Machine Learning - Modelling Uncertainy

There is several reasons why environments may be uncertain:

environment is only partially observable
sensors are unreliable
the results of actions are uncertin
high complexity

We deal with uncertainty using probabilities of propositions. Alternative: Model uncertainty using probalbilites of rules such as: $LawnSprinkler \vert\rightarrow_0.99 WetGrass$, $WetGras \vert\rightarrow_0.7 Rain$.

Probabilities summarize several factors:

missing knwoledge
incapability to devise complete models of complex domains
chance

Random variables

modeling uncertainty using random variables
types of random variable:
- Boolean random variable:
  - e.g. $Cavity$ (Is there a cavit in my tooth?)
  - Values: $<true, false>$
- Discrete random variable
  - e.g. $Weather$ has one of the values $<sun, rain, cloudy, snow>$
  - values must describe the domain sufficiently and be mutually exclusive
- Continous random variable
  - values are real numbers
  - e.g. $Length \in [1, 20]$

Propositions

a proposition is made by assigning a value to a random variable:
- \[Weather = sun\]
- \[Length = 2.4\]
complex propositions are made by using logical operators to connect simple propositions: $Weather = sun \vee Cavity = false$
notation:
- random variables with capital: $Weather$, but values: $sun$
- but: $cavity$ means $Cavity = true$; $\neg cavity$ means $Cavity = false$; $sun$ means $Weather = sun$

Atomic events

A complete specification of the state of the domain (the agent may be uncertain about the state). For example if the domain is fully descibed by the boolean variables $Cavity$ and $Toothache$ then there are 4 atomic events.

Atomic events are mutually exclusive and describe the domain completely.

Probability distribution

A-priori or unconditional probabilities of propositions: $P(Cavity=true)=0.1$ or $P(Weather=sum)=0.72$ denote the probability of guesses. The probabilities may change when new information becomes available.
the probability distribution $P$ comprises the probabilities of all values: $P(Weather)=<0.72, 0.1, 0.08, 0.1>$ for values $<sun, rain, cloudy, snow>$
$P$ is normalized, ie., sum = 1
Joint probability distribution for several random variables comprises all atomic states $P(Weather, Cavity)$ is a 4x2 matrix
the joint probability distribution holds the entire knowledge about the domain!

Conditional probability

conditional or posterior probability: $P(cavity\vert toothache)=0.8$, i.e., the information $toothache$ is known
notation for conditional distributions: $P(Cavity\vert Toothache)$ 2-component vector
if the additional information $cavity$ is known: $P(cavity\vert Toothache, cavity)=<1, 1>$
additional information may be irrelevant: $P(cavity\vert toothache, sun)=P(cavity\vert toothache)=0.8$ or more general: $P(Cavity\vert Toothache, sun)=P(Cavity\vert Toothache)$
domain knowledge of this kind facilitates finding the joint probability distribution
definition of conditional probability: $P(a\vert b)=P(a \wedge b)/P(b)$ uf $P(b) > 0$
product rule is an alternative formulation: $P(a \wedge b) = P(a\vert b)P(b)=P(b\vert a)P(a)$
general version for distributions: $P(Weather, Cavity) = P(Weather\vert Cavity)P(Cavity)$. This means 4 x 2 seperate equations, not matrix multiplication

Chain rule (derived by successive application of product rule): $P(X_1,....,X_n)=\Pi_{i=2^n}P(X_i\vert X_1,...,X_{i-1})P(X_1)$

Inference by enumeration

….

Independence

$A$ and $B$ are indendent if and only if $P(A\vert B)=P(A)$ or $P(B\vert A)=P(B)$. With this, the product rule leads to $P(A, B)=P(A\vert B)P(B)=P(A)P(B)$

Conditional Independence

Consider $P(Toothache, Cavity, Catch)$, which has $2^3 - 1 = 7$ independent proabilities. Miind $Catch$ is not independent of $Toothache$: In general $P(Catch\vert Toothache) \neq P(Catch)$. Rather, $P(Catch)$ does depend on the value of $Toothache$ - it’s far more likely finding a cavity provided there is $toothache$. But this holds only as long as the value of $Cavity$ is unknown.

Assume that:

for $Cavity = true$ the probability that the dentist finds the cavity does not depend on whether there is toothache or not $P(Catch\vert Toothache, cavity) = P(Catch\vert cavity)$
for $Cavity = false$ the probabilit for a catch is independent of $Toothache$ likewise $$P(Catch\vert Toothache, \neg cavity)=P(Catch\vert\neg cavity)

Summarizing 1 and 2, $Catch$ is conditionally indendent of $Toothache$ given the value of $Cavity$: $P(Catch\vert Toothache, Cavity) = P(Catch\vert Cavity)$. Likewise, $Toothache$ is conditionally indendent of $Catch$ given $Cavity$: $P(Toothache\ Catch, Cavity) = P(Toothache\vert Cavity)$