Handout Ch 4 實習

微積分複習第二波(1) Example

微積分複習第二波(2) 變數變換 Example 這是什麼鬼

微積分複習第二波(3) Ch 4積分補充

歸去來析 ( 乾脆去死,台語)

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.

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

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 Agei 18 19 20 21 25 ni 22 4 3 1

Solution E[X]=18*0.4+19*0.44+20*0.08+21*0.06+ 25*0.02=18.92 Agei 18 19 ni 22 4 3 1 Pi 0.4 0.44 0.08 0.06 0.02

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 n independent random variable such that each exists, then

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?

Solution

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 :

Properties of the Variance If X1 , …, Xn are independent random variables, then If X1,…, Xn are independent random variables, then

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

Solution

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,

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

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

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.

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

Solution

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).

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

Solution