Module #16: Probability Theory Rosen 5th ed., ch. 5 Let’s move on to probability, ch. 5.

Terminology A (stochastic) experiment is a procedure that yields one of a given set of possible outcomes The sample space S of the experiment is the set of possible outcomes. An event is a subset of sample space. A random variable is a function that assigns a real value to each outcome of an experiment Normally, a probability is related to an experiment or a trial. Let’s take flipping a coin for example, what are the possible outcomes? Heads or tails (front or back side) of the coin will be shown upwards. After a sufficient number of tossing, we can “statistically” conclude that the probability of head is 0.5. In rolling a dice, there are 6 outcomes. Suppose we want to calculate the prob. of the event of odd numbers of a dice. What is that probability?

Probability: Laplacian Definition First, assume that all outcomes in the sample space are equally likely This term still needs to be defined. Then, the probability of event E in sample space S is given by Pr[E] = |E|/|S|. Even though there are many definitions of probability, I would like to use the one from Laplace. The expression “equally likely” may be a little bit vague from the perspective of pure mathematics. But in engineering viewpoint, I think that is ok.

Probability of Complementary Events Let E be an event in a sample space S. Then, E represents the complementary event. Pr[E] = 1 − Pr[E] Pr[S] = 1

Probability of Unions of Events Let E1,E2  S Then: Pr[E1 E2] = Pr[E1] + Pr[E2] − Pr[E1E2] By the inclusion-exclusion principle.

Mutually Exclusive Events Two events E1, E2 are called mutually exclusive if they are disjoint: E1E2 =  Note that two mutually exclusive events cannot both occur in the same instance of a given experiment. For mutually exclusive events, Pr[E1  E2] = Pr[E1] + Pr[E2].

Exhaustive Sets of Events A set E = {E1, E2, …} of events in the sample space S is exhaustive if An exhaustive set of events that are all mutually exclusive with each other has the property that

Independent Events Two events E,F are independent if Pr[EF] = Pr[E]·Pr[F]. Relates to product rule for number of ways of doing two independent tasks Example: Flip a coin, and roll a die. Pr[ quarter is heads  die is 1 ] = Pr[quarter is heads] × Pr[die is 1] Now the question is: how we can figure out whether two events are independent or not

Conditional Probability Let E,F be events such that Pr[F]>0. Then, the conditional probability of E given F, written Pr[E|F], is defined as Pr[EF]/Pr[F]. This is the probability that E would turn out to be true, given just the information that F is true. If E and F are independent, Pr[E|F] = Pr[E]. Here is the most important part in the probability, the cond. prob. By the cond. prob., we can figure out whether there is a correlation or dependency between two probabilities.

Bayes’ Theorem Allows one to compute the probability that a hypothesis H is correct, given data D: Set of Hj is exhaustive

Bayes’ theorem: example Suppose 1% of population has AIDS Prob. that the positive result is right: 95% Prob. that the negative result is right: 90% What is the probability that someone who has the positive result is actually an AIDS patient? H: event that a person has AIDS D: event of positive result P[D] = P[D|H]P[H]+P[D|H]P[H ] = 0.95*0.01+0.1*0.99=0.1085 P[H|D] = 0.95*0.01/0.1085=0.0876

Expectation Values For a random variable X(s) having a numeric domain, its expectation value or expected value or weighted average value or arithmetic mean value E[X] is defined as

Linearity of Expectation Let X1, X2 be any two random variables derived from the same sample space. Then: E[X1+X2] = E[X1] + E[X2] E[aX1 + b] = aE[X1] + b

Variance The variance Var[X] = σ2(X) of a random variable X is the expected value of the square of the difference between the value of X and its expectation value E[X]: The standard deviation or root-mean-square (RMS) difference of X, σ(X) :≡ (Var[X])1/2.

Visualizing Sample Space 1. Listing S = {Head, Tail} 2. Venn Diagram 3. Contingency Table 4. Decision Tree Diagram

S Venn Diagram Tail Experiment: Toss 2 Coins. Note Faces. TH HT HH TT Event Other compound events could be formed: Tail on the second toss {HT, TT} At least 1 Head {HH, HT, TH} TH HT Outcome HH TT S Sample Space S = {HH, HT, TH, TT}

Contingency Table Experiment: Toss 2 Coins. Note Faces. 2 Coin 1 Coin nd 2 Coin st 1 Coin Head Tail Total Outcome To be consistent with the Berenson & Levine text, a simple event is shown. Typically, this is not considered an event since it is not an outcome of the experiment. Simple Event (Head on 1st Coin) Head HH HT HH, HT Tail TH TT TH, TT Total HH, TH HT, TT S S = {HH, HT, TH, TT} Sample Space

Event Probability Using Contingency Table Total 1 2 A P(A  B ) P(A  B ) P(A ) 1 1 1 1 2 1 A P(A  B ) P(A  B ) P(A ) 2 2 1 2 2 2 Total P(B ) P(B ) 1 1 2 Joint Probability Marginal (Simple) Probability

Marginal probability Let S be partitioned into m x n disjoint sets Ei and Fj where the general subset is denoted Ei  Fj . Then the marginal probability of Ei is

Tree Diagram H HH H T HT H TH T T TT Experiment: Toss 2 Coins. Note Faces. H HH H T HT Outcome H TH T T TT S = {HH, HT, TH, TT} Sample Space

Discrete Random Variable Possible values (outcomes) are discrete E.g., natural number (0, 1, 2, 3 etc.) Obtained by Counting Usually Finite Number of Values But could be infinite (must be “countable”)

Discrete Probability Distribution 1. List of All possible [x, p(x)] pairs x = Value of Random Variable (Outcome) p(x) = Probability Associated with Value 2. Mutually Exclusive (No Overlap) 3. Collectively Exhaustive (Nothing Left Out) 4. 0  p(x)  1 5.  p(x) = 1

Visualizing Discrete Probability Distributions Table Listing { (0, .25), (1, .50), (2, .25) } # Tails f(x ) p(x ) Count 1 .25 1 2 .50 Experiment is tossing 1 coin twice. 2 1 .25 p(x) Graph Equation .50 n ! .25 p ( x )  p x ( 1  p ) n  x x ! ( n  x ) ! x .00 1 2

Cumulative Distribution Function (CDF)

Binomial Distribution 1. Sequence of n Identical Trials 2. Each Trial Has 2 Outcomes ‘Success’ (Desired/specified Outcome) or ‘Failure’ 3. Constant Trial Probability Trials Are Independent # of successes in n trials is a binomial random variable

Binomial Probability Distribution Function p(x) = Probability of x ‘Successes’ n = Sample Size p = Probability of ‘Success’ x = Number of ‘Successes’ in Sample (x = 0, 1, 2, ..., n)

Binomial Distribution Characteristics Mean n = 5 p = 0.1 Distribution has different shapes. 1st Graph: If inspecting 5 items & the Probability of a defect is 0.1 (10%), the Probability of finding 0 defective item is about 0.6 (60%). If inspecting 5 items & the Probability of a defect is 0.1 (10%), the Probability of finding 1 defective items is about .35 (35%). 2nd Graph: If inspecting 5 items & the Probability of a defect is 0.5 (50%), the Probability of finding 1 defective items is about .18 (18%). Note: Could use formula or tables at end of text to get Probabilities. Standard Deviation n = 5 p = 0.5

Useful Observation 1 For any X and Y

One Binary Outcome Random variable X, one binary outcome Code success as 1, failure as 0 P(success)=p, P(failure)=(1-p)=q E(X) = p

Mean of a Binomial Independent, identically distributed X1, …, Xn; E(Xi)=p; Binomial X = By useful observation 1

Useful Observation 2 For independent X and Y

Useful Observation 3 For independent X and Y cancelled by obs. 2

Variance of Binomial Independent, identically distributed X1, …, Xn; E(Xi)=p; Binomial X =

Useful Observation 4 For any X

Continuous random variable

Continuous Prob. Density Function 1. Mathematical Formula 2. Shows All Values, x, and Frequencies, f(x) f(x) Is Not Probability 3. Properties (Value, Frequency) f(x)  f ( x ) dx  1 x All x a b (Area Under Curve) Value f ( x )  0, a  x  b

Continuous Random Variable Probability ( c  x  d )   f ( x ) dx c f(x) Probability Is Area Under Curve! X c d

Uniform Distribution f(x) x c d 1. Equally Likely Outcomes 2. Probability Density 3. Mean & Standard Deviation f(x) x c d Mean Median

Uniform Distribution Example You’re production manager of a soft drink bottling company. You believe that when a machine is set to dispense 12 oz., it really dispenses 11.5 to 12.5 oz. inclusive. Suppose the amount dispensed has a uniform distribution. What is the probability that less than 11.8 oz. is dispensed?

Uniform Distribution Solution f(x) 1.0 x 11.5 11.8 12.5 P(11.5  x  11.8) = (Base)(Height) = (11.8 - 11.5)(1) = 0.30

Normal Distribution 1. Describes Many Random Processes or Continuous Phenomena 2. Can Be Used to Approximate Discrete Probability Distributions Example: Binomial Basis for Classical Statistical Inference A.k.a. Gaussian distribution

Mean: 평균 Median: 중간값 Mode: 최빈값 Normal Distribution 1. ‘Bell-Shaped’ & Symmetrical 2. Mean, Median, Mode Are Equal 4. Random Variable Has Infinite Range Mean: 평균 Median: 중간값 Mode: 최빈값 * light-tailed distribution

Probability Density Function f(x) = Frequency of Random Variable x  = Population Standard Deviation  = 3.14159; e = 2.71828 x = Value of Random Variable (-< x < )  = Population Mean

Effect of Varying Parameters ( & )

Normal Distribution Probability Probability is area under curve!

Infinite Number of Tables Normal distributions differ by mean & standard deviation. Each distribution would require its own table. That’s an infinite number!

Standardize the Normal Distribution Standardized Normal Distribution One table!

Intuitions on Standardizing Subtracting  from each value X just moves the curve around, so values are centered on 0 instead of on  Once the curve is centered, dividing each value by >1 moves all values toward 0, pressing the curve

Standardizing Example Normal Distribution

Standardizing Example Normal Distribution Standardized Normal Distribution

Obtaining the Probability Standardized Normal Probability Table (Portion) .02 .0478 0.1 .0478 Shaded area exaggerated Probabilities