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Expectation and Properties of Random Variables

This handout provides a review of expectations, variances, moment generating functions, and properties of random variables. It includes examples and explanations for understanding key concepts in calculus.

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Expectation and Properties of Random Variables

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  1. Handout Ch 4 實習

  2. 微積分複習第二波(1) • Example

  3. 微積分複習第二波(2) • 變數變換 • Example 這是什麼鬼

  4. 微積分複習第二波(3) • Ch 4積分補充

  5. 歸去來析 ( 乾脆去死,台語)

  6. Expectation of a Random Variable • Discrete distribution • Continuous distribution • E(X) is called expected value, mean or expectation of X. • E(X) can be regarded as being the center of gravity of that distribution. • E(X) exists if and only if • E(X) exists if and only if Whenever X is a bounded random variable, then E(X) must exist.

  7. The Expectation of a Function • Let , then • Let , then • Suppose X has p.d.f as follows: • Let it can be shown that

  8. Example 1 (4.1.3) • In a class of 50 students, the number of students ni of each age i is shown in the following table: • If a student is to be selected at random from the class, what is the expected value of his age

  9. Solution • E[X]=18*0.4+19*0.44+20*0.08+21*0.06+ 25*0.02=18.92

  10. Properties of Expectations • If there exists a constant such that • If are n random variables such that each exists, then • For all constants • Usually Only linear functions g satisfy • If are nindependent random variable such that each exists, then

  11. Example 2 (4.2.7) • Suppose that on each play of a certain game a gambler is equally likely to win or to lose. Suppose that when he wins, his fortune is doubled; and when he loses, his fortune is cut in half. If he begins playing with a given fortune c, what is the expected value of his fortune after n independent plays of the game?

  12. Solution

  13. Properties of the Variance • Var(X ) = 0 if and only if there exists a constant c such that Pr(X = c) = 1. • For constant a and b, . Proof :

  14. Properties of the Variance • If X1 , …, Xnare independent random variables, then • If X1,…, Xn are independent random variables, then

  15. Example 3 (4.3.4) • Suppose that X is a random variable for which E(X)=μ and Var(X)=σ2. • Show that

  16. Solution

  17. Moment Generating Functions • Consider a given random variable X and for each real number t, we shall let . The function is called the moment generating function (m.g.f.) of X. • Suppose that the m.g.f. of X exists for all values of t in some open interval around t = 0. Then, • More generally,

  18. Properties of Moment Generating Functions • Let X has m.g.f. ; let Y = aX+b has m.g.f. . Then for every value of t such that exists, Proof: • Suppose that X1,…, Xn are n independent random variables; and for i = 1,…, n, let denote the m.g.f. of Xi. Let , and let the m.g.f. of Y be denoted by . Then for every value of t such that exists, we have Proof:

  19. The m.g.f. for the Binomial Distribution • Suppose that a random variable X has a binomial distribution with parameters n and p. We can represent X as the sum of n independent random variables X1,…, Xn. • Determine the m.g.f. of

  20. Uniqueness of Moment Generating Functions • If the m.g.f. of two random variables X1 and X2 are identical for all values of t in an open interval around t = 0, then the probability distributions of X1 and X2 must be identical. • The additive property of the binomial distribution Suppose X1 and X2 are independent random variables. They have binomial distributions with parameters n1 and p and n2 and p. Let the m.g.f. of X1 + X2 be denoted by . The distribution of X1 + X2 must be binomial distribution with parameters n1 + n2 and p.

  21. Example 4 (4.4.8) • Suppose that X is a random variable for which the m.g.f. is as follows: • Find the mean and the variance of X

  22. Solution

  23. Properties of Variance and Covariance • If X and Y are random variables such that and , then • Correlation only measures linear relationship. • Two random variables can be dependent, but uncorrelated. • Example: Suppose that X can take only three values –1, 0, and 1, and that each of these three values has the same probability. Let Y=X 2. So X and Y are dependent. E(XY)=E(X 3)=E(X)=0, so Cov(X,Y) = E(XY) – E(X)E(Y)=0 (uncorrelated).

  24. Example 5 (4.6.11) • Suppose that two random variables X and Y cannot possibly have the following properties: E(X)=3, E(Y)=2, E(X2)=10. E(Y2)=29, and E(XY)=0

  25. Solution

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