4.1 Expected value of a discrete random variable. Suppose we shall win $i if the die lands on i. What is our ideal average winnings per toss?

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1 Chapter 4 Mathematical Expectation 4.1 Expected value of a discrete random variable Examples: (a) Given 1, 2, 3, 4, 5, 6. What is the average? (b) Toss an unbalanced die 1 times. Lands on Probab Suppose we shall win $i if the die lands on i. What is our ideal average winnings per toss? Let X be our winning per toss. x f(x) The average of X is = = 3.4 Definition 4.1 If X is a discrete r.v. taking on values x 1, x 2,... and f(x) is the probability density function of X, the expected value ( or mathematical expectation or mean ) of 4-1

2 X, denoted by E(X), or µ X, is defined by E(X) = i x i f(x i ) = x xf(x) The expected value of X is a weighted average of the possible values that X can take on, each value being weighted by the probability that X assumes. Example 4.1 The probability density function of X is given by Find EX EX = (1 p) + 1 p = p. x 1 f(x) 1 p p Example 4.2 The indicator function of an event A, denoted by I A or I(A), is defined by { 1 if A occurs I A = if A doesn t occur Find E(I A ). Since I A 1 f(x) P (A ) P (A) we have E(I A ) = P (A) 4-2

3 Example 4.3 Chuck-a-luck is a popular game in which three fair dice are rolled. A bet placed on one of the numbers 1 through 6 has a payoff that depends on the number of times the number appears on the three dice. Suppose one bets $ 1 on the number 4. Let X be the payoff. Then X equals -$1 if none of the three dice shows a 4; otherwise X = $k, where k is the number of dice that show the number 4. Find E(X). P (X = 1) = (5/6) 3 ( ) 3 P (X = 1) = (1/6)(5/6) 2 1 ( ) 3 P (X = 2) = (1/6) 2 (5/6) 2 ( ) 3 P (X = 3) = (1/6) 3 3 E(X) = Example 4.4 Over the weekend of Feb. 1995, the financial world was startled to learn the Baring Brothers, a highly respected merchant bank that had been in business for 233 years, collapsed into bankruptcy as a result of a series of highly imprudent financial transactions by one of its junior bond traders. It was revealed that the trader was attempting to recoup his previous losses by committing ever-increasing sums of the bank s capital to high-risk bets on future value of several indices based on the prices of a complex hodgepodge of stocks and bonds. Here is a mathematical model. Consider a gambler who has a probability 1/2 of winning on any one of s sequence of bets. Suppose that the gambler s initial bet is $1 and that he doubles his bet each time until he wins for the first time. Doubling the bet means that he bets 2 x 1 dollars on the x th bet. Let X be the amount of capital the gambler needs to play this game until he wins for the first time. What is E(X)? E(X) = k=1 (1 2 k ) = P (X = 2 k 1) = (1/2) k, k = 1, 2, 3,

4 4.2 Expected value of a continuous random variable Let X be a continuous r.v. with p.d.f. f(x). Then P (x X x + dx) = x+dx x f(t)dt f(x) dx 2 Definition 4.2 The expected value of X is defined by EX = xf(x) dx Example 4 The p.d.f. of X is given by 1 if x 1 1 f(x) = otherwise Find E(X). Example 4.6 Find E(X). E(X) = f(x) = = 1 xf(x) dx x dx = 1 2 { 1 β e x/β for x > otherwise E(X) = x 1 β e x/β dx = β x n e x dx = n! 4-4

5 4.3 Expectation of a function of a r.v Example 4.7 Suppose the probability density function of X is given by x 1 1 f(x).2.3 Compute E(X 2 ) Let Y = X 2. x -1 1 f(x).2.3 = y = x2 1 1 f Y (y).2.3 = y 1 p Y (y) Hence, EY = To compute E(X 2 ), actually no need to find f Y!! If X is a discrete r.v., then for any real-valued function g E[g(X)] = x g(x)f(x) provided that x g(x) f(x) <. If X is a continuous r.v., then the expected value of g(x) is given by provided that E[g(X)] = g(x) f(x)dx <. 4-5 g(x)f(x)dx

6 Properties: E (ag(x) + bh(x) + c) = aeg(x) + beh(x) + c E(aX + b) = ae(x) + b E(X EX) = Example 4.8 Find E(X 2 ). x f(x) E(X 2 ) = ( 2) = 29 8 Example 4.9 Find E(X). f(x) = { 1 β e x/β for x > otherwise E(X 2 ) = x 2 1 β e x/β dx = β 2 t 2 e t dt = 2β 2 t = x/β 4.4 Variance The probability density functions of X, Y and Z are given by, respectively x f X (x) 1 1 y -1 1 f Y (y) z f Z (z)

7 It is easy to see that E(X) = E(Y ) = E(Z) =. Although EX yields the average of the possible values of X, it does not tell us anything about the variation, or spread, of these values. One way of measuring the possible variation of X is to look at how far apart X is from its mean on the average. Definition 4.3 The variance of X, denoted by Var(X), or σ 2, is defined by X Var(X) = E(X EX) 2 A useful property: Var(X) = E(X 2 ) (EX) 2 a, b R, Var(a X + b) = a 2 Var(X) Let X be a r.v. with mean µ and variance σ 2, and put Y = X µ. Then σ EY =, Var(Y ) = 1 Example 4.1 If E(X) = 1 and Var(X) = 5, find (a) E[(2 + X) 2 ] (b) Var(1 3X) (a) E(2 + X) 2 = 4 + 4E(X) + E(X 2 ) = = 14 (b) Var(X) = ( 3) 2 5 = 45 Definition 4.4 The standard deviation of X, denoted by σ X, is defined to be Var(X). Definition 4 (a) The k th moment about the origin of a r.v. X is defined by µ k = E(X k ), r = 1, 2,... (b) The k th moment about the mean of a r.v. X is defined by µ k = E[(X µ) k ], r = 1, 2,

8 4 Moment-generating functions Definition 4.6 The moment generating function of X is defined by M(t) = M X (t) = E(e tx ), t R 1 If there is a δ >, such that M X (t) = E[e tx ] < for every t < δ then for k = 1, 2,... E[X k ] = M (k) () = dk dt M(t) k t= Let X and Y be two r.v. s. If there exists a δ >, such that then X and Y have the same distribution. M X (t) = M Y (t) < for all t < δ Example 4.11 Find M(t). f(x) = { 1 θ e x/β for x > otherwise M(t) = = = e tx 1 β e x/β dx e x(1/β t) 1 β dx { 1 1 tβ, t < 1/β t 1/β 4-8

9 Example 4.12 Let X have the p.d.f. ( ) n f(k) = p k (1 p) n k, k k =, 1,..., n Find the moment generating function, mean and variance of X. M(t) = (pe t + 1 p) n, EX = np, Var(X) = np(1 p) 4.6 Chebyshev s inequality Let X be any random variable with finite mean. Then a >, P ( X a) E X a In particular, if X has mean µ and variance σ 2 x >, P ( X µ x) σ2 x 2 4-9