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# S520, January 22

Here’s another edition of my notes from my S520 class.  (Sorry, I’m a little behind.) This time, I wrote all the notes in $LaTeX$, both for the challenge and for the practice.  Therefore, today’s entry is a PDF file, not a real blog post.  (Sorry.)  Here are the notes.  Let me know if you find any major errors.

Notes for January 22, 2010

By the way, I am still working on the RAMMCAP post I promised last time–it’s just turning out to be a lot longer than I’d originally intended.  I’ll probably break it up into multiple posts next week.

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# S520, January 20

Brief introduction into the class:  S520 is an introductory course in statistics for graduate students taught by Professor Michael Trosset.  It aims for depth in each of the subjects we talk about, explaining the why’s and wherefore’s of each topic instead of trying to cram as many topics in as possible (which is a major perk, in my opinion).  It’s actually taught in conjunction with S320, the undergraduate version of the same course, so I think most of the other students in the course are undergrads (not a bad thing, just an observation).  You can read more about it on the pages linked above.  Ok, on with the notes.

# Chapter 3, Probability

### Section 3.2

• Basics of modern probability devised by Kolmogorov as discussed last time.
• Kolmogorov’s probability space has three major components
1. $S$, the sample space – the universe of possible experimental outcomes
2. $C$, the collection of subsets of $S$ called “events”- events have well-defined probabilities
3. $P$, the probability measure-a function that assigns real numbers (called “probabilities”) to events-in proper notation, that looks like:$P : C rightarrow R$
1. There are several criteria the probability measure has to satisfy, though many of the are by convention only.
1. Convention – we require $0 leq P(E) leq 1$, where $P(E)$ is the probability of any given event.
• If $P(E) = 0 Rightarrow$ not going to happen, generally speaking.  (Apparently there are exceptions I don’t know about.)
• If $P(E) = 1 Rightarrow$ definitely going to happen.
2. Convention – $P(S) = 1 Rightarrow S$ has to be an event in $S$
3. Crucial property: If $A$ and $B$ are disjoint $(A cap B = emptyset )$ events, then the probablility of their union is:$P(Acup B) = P(A) + P(B)$
• Example: Drawing a card$S = mbox{deck of 52 cards }$
$P(mbox{A = card is ace}) = frac{4}{52} = frac{1}{13}$
$P(B = mbox{card is king}) = frac{4}{52} = frac{1}{13}$
$P(A cup B) = frac{1}{13} + frac{1}{13} = frac{2}{13}$
2. Because of those conventions, we can learn these things:
1. $P(E^C) = 1 - P(E)$
$E, E^C$ are disjoint, then by “crucial property”:

$P(S) = P(E) + P(E^C) ;;; since , E + E^C = S$
$1 = P(E) + P(E^C) ;;; mbox{by convention 2 above}$
$P(E^C) = 1 - P(E)$

2. If $A subset B$, then $P(A) leq P(B)$
3. $P(A cup B) = P(A) + P(B) - P(Acap B)$
$P(mbox{Overlapping sets}) = Sigma P(mbox{each set}) - P(mbox{intersection of sets})$

### Section 3.3 Finite Sample Space

• $S$ has a finite number $N$ of outcomes, represented by$S = {s_1, s_2, dots, s_N}$
• For example:
• $E = {s_2, s_3, s_7, s_9} = {s_2} cup {s_3} cup {s_7} cup {s_9}$
• So, $P(E) = ?$
• $P(E) = P({s_2}) + P({s_3}) + P({s_7}) + P({s_9})$
• If we know the probability of the individual outcomes, we can compute the probability of any event by adding the probabilities of the relevant outcomes.
• Core assumption:  individual events in the collection are separate by convention.
• Important special case (limit of most high school probability): Suppose that each of the $N$ outcomes is “equally likely” (recurring phrase we’ll hear a lot).  In this case, each outcome has probability:$P({s_i}) = frac{1}{N} mbox{for } i = 1, 2, dots, N$

This means that with equally likely outcomes, figuring probabilities is basically just counting.

• Example:

$E = { s_1, s_2, dots, s_k}$ for $k$ outcomes in $E$, then
$P(E) = P({s_1}) + P({s_2}) + dots + P({s_k})$
$P(E) = frac{1}{N} + frac{1}{N} + dots + frac{1}{N}$
$P(E) = frac{displaystyle k}{displaystyle N}= frac{displaystyle #(mbox{outcomes in E})}{ displaystyle#(mbox{outcomes in S})}$

• 3 of a kind example:  Draw 5 cards from a standard deck of cards.  What is $P(mbox{3 of a kind})$?

A: $frac{displaystyle #(mbox{3 of a kind})}{displaystyle #(mbox{5 cards})} = frac{displaystyle formula , derived , previously}{displaystyle {52 choose 5}}$

• Counting example (Exercise 3.7.2 in book):  Assume the probabilities of throwing an astragalus are:

$P(1) = 0.1, P(3) = 0.4, P(4) = 0.4, P(6) = 0.1$

• So we have:

$p_1 = P({s_1}) = 0.1$
$p_2 = P({s_3}) = 0.4$
$p_3 = P({s_4}) = 0.4$
$p_4 = P({s_6}) = 0.1$

• Make an urn model with equally likely outcomes that mimics an astragalus.
• First (wrong) idea:
• 4 slips of paper with one of the four outcomes on it
• Why wrong?  Doesn’t have the right probability.
• Second (right) model:
• 10 slips of paper with the values ${1, 3, 3, 3, 3, 4, 4, 4, 4, 6}$ on them
• $P(1) = 0.1, P(3) = 0.4, P(4) = 0.4, P(6) = 0.1$

### Friday – Conditional  probabilities

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