1  Probability Refresher

1.1 Notations

\(\Omega\): A sample space, a set of possible outcomes of a random experiment.

\(X\): A random variable, a function from the sample space to the real numbers: \(X: \Omega \to \R\).

Stochastic Process

A stochastic process is a family of random variables, \(\{X(t): t\in T\},\) where \(t\) usually denotes time. That is, at every time \(t\) in the set \(T\), a random number \(X(t)\) is observed.

  • Discrete-time process: \(T=\{0,1,2,3\}\), the discrete process is \(\{X(0), X(1), X(2), \dots\}\)
  • Continuous-time process: \(T=[0, \infty]\) or \(T=[0, K]\) for some \(K\).

The state space, \(S\), is the set of real values that \(X(t)\) can take.

You can think of “conditioning” as “changing the sample space.”

  • From unconditional to conditional

    \[ \P (B) = \P(B\mid \Omega) \]

    \(\Omega\) denotes the sample space, \(\P (B) = \P(B\mid \Omega)\) just means that we are looking for the probability of the event \(B\), out of all possible outcomes in the set \(\Omega.\)

  • Partition Theorem

    \[ \P(A) = \sum_{i=1}^m \P(A\cap B_i) = \sum_{i=1}^m \P(A\mid B_i) \P(B_i) \]

    where \(B_i, i=1,\dots,m,\) are a partition of \(\Omega.\) The intuition behind the Partition Theorem is that the whole is the sum of its parts.

    A partition of \(\Omega\) is a collection of mutually exclusive events whose union is \(\Omega.\)

    That is, sets \(B_1, B_2, \dots, B_m\) form a partition of \(\Omega\) if

    \[ \begin{split} B_i \cap B_j &= \emptyset \;\text{ for all $i, j$ with $i\ne j,$} \\ \text{and } \bigcup_{i=1}^m B_i &= B_1 \cup B_2 \cup \dots \cup B_m = \Omega. \end{split} \]

  • Bayes’ Theorem

    Bayes’ Theorem allows us to invert a conditional statement, i.e., the express \(\P(B\mid A)\) in terms of \(\P(A\mid B).\)

    For any events \(A\) and \(B\):

    \[ \P(B\mid A) = \frac{\P(A\cap B)}{\P(A)} = \frac{\P(A\mid B)\P(B)}{\P(A)} \]

  • Generalized Bayes’ Theorem

    For any partition member \(B_j\),

    \[ \P(B_j\mid A) = \frac{\P(A\mid B_j)\P(B_j)}{\P(A)} = \frac{\P(A\mid B_j)\P(B_j)}{\sum_{i=1}^m\P(A\mid B_i)\P(B_i)} \]