RL 001

July 28, 2026 · 2 min read

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Reinforcement Learning

Learn how to make a good sequence of decisions by interacting with the environment.

Trial and Error: Based learning approach.
Optimization: Find good sequences of actions or decisions.
Delayed Consequences/Rewards: Takes time to relize the actions or decisions are good or bad.
Exploration: Learn by making decisions or forming actions or through experiences. Trying new actions to discover their effects.
Exploitation: Choosing actions with the highest expected reward, based on current knowledge.
Generalization: Use previous experiences or knowledge to new or unseen situations effectively.

$P(A \cap B) \geq P(A) + P(B) - 1$ $P(\cap^{n}_{i=1} A_i) \geq \sum_{i=1}^{n} P(A_i) - (n-1)$

Gives a lower bound on the intersection probability which is useful when this probability is difficult to compute directly.
It is useful when the probabilities of individual events are sufficiently large.

$P(\cup^{n}_{i=1} A_i) \leq \sum_{i=1}^{n} P(A_i)$

for any sets $A_1, A_2, ..., A_n$ .

It is useful when finding an upper bound for the probabilities of the union of events.

$P(A|B) = \frac{P(A \cap B)}{P(B)}$ $P(A\cap B) = P(B|A) P(A) = P(A|B) P(B)$ $P(A|B)P(B) = P(B|A)P(A)$ $P(A|B) = \frac{P(B|A) P(A)}{P(B)}$

It allows us to compute the conditional probability $P(A|B)$ from the inverse conditional probability $P(B|A)$ .
Let $A_1, A_2, ..., A_n$ be a partition of the sample space $S$ . Then let $B$ be any subset of $S$ we have:

$P(A_i|B) = \frac{P(B|A_i) P(A_i)}{\sum_{j=1}^{\infty} P(B|A_j) P(A_j)}$

$P(A \cap B) = P(A) P(B)$

A family $A_i :i \in I$ of events is independent if for every finite subset $J \subseteq I$ we have: $P\left(\bigcap_{i \in J} A_i\right) = \prod_{i \in J} P(A_i)$
The pair-wise independence of events does not imply their mutual independence.

$P(A \cap B|C) = P(A|C) P(B|C)$