and environments

Why probability in robotics?

State: position, orientation, velocity, angular rate

Sensors: GPS : noisy estimate of position (sometimes

also velocity) Inertial sensing unit: noisy measurements from

(i) 3-axis gyro [=angular rate sensor], (ii) 3-axis accelerometer [=measures

acceleration + gravity; e.g., measures (0,0,0) in free-fall],

(iii) 3-axis magnetometer Dynamics:

Noise from: wind, unmodeled dynamics in engine, servos, blades

Example 1: Helicopter

State: position and heading Sensors:

Odometry (=sensing motion of actuators): e.g., wheel encoders

Laser range finder: Measures time of flight of a laser beam

between departure and return Return is typically happening when hitting a

surface that reflects the beam back to where it came from

Dynamics: Noise from: wheel slippage, unmodeled

variation in floor

Example 2: Mobile robot inside building

5

Axioms of Probability Theory

1)Pr(0 A

Pr(A) denotes probability that the outcome ω is an element of the set of possible outcomes A. A is often called an event. Same for B.

Ω is the set of all possible outcomes.ϕ is the empty set.

6

A Closer Look at Axiom 3

A B

7

Using the Axioms

8

Discrete Random Variables

X denotes a random variable. X can take on a countable number of values

in {x1, x2, …, xn}. P(X=xi), or P(xi), is the probability that the

random variable X takes on value xi. P( ) is called probability mass function.

E.g., X models the outcome of a coin flip, x1 = head, x2 = tail, P( x1 ) = 0.5 , P( x2 ) = 0.5

.

x1

x2

x4

x3

9

Continuous Random Variables

X takes on values in the continuum. p(X=x), or p(x), is a probability density

function.

E.g.

b

a

dxxpbax )()),(Pr(

x

p(x)

10

Joint and Conditional Probability P(X=x and Y=y) = P(x,y)

If X and Y are independent then P(x,y) = P(x) P(y)

P(x | y) is the probability of x given yP(x | y) = P(x,y) / P(y)P(x,y) = P(x | y) P(y)

If X and Y are independent thenP(x | y) = P(x)

Same for probability densities, just P p

11

Law of Total Probability, Marginals

y

yxPxP ),()(

y

yPyxPxP )()|()(

x

xP 1)(

Discrete case

1)( dxxp

Continuous case

dyypyxpxp )()|()(

dyyxpxp ),()(

12

Bayes Formula

evidenceprior likelihood

)()()|()(

)()|()()|(),(

yPxPxyPyxP

xPxyPyPyxPyxP

13

Normalization

)()|(1)(

)()|()(

)()|()(

1

xPxyPyP

xPxyPyP

xPxyPyxP

x

yx

xyx

yx

yxPx

xPxyPx

|

|

|

aux)|(:

aux1

)()|(aux:

Algorithm:

14

Conditioning Law of total probability:

dzyzPzyxPyxP

dzzPzxPxP

dzzxPxP

)|(),|()(

)()|()(

),()(

15

Bayes Rule with Background Knowledge

)|()|(),|(),|(

zyPzxPzxyPzyxP

16

Conditional Independence

)|()|(),( zyPzxPzyxP

),|()( yzxPzxP

),|()( xzyPzyP

equivalent to

and

17

Simple Example of State Estimation

Suppose a robot obtains measurement z What is P(open|z)?

18

Causal vs. Diagnostic Reasoning P(open|z) is diagnostic. P(z|open) is causal. Often causal knowledge is easier to obtain. Bayes rule allows us to use causal knowledge:

)()()|()|(

zPopenPopenzPzopenP

count frequencies!

19

Example P(z|open) = 0.6 P(z|open) = 0.3 P(open) = P(open) = 0.5

67.032

5.03.05.06.05.06.0)|(

)()|()()|()()|()|(

zopenP

openpopenzPopenpopenzPopenPopenzPzopenP

• z raises the probability that the door is open.

20

Combining Evidence Suppose our robot obtains another observation z2. How can we integrate this new information? More generally, how can we estimate

P(x| z1...zn )?

21

Recursive Bayesian Updating

),,|(),,|(),,,|(),,|(

11

11111

nn

nnnn

zzzPzzxPzzxzPzzxP

Markov assumption: zn is independent of z1,...,zn-1 if we know x.

22

Example: Second Measurement

P(z2|open) = 0.5 P(z2|open) = 0.6 P(open|z1)=2/3

625.085

31

53

32

21

32

21

)|()|()|()|()|()|(),|(

1212

1212

zopenPopenzPzopenPopenzPzopenPopenzPzzopenP

• z2 lowers the probability that the door is open.

23

A Typical Pitfall Two possible locations x1 and x2

P(x1)=0.99 P(z|x2)=0.09 P(z|x1)=0.07

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

5 10 15 20 25 30 35 40 45 50

p( x

| d)

Number of integrations

p(x2 | d)p(x1 | d)