1. Probability: Review TexPoint fonts used in EMF.
Pieter Abbeel
UC Berkeley EECS
Many slides adapted from Thrun, Burgard and Fox, Probabilistic Robotics
TexPoint fonts used in EMF.
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2. Why probability in 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

3. Example 1: Helicopter
State: position, orientation, velocity, angular rate
Sensors:
GPS : noisy estimate of position (sometimes also velocity)
Inertial sensing unit: noisy measurements from
3-axis gyro [=angular rate sensor],
3-axis accelerometer [=measures acceleration + gravity; e.g., measures (0,0,0) in free-fall],
3-axis magnetometer
Dynamics:
Noise from: wind, unmodeled dynamics in engine, servos, blades

4. Example 2: Mobile robot inside building
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

5. Axioms of Probability Theory
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

7. Using the Axioms

8. Discrete Random Variablesx1 x2 x3 x4
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 .

9. Continuous Random Variables
X takes on values in the continuum.
p(X=x), or p(x), is a probability density function.
E.g.
p(x) 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 y P(x | y) = P(x,y) / P(y) P(x,y) = P(x | y) P(y)
If X and Y are independent then P(x | y) = P(x)
Same for probability densities, just P  p

11. Law of Total Probability, Marginals
Discrete case
Continuous case

12. Bayes Formula

13. Normalization
Algorithm:

14. Conditioning
Law of total probability:

15. Bayes Rule with Background Knowledge

16. Conditional Independence
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:
count frequencies!

19. Example P(z|open) = 0.6 P(z|open) = 0.3 P(open) = P(open) = 0.5
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
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
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

24. Probability review For any random variables X, Y we have:
Definition of conditional probability:
P(X=x | Y=y) = P(X=x, Y=y) / P(Y=y)
Chain rule: (follows directly from the above)
P(X=x, Y=y) = P(X=x) P(Y=y | X=x )
= P(Y=y) P(X=x | Y=y)
Bayes rule: (really just a re-ordering of terms in the above)
P(X=x | Y=y) = P(Y=y | X=x) P(X=x) / P(Y=y)
Marginalization:
P(X=x) = y P(X=x, Y=y)
Note: no assumptions beyond X, Y being random variables are made for any of these to hold true (and when we divide by something, that something is not zero)

25. Probability review For any random variables X, Y, Z, W we have:
Conditional probability: (can condition on a third variable z throughout)
P(X=x | Y=y, Z=z) = P(X=x, Y=y | Z=z) / P(Y=y | Z=z)
Chain rule:
P(X=x, Y=y, Z=z, W=w) = P(X=x) P(Y=y | X=x ) P(Z=z | X=x, Y=y) P(W=w| X=x, Y=y, Z=z)
Bayes rule: (can condition on other variable z throughout)
P(X=x | Y=y, Z=z) = P(Y=y | X=x, Z=z) P(X=x | Z=z) / P(Y=y | Z=z)
Marginalization:
P(X=x | W=w) = y,z P(X=x, Y=y, Z=z | W=w)
Note: no assumptions beyond X, Y, Z, W being random variables are made for any of these to hold true (and when we divide by something, that something is not zero)

26. Independence Two random variables X and Y are independent iff
for all x, y : P(X=x, Y=y) = P(X=x) P(Y=y)
Representing a probability distribution over a set of random variables X1, X2, …, XT in its most general form can be expensive.
E.g., if all Xi are binary valued, then there would be a total of 2T possible instantiations and it would require 2T-1 numbers to represent the probability distribution.
However, if we assumed the random variables were independent, then we could very compactly represent the joint distribution as follows:
P(X1=x1, X2=x2, …, XT=xT) = P(X1=x1) P(X2=x2) … P(XT=xT)
Thanks to the independence assumptions, for the binary case, we went from requiring 2T-1 parameters, to only requiring T parameters!
Unfortunately independence is often too strong an assumption …

27. Conditional independence
Two random variables X and Y are conditionally independent given a third random variable Z iff
for all x, y, z : P(X=x, Y=y | Z=z) = P(X=x | Z=z) P(Y=y | Z=z)
Chain rule (which holds true for all distributions, no assumptions needed):
P(X=x,Y=y,Z=z,W=w) = P(X=x)P(Y=y|X=x)P(Z=z|X=x,Y=y)P(W=w|X=x,Y=y,Z=z)
 For binary variables the representation requires 1 + 2*1 + 4*1 + 8*1 = 24-1 numbers (just like a full joint probability table)
Now assume Z independent of X given Y, and assume W independent of X and Y given Z, then we obtain:
P(X=x,Y=y,Z=z,W=w) = P(X=x)P(Y=y|X=x)P(Z=z|Y=y)P(W=w|Z=z)
For binary variables the representation requires 1 + 2*1 + 2*1 + 2*1 = 1+(4-1)*2 numbers -- - significantly less!!