From 1e3c434c8b108a5abd9f6810d629c3ae83face98 Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Tue, 31 Aug 2021 17:06:06 -0400 Subject: added notes for math classes and the first non-computing homework --- zhilova/03_probability_function | 59 +++++++++++++++++++++++++++++++++++++++++ 1 file changed, 59 insertions(+) create mode 100644 zhilova/03_probability_function (limited to 'zhilova/03_probability_function') diff --git a/zhilova/03_probability_function b/zhilova/03_probability_function new file mode 100644 index 0000000..218e941 --- /dev/null +++ b/zhilova/03_probability_function @@ -0,0 +1,59 @@ + The Probability Set Function + +P: B -> R + +B is a sigma-algebra on C. + +Properties: +P(A) >= 0 \forall A in B +P(C) = 1 +\forall A1, A2, A3, ... in B, if A_i \cap A_j = \empty, +P(infinite union of A1, A2, ...) = sum over all j of P(A_j) + +Useful inequalities: +Boole's inequality (th 1.3.7) P(union of A1, A2, ...) = P(A1) + P(A2) + ... +(Derives from the inclusion-exclusion formula) + + Conditional Probability + +Let A, B be sets in \B (Borel Algebra) +Assume P(B) > 0 [because it wouldn't make sense to condition on an +impossible event] + +P(A | B) = P(A \cap B) / P(B) + +P(* | B) : B -> R [that's a new notation] + +Gives similar properties to the main probability function because it is +a probability set function. + +P(A | B) >= 0 +P(C | B) = 1 [and P(B | B) = 1 ] +P(* | B) is z-additive + +Sometimes, it's simpler to define P(A \cap B) = P(A | B) * P(B) like in +a Markov chain. +P(A \cap B_1 \cap B_2) = P(A | B_1 \cap B_2)P(B_1 | B_2)P(B_2). + Trivially proved by induction. + +The law of total probability. + +Consider B_1, B_2, ... in B such that any B_i, B_j are disjoint and the +union of all B_1 to B_\infty = C. + +If P(B_i) > 0, P(A) = \sum^infty P(A | B_i) * P(B_i) + + Proof + +For any i >= 1, +P(A | B_i) * P(B_i) = P(A \cap B_i) [basic property of conditionals] +A = A \cap C = A \cap (countable union of B_i) = (countable union of A +\cap B_i). +\to P(A) = P(countable union of A \cap B_i) +\to P(A) = (countable sum of P(A | B_i)*P(B_i)) + + Bayes' Theorem +P(B_i | A) = P(A | B_i) * P(B_i) / (sum over all B_j P(A | B_j)*P(B_j)) + +Applies the law of total probability and the definition of conditional +probability. -- cgit