Review Sheet for Statistical Theory I - Fall 2007 | BST 631, Study notes of Biostatistics

Download Review Sheet for Statistical Theory I - Fall 2007 | BST 631 and more Study notes Biostatistics in PDF only on Docsity! Lecture 3 on BST 631: Statistical Theory I – Kui Zhang, 08/30/2007 Review for the previous lecture Theorem: How to calculate probabilities of events (1.2.8, 1.2.9, 1.2.11) Definition: , ⎟ . !n n r ⎛ ⎞ ⎜ ⎝ ⎠ Theorem: How to count. Number of possible arrangements of size from subjectsr n Without Replacement ( r n≤ ) With Replacement Ordered ! ( )! n n r− rn Unordered n r ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 1n r r + −⎛ ⎞ ⎜ ⎟ ⎝ ⎠ Example: Enumerating Outcomes Chapter 1 – Probability Theory 1.2.4 Enumerating Outcomes 1 Lecture 3 on BST 631: Statistical Theory I – Kui Zhang, 08/30/2007 Example 1.2.19 (Sampling with replacement): Consider sampling items from 2r = 3n = objects, with replacement. The outcomes in the ordered and unordered sample spaces are these: Unordered {1,1} {2,2} {3,3} {1,2} {1,3} {2,3} Ordered (1,1) (2,2) (3,3) (1,2),(2,1) (1,3),(3,1) (2,3),(3,2) Probability 1/9 1/9 1/9 2/9 2/9 2/9 Example (Sampling without replacement): Consider sampling items from 2r = 3n = objects, without replacement. The outcomes in the ordered and unordered sample spaces are these: Unordered {1,2} {1,3} {2,3} Ordered (1,2),(2,1) (1,3),(3,1) (2,3),(3,2) Probability 1/3 1/3 1/3 Note: 1. To calculate the right probability, you must use the right sample space to count the number of elements in the event and the sample space. The right sample space means that all elements in the sample space have the same probabilities. 2. For sampling without replacement, if we want to calculate the probability of an event that does not depend on the order, we can use either the ordered space or the unordered space. 3. For sampling with replacement, if we want to calculate the probability of an event that does not depend on the order, we must use the ordered space. This is corresponds to the common interpretation of “sampling with replacement”. For example 1.2.19, that means one of three objects is chosen, each with probability of 1/3, the object is noted and replaced; three objects are mixed and again one of them is chosen, each with 2 Lecture 3 on BST 631: Statistical Theory I – Kui Zhang, 08/30/2007 1. ( | ) ( | ) ( ) /( ( ) ( ) ( )) ( | ) ( ) /( ( | ) ( ) ( | ) ( ) ( | ) ( )) P A W P W A P A P W A P W B P W C P W A P A P W A P A P W B P B P W C P C = ∩ + ∩ + ∩ = + + . 2. ) . ( ) ( | ) ( ) ( | ) (P A B P A B P B P B A P A∩ = = 3 ) 1 1 ( ) ( ) ( | ) (i i ii iP B P B A P B A P A ∞ ∞ = = = ∩ =∑ ∑ Theorem 1.3.5 (Bayes’ Rule) Let be a partition of the sample space and let 1 2, ,A A B be any set. Then, for each i , 1,2,= 1 1 ( ) ( | ) ( ) ( | ) ( )( | ) ( ) ( ) ( | ) ( ) i i i i i i k k kk k P A B P B A P A P B A P AP A B P B P B A P B A P A∞ ∞ = = ∩ = = = ∩∑ ∑ The meaning of Bayes’ Rule: can be considered as the reasons that results the event 1 2, ,A A B . So ) is generally called as the prior probability and determined by previous experiments. If the event ( | iP B A B occurred, this formula can help us to find which reason results the event B . is generally called the “posterior probability”. ( | )iP A B Example 1.3.6: Morse Codes. Definition 1.3.7: Two events, A and B are statistically independent if )( ) ( ) (P A B P A P B∩ = . Equivalently, A and B are statistically independent if ( | ) ( ) where ( ) 0P A B P A P B= > or )( | ) (P B A P B= where ( ) 0P A > . Theorem 1.3.9: If A and B are independent events, then the following pairs are also independent: (a) A and cB ; (b) cA and B ; (c) cA and cB . 5 Lecture 3 on BST 631: Statistical Theory I – Kui Zhang, 08/30/2007 Definition 1.3.12: A collection of events are mutually independent if for any subcollection 1 2, ,A A 1 2, , , ki i iA A A , we have 11 ( ) ( ) j j k k i i jj P A P A == =∏∩ . Remark: If a collection of events are mutually independent, then all the pairs of them are independent. However, even all of pairs are independent; they may not be mutually independent. 1 2, ,A A Example: If a collection of events are mutually independent, 1 2, ,A A ( )i iP A p= ( 1, ,i n= ), then calculate the probability of the following events: (1) 1 n c ii B A = =∩ (none of iA occurs): ) 1( ) (1 n ii P B p = = −∏ (2) A (at least of 1 n ii C = =∪ iA occurs): ) 1( ) 1 (1 n ii P C p = = − −∏ (3) ))ji1( ( n c ii j D A = ≠ = ∩∪ ∩ A (exactly one iA occurs): ) 1( ) ( (1 ) n i ji j i P D p p = ≠ = −∑ ∏ Example: A large machine consists of 50 components. Past experience has shown that a particular component has a probability of 0.1% to fail. Suppose that the performance of one component will not affect another. The equipment will work if no more than one component fails. What is the probability that the machine will work? Solution: Define iA = {the th component fails}, then i ( ) 0.1%iP A = and ( 1, ,50)iA i = are mutually independent. Define D ={the equipment will work} ={exactly one component fails } {none of component fails}, therefore, ∪ 6 Lecture 3 on BST 631: Statistical Theory I – Kui Zhang, 08/30/2007 7 We can get and ( ) 97.38%P D = ( ) 91.08%P D = if ( ) 0.5%iP A = and ( ) 1%iP A = , respectively. 50 4950( ) (1 0.1%) 0.1% *(1 0.1%) 99.88% 1 P D ⎛ ⎞= − + − =⎜ ⎟ ⎝ ⎠ .