Alex Psomas: Lecture 16. Random Variables Questions about outcomes - PowerPoint PPT Presentation

Alex Psomas: Lecture 16. Random Variables Questions about outcomes ... Experiment: roll two dice. Sample Space: { ( 1 , 1 ) , ( 1 , 2 ) ,..., ( 6 , 6 ) } = { 1 ,..., 6 } 2 How many dots? Random Variables Experiment: flip 100 coins. Sample Space: { HHH ··· H , THH ··· H ,..., TTT ··· T } 1. Random Variables. How many heads in 100 coin tosses? 2. Distributions. I Regrade requests open. Experiment: choose a random student in cs70. I Quiz due tomorrow. 3. Combining random variables. Sample Space: { Peter , Phoebe ,..., } 4. Expectation I Quiz coming out today. What midterm score? I Non-technical office hours tomorrow 1-3pm. Experiment: hand back assignments to 3 students at random. I Anonymous questionnaire tonight or tomorrow. Sample Space: { 123 , 132 , 213 , 231 , 312 , 321 } How many students get back their own assignment? In each scenario, each outcome gives a number. The number is a (known) function of the outcome. Random Variables. Example 1 of Random Variable Example 2 of Random Variable A random variable, X , for an experiment with sample space Ω is a function X : Ω ! ℜ . Thus, X ( · ) assigns a real number X ( ω ) to each ω 2 Ω . Experiment: roll two dice. Sample Space: { ( 1 , 1 ) , ( 1 , 2 ) ,..., ( 6 , 6 ) } = { 1 ,..., 6 } 2 Experiment: flip three coins Random Variable X : number of pips. Sample Space: { HHH , THH , HTH , TTH , HHT , THT , HTT , TTT } X ( 1 , 1 ) = 2 Winnings: if win 1 on heads, lose 1 on tails: X X ( 1 , 2 ) = 3, X ( HHH ) = 3 X ( THH ) = 1 X ( HTH ) = 1 X ( TTH ) = � 1 . . X ( HHT ) = 1 X ( THT ) = � 1 X ( HTT ) = � 1 X ( TTT ) = � 3 . X ( 6 , 6 ) = 12, X ( a , b ) = a + b , ( a , b ) 2 Ω . The function X ( · ) is defined on the outcomes Ω . A random variable X is not random, not a variable! What varies at random (from experiment to experiment)? The outcome!

Number of dots in two dice. Distribution Handing back assignments “What is the likelihood of seeing n dots?” The probability of X taking on a value a . Experiment: hand back assignments to 3 students at random. Definition: The distribution of a random variable X , is Sample Space: Ω = { 123 , 132 , 213 , 231 , 312 , 321 } { ( a , Pr [ X = a ]) : a 2 A } , where A is the range of X . How many students get back their own assignment? Random Variable: values of X ( ω ) : { 3 , 1 , 1 , 0 , 0 , 1 } Distribution: 8 0 , w.p. 1 / 3 < X = 1 , w.p. 1 / 2 0 . 4 3 , w.p. 1 / 6 : 0 . 2 Pr [ X = a ] := Pr [ X � 1 ( a )] where X � 1 ( a ) := { ω | X ( ω ) = a } . 0 0 1 2 3 Pr [ X = 10 ] = 3 / 36 = Pr [ X � 1 ( 10 )] = ∑ ω 2 X � 1 ( 10 ) Pr [ ω ] Pr [ X = 8 ] = 5 / 36 = Pr [ X � 1 ( 8 )] . Flip three coins Number of dots. The Bernoulli distribution Experiment: flip three coins Sample Space: { HHH , THH , HTH , TTH , HHT , THT , HTT , TTT } Experiment: roll two dice. Flip a coin, with heads probability p . Winnings: if win 1 on heads, lose 1 on tails. X Random variable X : 1 is heads, 0 if not heads. Random Variable: { 3 , 1 , 1 , � 1 , 1 , � 1 , � 1 , � 3 } X has the Bernoulli distribution. Distribution: We will also call this an indicator random variable . It indicates whether the event happened. 8 � 3 , w. p. 1 / 8 > > � 1 , w. p. 3 / 8 Distribution: < 0 . 4 X = 1 , w. p. 3 / 8 > 0 . 3 ( > 3 w. p. 1 / 8 1 w.p. p : X = 0 . 2 0 w.p. 1 � p 0 . 1 0 � 3 � 2 � 1 0 1 2 3

The binomial distribution. The binomial distribution. Combining Random Variables. Let X and Y be two RV on the same probability space. Flip n coins with heads probability p . That is, X : Ω ! ℜ assigns the value X ( ω ) to ω . Also, Random variable: number of heads. Y : Ω ! ℜ assigns the value Y ( ω ) to ω . Binomial Distribution: Pr [ X = i ] , for each i . Then Z = X + Y is a random variable: It assigns the value How many sample points in event “ X = i ”? � n Z ( ω ) = X ( ω )+ Y ( ω ) i heads out of n coin flips = � ) i Sample space: Ω = { HHH ... HH , HHH ... HT ,... } to outcome ω . What is the probability of ω if ω has i heads? Experiment: Roll two dice. X = outcome of first die, Y = Probability of heads in any position is p . outcome of second die. Probability of tails in any position is ( 1 � p ) . So, we get Pr [ ω ] = p i ( 1 � p ) n � i . X ( a , b ) = a and Y ( a , b ) = b for ( a , b ) 2 Ω = { 1 ,..., 6 } 2 . Probability of “ X = i ” is sum of Pr [ ω ] , ω 2 “ X = i ”. Then Z = X + Y = sum of two dice is defined by ✓ n ◆ p i ( 1 � p ) n � i , i = 0 , 1 ,..., n : B ( n , p ) distribution Pr [ X = i ] = i Z ( a , b ) = X ( a , b )+ Y ( a , b ) = a + b . Combining Random Variables Expectation. Expectation - Intuition How did people do on the midterm? Flip a loaded coin with Pr [ H ] = p a large number N of times. Distribution. We expect heads to come up a fraction p of the times and tails a fraction 1 � p . Summary of distribution? Say that you get 5 for every H and 3 for every T . Average! Other random variables: If there are N H outcomes equal to H and N T outcomes equal to I X k : Ω ! ℜ is defined by X k ( ω ) = [ X ( ω )] k . T , you collect In the dice example, X 3 ( a , b ) = a 3 . 5 ⇥ N H + 3 ⇥ N T . I ( X � 2 ) 2 + 4 XY assigns the value Your average gain per experiment is ( X ( ω ) � 2 ) 2 + 4 X ( ω ) Y ( ω ) to ω . I g ( X , Y , Z ) assigned the value g ( X ( ω ) , Y ( ω ) , Z ( ω )) to ω . 5 N H + 3 N T . N Since N H N ⇡ p = Pr [ X = 5 ] and N T N ⇡ 1 � p = Pr [ X = 3 ] , we find that the average gain per outcome is approximately equal to 5 Pr [ X = 5 ]+ 3 Pr [ X = 3 ] . We use this frequentist interpretation as a definition.

Expectation - Definition Expectation: A Useful Fact An Example Theorem: Definition: The expected value of a random variable X is Flip a fair coin three times. E [ X ] = ∑ E [ X ] = ∑ X ( ω ) ⇥ Pr [ ω ] . a ⇥ Pr [ X = a ] . Ω = { HHH , HHT , HTH , THH , HTT , THT , TTH , TTT } . ω 2 Ω a X = number of H ’s: { 3 , 2 , 2 , 2 , 1 , 1 , 1 , 0 } . a in the range of X . Proof: = ∑ The expected value is also called the mean. E [ X ] a ⇥ Pr [ X = a ] Thus, a According to our intuition, we expect that if we repeat an = ∑ X ( ω ) Pr [ ω ] = { 3 + 2 + 2 + 2 + 1 + 1 + 1 + 0 } ⇥ 1 ∑ Pr [ ω ] a ⇥ ∑ 8 . experiment a large number N of times and if X 1 ,..., X N are the a ω : X ( ω )= a successive values of the random variable, then ω = ∑ ∑ a ⇥ Pr [ ω ] Also, X 1 + ··· + X N a ω : X ( ω )= a ⇡ E [ X ] . = ∑ ∑ N X ( ω ) Pr [ ω ] a ⇥ Pr [ X = a ] = 3 ⇥ 1 8 + 2 ⇥ 3 8 + 1 ⇥ 3 8 + 0 ⇥ 1 ∑ 8 . a ω : X ( ω )= a That is indeed the case, in the same way that the fraction of a = ∑ X ( ω ) Pr [ ω ] times that X = x approaches Pr [ X = x ] . ω This (nontrivial) result is called the Law of Large Numbers. Expectation and Average. Handing back assignments Win or Lose. There are n students in the class; We give back assignments randomly to three students. What is the expected number of students that get their own X ( m ) = score of student m , for m = 1 , 2 ,..., n . assignment back? “Average score” of the n students: add scores and divide by n : Expected winnings for heads/tails games, with 3 flips? The expected number of fixed points in a random permutation. Every time it’s H ,I get 1,. Every time it’s T , I lose 1. Average = X ( 1 )+ X ( 1 )+ ··· + X ( n ) Expected value of a random variable: . n E [ X ] = 3 ⇥ 1 8 + 1 ⇥ 3 8 � 1 ⇥ 3 8 � 3 ⇥ 1 8 = 0 . E [ X ] = ∑ a ⇥ Pr [ X = a ] . Experiment: choose a student uniformly at random. a Uniform sample space: Ω = { 1 , 2 , ··· , n } , Pr [ ω ] = 1 / n , for all ω . Can you ever win 0? Random Variable: midterm score: X ( ω ) . For 3 students (permutations of 3 elements): Expectation: Apparently: expected value is not a common value, by any X ( ω ) 1 E ( X ) = ∑ X ( ω ) Pr [ ω ] = ∑ n . means. Pr [ X = 3 ] = 1 / 6 , Pr [ X = 1 ] = 3 / 6 , Pr [ X = 0 ] = 2 / 6 . ω ω Hence, E [ X ] = 3 ⇥ 1 6 + 1 ⇥ 3 6 + 0 ⇥ 2 6 = 1 . Average = E ( X ) . Our intuition matches the math.