Example. A casino offers the following bets (the fairest bets in the casino!) 1 You get $0 (i.e., you can walk away)


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1 : Three bets Math 45 Introduction to Probability Lecture 5 Kenneth Harris Department of Mathematics University of Michigan February, 009. A casino offers the following bets (the fairest bets in the casino!) You get $0 (i.e., you can wal away) You get $0 with probability, and otherwise pay $0. You get $0, 000 with probability 0, and otherwise pay $0. All three bets have the same expectation: 0, but they differ in how spread out they are about their mean: This bet is not spread out at all. This bet is symmetrically disposed not too far from its mean. This bet is very spread out. Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, 009 / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, 009 / Definition: Standard Deviation is a measure of the spread of a random variable. The standard deviation is more lie a distance function than variance. Definition Let X be a random variable with distribution p X and mean µ. The variance of X, denoted by Var(X), is defined as Var(X) (r µ) p X (r). The variance is a weighted average of the squared distance from the mean. The variance of X is standardly written σ (X), where σ(x) is often used as a measure of the spread of X. Definition The standard deviation of a random variable X is the positive square root of the variation: SD(X) Var(X). Equivalently, if the mean of X has distribution p X and mean µ, its pmf p(r) then SD(X) (r µ) p X (r). The standard deviation of X is standardly written σ(x). Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, /
2 Three bets revisited Throwing Dice revisted. Our three bets with mean 0: You get $0 (X ). You get $0 with probability, and otherwise pay $0 (X ). You get $0, 000 with probability 0, and otherwise pay $0 (X ). They differ dramatically in the variance: Var(X ) 0 Var(X ) (0 0) + ( 0 0) 00 Var(X ) ( 0 )(0 4 0) + 0 ( 0 0) 0 8 and so too in standard variation, SD(X ) 0 SD(X ) 0 SD(X ) Let X be the outcome of the throw of a single fair die. The distribution of X is So, E[X] Var(X) p X () ( 7 ) SD(X).7 Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Caveat on Caveat on There is a tendancy to believe the standard deviation gives a good measure of dispersion: expected deviation from the mean. However, squaring big deviations tend to dominate the sum. Consider the random variable X with distribution: So, p X () 9 0 p X (0) 0 E[X] Var(X) (.9) (8.) SD(X).7 If deviation from the mean varies alot, then standard deviation is NOT a good measure of the expected deviation. A closer formula might be r µ p X (r). In this case, our previous example gives an average deviation SD(X).7 0 While this version of dispersion of values has good statistical properties, it is much more difficult mathematical properties, so is little used. Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, /
3 . A roulette wheel is divided into 8 slotted sectors. 8 are red, 8 are blac and are green, 6 are numbered to 6, together with one mared 0 and one mared 00. A croupier spins the wheel and throws an ivory ball. The success of a bet depends into which slot the ball falls. Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, 009 / Compare the following two bets: Bet $ on red at even money (X). Bet $ on the number 7 at 5 : (Y ). You win $6 on a $ bet. We compare the expected values E[X] ( ) E[Y ] ( ) You can expect to lose about 5 cents a bet in either case. The variance is dramatically different though Var[X] ( + 9 ) ( + 9 ) Var[Y ] ( + 9 ) (5 + 9 ) 8. Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, 009 / I played both bets 0,000 times. The standard mean of the outcome of the bet and the variance of the outcomes based on the computed mean is as follows: Even money on red: Mean: (E[X] 0.05) : (Var(x) 0.997) 5 : on the number 7: Mean: (E[X] 0.05) :.5 (Var(x).) I placed a 000 bets at the roulette table and computed my total winnings. Here is the mean and variance over 00 plays of each type of bet. Even money on red for 000 bets: Mean winnings: 5.5 (000E[X] 5) : 70.5 (000Var(x) 997) 5 : on the number 7 for 000 bets: Mean: 5.9 (000E[X] 5) : 9, 48 (000Var(x), 0) This suggests that running these independent trials 000 times leads to a 000fold increase in expectation and variance. This is the case, and we will loo into this relationship in Chapter 7. Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, 009 / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, /
4 Linearity properties Proof of Theorem Let X be a random variable with distribution p X and let Y ax + b with distribution p Y. See Ross, Corollary 4. and page 50. Theorem Let a and b be constants. For any random variable X, E[aX + b] ae[x] + b Var(aX + b) a Var(X) Since ar + b ar + b if and only if r r, it follows: p X (r) p Y (ar + b). So, E[aX + b] E[Y ] (ar + b) p Y (ar + b) a r:p Y (ar+b)>0 (ar + b) p X (r) r p X (r) + b p X (r) ae[x] + b. Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Proof of Theorem continued : Temperature readings Let µ E[X], so that E[Y ] E[aX + b] a µ + b. Since Var(aX + b) Var(Y ), Var(aX + b) (ar + b (a µ + b)) p Y (ar + b) r:p Y (ar+b)>0 (ar a µ) p X (r) a (r µ) p X (r) a Var(X). In a certain manufacturing process, the (Fahrenheit) temperature never varies more than two degrees from 6 F. The temperature is a random variable X with distribution Find the following temp ( F ) prob temp ( C) Compute E[X], Var(X) and SD(X). It is decided to convert the temperature readings to Celcius, so that Y 5 9 (X ). Compute E[Y ], Var(Y), and SD(Y ). Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, /
5 : Temperature readings In Fahrenheit E[X] Var(X) ( ) 0 + ( ) (0) 0 + () 0 + () SD(X) 5.. In Celcius E[Y ] E[ 5 9 (X )] 5 60 E[X] Var(Y ) Var( 5 9 SD(Y ) 5 0 (X )) Var(X) Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Definition We generalize the previous result to arbitrary functions. Definition Let X be a random variable over sample space S, and g : R R. We write g(x) for the random variable Y on S defined by Y (s) g(x(s)) for all s S. ax + b is the random variable: ax(s) + b for s S, X is the random variable: ( X(s) ) for s S. Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, 009 / Proposition Proof of Proposition See Ross, Proposition 4. Proposition Let X be a random variable with distribution p X. For any realvalued function g, E[g(X)] g(r)p(r). Let X be a random variable with distribution, and let Y g(x) be the random variable with distribution p Y. It is not necessarily true that p Y (g(r)) p X (r), since g may map several numbers to the same value. For example, let X be the random variable with distribution p X ( ) p X (0) p X (). Then Y X is the random variable with distribution p Y (0) p Y (). So, the distributions for X and Y are different. Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, 009 / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, 009 /
6 Proof of Proposition Corollary However, the following is true for each real number q: p Y (q) p X (r). r:g(r)q By regrouping the summation: g(r) p X (r) g(r) p X (r) q:p Y (q)>0 r:g(r)q q:p Y (q)>0 q:p Y (q)>0 q r:g(r)q q p Y (q) E[Y ] E[g(X)]. p X (r) The following corollary maes computation of variance much easier. Corollary Let X be a random variable with expected value µ. Then Var(X) E[(X µ) ] E[X ] µ That is, Var(X) E[X ] ( E[X] ) Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Proof of Corollary We defined the variance of a random variable X with distribution p X and expected value µ by Var(X) (r µ) p X (r) E[(X µ) ], by the previous theorem with g(x) (x µ). Var(X) (r µ) p X (r) (r rµ + µ ) p X (r) r p X (r) µ r p X (r) + µ p X (r) E[X ] µ + µ E[X ] µ E[X ] ( E[X] ). : Algebra. Let X be a random variable with E[X] 0 and Var(X) 4. Compute the following (a) E[X ] (b) E[X + 0] and E[ X] (c) Var(X + 0) and Var( X) (d) SD(X), SD(X + 0) and SD( X). Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, /
7 : Algebra : children Given: E[X] 0 and Var(X) 4 (a). Since Var(X) E[X ] ( E[X] ), (b). (c). (c). E[X ] Var(X) + ( E[X] ) E[X + 0] E[X] E[ X] E[X] 0 Var(X + 0) 9Var(X) 6 V ( X) V (X) 4 SD(X) 4 SD(X + 0) 6 SD( X) 4 Find the expected value and the variance for the number of boys and girls in a royal family that has children until there is a boy or until there are three children, whichever comes first. Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / continued Solution. Let B (G) be the random variable which counts the boys (girls). Then, the distributions are given by the table: B p B G p G G p G A die is loaded so that the probability of a face coming up is proportional to the number on the face. Find the expected value, variance and standard deviation of the face value. Solution. Let X be the random value of the face value. X has distribution given by E[B] 7 8 E[G] 7 8 E[B ] 7 8 E[G ] 5 8 Var[B] 7 64 Var[G] 7 64 p X () since Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, 009 / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, 009 /
8 continued : heads Compute. p X () Recall, for a fair die these values are 6. E[X] 4. E[X ] 49 Var(X) 4 6. SD(X).46 E[X].5 Var(X).9 SD(X).7 What is the expected number of heads in 4 tosses of a fair coin? What is the standard deviation? Solution. Let X be the random variable counting heads. X has distribution ( ) 4 p X () So, E[X] E[X ] 4 ( ) ( ) Var(X) SD(X) Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, 009 / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / 4: heads 4: heads What is the expected number of heads in 6 tosses of a fair coin? What is the standard deviation? Solution. Let X be the random variable counting heads. X has distribution ( ) 6 p X () So, E[X] E[X ] Var(X) 6 ( ) ( ) SD(X).5 Random variables for counting heads for various numbers of tosses. X E[X] Var(X) We explore this trend further on Monday. Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, /
9 5: Jac and Jill, redoux 5 continued Solution. Let X count the number of rounds played. The probability of a winning throw in any given round is Jac and Jill are playing the game of Heads. Jill s coin is biased of the time heads, Jac s coin is a fair coin. Jac, always the gentleman, allows Jill to toss first. They agree to stop after four rounds. What is the expected number of rounds? What is the variance? +. So, the probability that play goes rounds (when < 4) is p X () ( ) () The last round also has the possibility of no winner: p X (4) ( ) 4 ( ) + ( ) 4 () 4 Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, / 5 continued The probabilities for X are n p X n p X The expected number of rounds of the game: E[X] E[X ] 76 7 Var(X) 0.94 SD(X) 0.97 Kenneth Harris (Math 45) Math 45 Introduction to Probability Lecture 5 February, /
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