lecture_04

Wed Jun 13 09:02:27 CDT 2012

Homework

Friday, Read axiomatics and assorted examples in chapter 2 and do p.50 # 3, 8, 15c, 16abc, 25, 36, and ?

Recall

• [ (x + y)^n = \sum\limit_{k=0}^n { \frac{ n!}{ k!(n - k)!} x^k y^{n-k} ]
  • [ = \sum\limits_{ a + b - n , a \ge 0, b \ge 0}{ \frac{ n!}{ a! b!} x^a y^b ]
  • therefore [ (x_1 + x_2)^n = \sum\limits_{ n_1 + n_2 = n,n_1 \ge 0, n_2 \ge 0}{ \frac{ n!}{ n_1! n_2!} x_1^{n_1} x_2^{n_2} ]
• Trinomial Theorem
  • [ (x_1 + x_2 + x_3)^n = \sum\limits_{ n_1 + n_2 + n_3 = n, n_i \ge 0}{ \frac{ n!}{ n_1! n_2! n_3!} x_1^{n_1} x_2^{n_2} x_3^{n_3} } ]
• Multinomial Theorem
  • [ \left( \sum\limits_{ i = 1}^I{ x_i} \right)^n = \sum\limits_{ n_1 + \cdots n_I = n, n_i \ge 0}{ \frac{ n!}{ n_1! n_2! \cdots n_I!} x_1^{n_1} \cdots x_I^{n_I} ]

Axromatic Probability

• [ { A \cup B}: \forall x \in S s.t. x \in A or x \in B ]

Facts from set Theory

• [ A \cap \left( B \cup C \right) = \left( A \cap B \right) \cup \left( A \cap B \right)]
• [ A \left( B \cup C \right) = AB \cup AC]
• [ A \cup \left( B \cap C \right) = \left( A \cup B \right) \cap \left( A \cup C \right)]
• [ \left( \bigcup\limits_{ i\in I}{ A_i} \right) = \left( \bigcap\limits_{ i \in I}{ A_i^c \right) ]
• [ \left( A \cup B \right)^c = A^c \cap B^c ]

A probability theory is a triple

• [ \left( S, { E_i }, P \right) ]
• satisfying 3 axioms:
  • A.1 [ 0 \le P(E) \le 1 ]
  • A.2 [ P(S) = 1 ]
  • A.3 [ P\left( \bigcup\limits_{ i = 1}^{ \infty}{ E_i} \right) = \sum\limits_{ i = 1}{ \infty}{ P\left( E_i \right) }] whenever [ E_i \cap E_2 = \emtyset \forall i \notequal j]

Thereom

• [ P\left( A \cup B \right) = P( A) + P( B) - P( A \cap B) ]
  • PF: [ P( A \cup B) = P( A \cup BA^c) = P( A) + P( B A^c) ]
    • [ P( A) + P( B A^c) = P( A) + P( B) - P( AB) ]
  • [ P( B) = P( BA \cup BA^c) = P(AB) + P( BA^c) ]

Somethign else

• [ P( A \cup (B \cup C)) \ = P( A) + P( B \cup C) - P( AB \cup AC) \ = P( A) + P( B) + P( C) - P( BC) - P( AB \cup AC) \ = P( A) + P( B) + P( C) - P( BC) - \left( P( AB) + P(AC) - P( ABC)\right) ]

prop.

• [ P( \emptyset) = 0 ]
  • PF: [ P( S) = P( S \cup \emptyset) \ = P( S) + P( \emptyset)] * \therefore [ P( \emptyset) = 0 ]