ECSE-2500, Engineering Probability, Spring 2010, Rensselaer Polytechnic Institute

# Lecture 3

Starting on page 28.

- iClicker exercises: P={2,3,5,7}, Q={2,4,6,8}.
- What is P{$\cap Q$}?
- A: {3,5,7}
- B: {4,6,8}
- C: {2}
- D: {2,3,4,5,6,7,8}
- E: {3,4,5,6,7,8}

- What is P{$\cup Q$}?
- What is A\B?
- What is B\A?

- What is P{$\cap Q$}?
- Note that this proof of de Morgan uses the fact that A equals B iff A is a subset of B and B is a subset of A.
- Equivalently, just look at the Venn diagram; there are only 4 cases.
- Countable vs non countable infinite
- A countably infinite set means you can put the its elements in 1-1 correspondance with the natural numbers; you can count them.
- E.g., integers, rational numbers
- Uncountably infinite means you cannot.
- E.g., real numbers.
- That was proved in the 19 century.
- Don't worry about this for this course.

- 2.1.4 Event classes
- This is a technical detail I'm not going to worry about in this course.
- The problem occurs when the outcome is a real number, say a voltage.
- There are an uncountably infinite number of real numbers, even in a finite interval [0,1].
- We cannot assign them probabilities; it just doesn't work.
- It does work to use intervals of real numbers, like [.5,.6] and [.7,.8] and unions and complements of them.
- This matches the real world. You can't measure a voltage as 3.14159265...; you measure it in the range [3.14,3.15].
- {$\cal F$} class of events of interest: those sets of intervals.

- 2.2 Axioms of probability
- I: 0<=P[A}
- II: P[S]=1
- III: {$ A\cap B=\emptyset \rightarrow P[A\cup B] = P[A]+P[B] $}
- III': For {$ A_1, A_2, .... $} if {$ \forall_{i\ne j} A_i \cap A_j = \emptyset$} then {$P[\bigcup_{i=1}^\infty A_i] = \sum_{i=1}^\infty P[A_i] $}

- Example: cards. Q=event that card is queen, H=event that card is heart.
These events are not disjoint. Probabilities do not sum.
- {$Q\cap H \ne\emptyset $}
- P[Q] = 1/13=4/52, P[H] = 1/4=13/52, P[Q {$\cup$} H] = 16/52!=17/52.

- Example C=event that card is clubs. H and C are disjoint. Probabilities
do sum.
- {$C\cap H \ne\emptyset $}
- P[C] = 13/52, P[H] = 1/4=13/52, P[Q {$\cup$} H] = 26/52.

- Example. Flip a fair coin {$A_i$} is the event that the first time you see
heads is the i-th time, for {$i\ge1$}.
- We can assign probabilities to these countably infinite number of events.
- {$ P[A_i] = 1/2^i $}
- They are disjoint, so probabilities sum.
- Probability that the first head occurs in the 10th or later toss = {$ \sum_{i=10}^\infty 1/2^i $}

- Corollory 1
- {$ P[A^c] = 1-P[A] $}
- E.g., P[heart] = 1/4, so P[not heart] = 3/4

- Corollory 2: P[A] <=1
- Corollory 3: P[{$ \emptyset $}] = 0
- Corollory 4:
- For {$ A_1, A_2, .... A_n $} if {$ \forall_{i\ne j} A_i \cap A_j = \emptyset$} then {$P\left[\bigcup_{i=1}^n A_i\right] = \sum_{i=1}^n P[A_i] $}
- Proof by induction from axiom III.

- Corollory 5:
{$ P[A\cup B] = P[A] + P[B] - P[A\cap B] $}
- Proof from the fact that {$ P[A], P[B], P[A\cap B] $} are disjoint.
- Example: Queens and hearts. P[Q]=4/52, P[H]=13/52, P[Q {$\cup$} H]=16/52, P[Q {$\cap$} H]=1/52.
- {$ P[A\cup B] \le P[A] + P[B] $}

- Corollory 6:
{$P\left[\cup_{i=1}^n A_i\right] = \sum_{i=1}^n P[A_i] - \sum_{i<j} P[A_i\cap A_j] + \sum_{i<j<k} P[A_i\cap A_j\cap A_k] \cdots + (-1)^{n+1} P[\cap_{i=1}^n A_i] $}
- Example Q=queen card, H=heart, F=face card
- P[Q]=4/52, P[H]=13/52, P[F]=12/52,
- P[Q {$\cap$} H]=1/52, P[Q {$\cap$} F] =
*you tell me* - P[H {$\cap$} F]=
*you tell me* - P[Q {$\cap$} H {$\cap$} F] =
*you tell me* - So P[Q {$\cup$} H {$\cup$} F] = ?

- Example from [([http://en.wikipedia.org/wiki/)Roulette]]:
- R=red, B=black, E=even, A=1-12
- P[R] = P[B] = P[E] = 16/38. P[A]=12/38
- {$ P[R\cup E \cup A] $} = ?
- You could try at http://www.roulette4fun.com/online_game.php but it's slow.

- Example Q=queen card, H=heart, F=face card
- Corollory 7: if {$A\subset B$} then P[A] <= P[B] Example: Probability of a repeated coin toss having its first head in the 2nd-4th toss (1/2+1/4+1/8) <= Prob of it happening in the 3rd toss (1/4).
- 2.2.1 Discrete sample space
- If sample space is finite, probabilities of all the outcomes tell you everything.
- sometimes they're all equal.
- Then P[event] = (# outcomes in event) / (total # outcomes)
- For countably infinite sample space, probabilities of all the outcomes also tell you everything.
- E.g. fair coin. P[even] = 1/2
- E.g. example 2.9. Try numbers from http://random.org/.
- What probabilities to assign to outcomes is a good question.
- Example 2.10. Toss coin 3 times.
- Choice 1: outcomes are TTT ... HHH, each w.p. 1/8
- Choice 2: outcomes are # heads: 0...3, each w.p. 1/4.
- Incompatible. What are probabilities of # heads for choice 1?
- Which is correct? Why?

- Example 2.11: countably infinite sample space.
- Toss fair coin, outcome is # tosses until 1st head.
- What are reasonable probabilities?
- Do they sum to 1?

- 2.2.2 Continuous sample spaces
- Can't assign probabilities to points on real line. (It just doesn't work.)
- Work with set of intervals, and Boolean operations on them.
- Set may be finite or countable.
- This set of events is a
*Borel set*. - Notation:
- [a,b] closed. includes both. a<=x<=b
- (a,b) open. includes neither. a<x<b
- [a,b) includes a but not b, a<=x<b
- (a,b] includes b but not a, a<x<=b

- Assign probabilities to intervals (open or closed).
- E.g., uniform distribution on [0,1] P[a<=x<=b] = 1/(b-a).
- Nonuniform distributions are common.