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Random Variables Math 480 Lecture By Sara Billey Outline • Last time: Defined sample spaces, probability distribution functions independence, uniform distribution, Bernoulli process, and histograms. • Discussed Law of Large Numbers: Roughly, histograms with enough data will approach the graph of the probability distribution function P: S -> [0,1]. • Today: Random variables and Expectations. • Model the game Pass the Pigs Random Variables • Def: Given a random process with sample space S and probability P: S [0,1], a random variable N is any function N: S Set of Numbers such as the integers, reals, or complexes. • Every random variable has associated probabilities for each event of the form (N=i) = { w in S : N(w)=i}. P(N=i) =P({ w in S : N(w)=i}) = S P(w). Random Variables • Def: Given a random process with sample space S and probability P: S [0,1], a random variable N is any function N: S Set of Numbers such as the integers, reals, or complexes. • Example: Nbox = height in mm of object Height of tennis ball = 68 mm Height of ping pong ball = 40 mm Height of marble= 22 mm Height of zometool= 9 mm Random Variables • Example: Nbox = height in mm of object : S {9,22,40,68}. P(Nbox = P(Nbox = P(Nbox = P(Nbox = 68) “=“ 16/20 40) “=“ 2/20 22) “=“ 1/20 9) “=“ 1/20 Random Variables • What other random variables (RV’s) could we measure on the same random process? Expectation • Def: The expectation of a random variable X: S {x1, x2, …, xk} with P(X = xi) = pi is • Ex: E(Nbox) “=“ 68 (16/20) + 40 (2/20) + 22 (1/20) + 9(1/20) = 59.95 Expectation • Def: The expectation of a random variable X: S {x1, x2, …, xk} with P(X = xi) = pi is • Ex: E(Nbox) “=“ 59.95 = (68 *16 + 40 *2+ 22 + 9) /20. By the Law of Large Numbers, E[X] should be close to the average value of X on a large number of samples from the random process. Basic Fact of Expectations • Basic Fact: If we have two random variables on the same random process X: S {n1, n2, …, nk} Y: S {n1, n2, …, nk}, then E[X+Y] = E[X]+E[Y]. Proof: First, E[X] = S X(w) P(w) and E[Y] = S Y(w) P(w) as a consequence of the definition. Therefore, E[X+Y] = S (X(w)+Y(w)) P(w) = S X(w) P(w) + S Y(w) P(w) = E[X]+E[Y]. Doesn’t even matter if X and Y are independent! (Try E[X + X2]) Expectations in Modeling Decision Problem: Should we buy a megamillions lottery ticket? Cost = $1 to play. Pick 5 numbers between 1 and 75 plus one powerball number between 1 and 15. Jackpot = $161,000,000. Expectations in Modeling Decision Problem: Should we buy a Megamillions Lottery ticket? Next drawing is at 11pm tonight! Let W = net profit from one ticket. E[W] = $161M (1/258890850) + $1M (1/18492204) + $5K(1/739688) + $500 (1/52835) + 50(1/10720) + 5(1/766) + 5(1/473) + 2(1/56) + 1(1/21) -1 = -0.20272 Check That! Expectations in Modeling Decision Problem: If Megamillions gets up to $686M again, how many tickets should I buy? Let W = net profit from one ticket. E[W] = $686M (1/258890850) + $1M (1/18492204) + $5K(1/739688) + $500 (1/52835) + 50(1/10720) + 5(1/766) + 5(1/473) + 2(1/56) + 1(1/21) -1 = 1.82516 Expectations in Modeling Decision Problem: If Megamillions gets up to $686M again, how many tickets should I buy? If E[W]= 1.82516, then what is the expected net value of 5 tickets? E[W+W+W+W+W] = 5 * 1.82516 = 9.12580. E[50* W] = 50 * 1.82516 = 91.2580. E[500* W] = 500 * 1.82516 = 912.580. (Don’t forget to keep in mind the probability of winning!) Pass the Pigs