*This preview shows
pages
1–4. Sign up
to
view the full content.*

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **MIT OpenCourseWare http://ocw.mit.edu 14.30 Introduction to Statistical Methods in Economics Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 2 Definition 1 A probability distribution on a sample space S is a collection of numbers P ( A ) which satisfies the axioms (P1)-(P3). Note that the axioms (P1)-(P3) do not pin down a unique assignment of probabilities to events. Instead, these axioms only give minimal requirements which any probability distribution should satisfy in order to be consistent with our basic intuitions of what constitutes a probability (well actually check that below). In principle any function P ( ) satisfying these properties constitutes a valid probability, but wed have to see separately whether its actually a good description of the random experiment at hand, which is always a hard question. In part 5 of this class (Special Distributions), well discuss a number of popular choices of P ( ) for certain standard situations. Some Properties of Probabilities Now we still have to convince ourselves that the axioms (P1)-(P3) actually are sucient to ensure that our probability function has the properties we would intuitively expect it to have, i.e. (1) the probability that an event happens plus the probability that it doesnt happen should sum to one, (2) the probability that the impossible event, , happens should equal zero, (3) if an event B is contained in an event A , its probability cant be greater than P ( A ), and (4) the probability for any event should be in the interval [0 , 1]. Well now prove these properties from the basic axioms. Proposition 1 P ( A C ) = 1 P ( A ) Proof: By the definition of the complement A C , ( P 2) Defn. A C ( P 3) 1 = P ( S ) = P ( A A C ) = P ( A ) + P ( A C ) where the last step uses that A A C = , i.e. that A and its complement are disjoint. Rearranging this, we get P ( A C ) = 1 P ( A ) which is what we wanted to show Proposition 2 P ( ) = 0 Proof: Since C = S , we can use the previous proposition to show that Prop.1 ( P 2) P ( ) = P ( S C ) = 1 P ( S ) = 1 1 = 0 Proposition 3 If B A , then P ( B ) P ( A ) . As an aside, cognitive psychologists found out that even though this rule seems very intuitive, people often violate it in everyday probabilistic reasoning. 2 Proof: In order to be able to use the probability 2 E.g. in a study by the psychologists Daniel Kahneman and Amos Tversky, several people were given the following description of Linda: Linda is 31 years old, single, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations....

View
Full
Document