Expected Value and Variance

Transcription

1 Expected Value and Variance MATH 472 Financial Mathematics J Robert Buchanan 2018

2 Objectives In this lesson we will learn: the definition of expected value, how to calculate the expected value of a random variable, the properties of expected value, the definition of variance, the properties of variance.

3 Expected Value of a Random Variable Definition If X is a discrete random variable with probability mass function f X (x) then the expected value of X is denoted E (X) and defined as E (X) = x Ω(x f X (x)).

4 Example What is the expected value of the number of female children in a family of 5 children?

5 Example What is the expected value of the number of female children in a family of 5 children? If X represents the number of female children in a family with five children then E (X) = x Ω (x f X (x)) = ( 5 x x=0 = 5! 32 = x=0 ( 5 x ) ( 1 2 x (x!)(5 x)! ) x ( ) ) 1 5 x 2

6 Expected Value of a Function of a Random Variable Definition If G is a function of the random variable X, then the expected value of G is E (G(X)) = x Ω G(x)f X (x).

7 Example What is the expected value of the square of the number of female children in a family of 5 children?

8 Example What is the expected value of the square of the number of female children in a family of 5 children? If X represents the number of female children in a family with five children then ( E X 2) = ( ) x 2 f X (x) x Ω = ( 5 x 2 x=0 = 7.5 ( 5 x ) ( 1 2 ) x ( ) ) 1 5 x 2

9 Linearity of Expected Value Theorem If X is a random variable and a is a constant, then E (a X) = a E (X).

10 Linearity of Expected Value Theorem If X is a random variable and a is a constant, then E (a X) = a E (X). Proof. E (a X) = x Ω ((ax) f X (x)) = a x Ω (x f X (x)) = a E (X).

11 Joint Probability Function Definition If X and Y are discrete random variables the joint probability mass function of X and Y is denoted f X,Y (x, y) where f X,Y (x, y) = P (X = x, Y = y) = P ((X = x) (Y = y)).

12 Joint Probability Function Definition If X and Y are discrete random variables the joint probability mass function of X and Y is denoted f X,Y (x, y) where f X,Y (x, y) = P (X = x, Y = y) = P ((X = x) (Y = y)). A joint probability mass function has the properties: 1. 0 f X,Y (x, y) 1, 2. f X,Y (x, y) = 1. (x,y) Ω

13 Marginal Probability Definition If the joint probability mass function of (X, Y ) is f X,Y (x, y) then the function f X (x) = f X,Y (x, y) y Ω is called the marginal probability distribution of X. We may define a marginal probability distribution for Y similarly.

14 Expected Value of a Sum (1 of 2) Theorem If X 1, X 2,..., X k are random variables then E (X 1 + X X k ) = E (X 1 ) + E (X 2 ) + + E (X k ).

15 Expected Value of a Sum (2 of 2) Proof. If k = 2 then ( (x + y)fx,y (x, y) ) E (X + Y ) = x = x = x y x f X,Y (x, y) + y y x f X,Y (x, y) + y y y f X,Y (x, y) x y f X,Y (x, y) x = x x f X (x) + y y f Y (y) = E (X) + E (Y ). The general case then holds by induction on k.

16 Expected Value of a Sum of Functions Corollary Let X 1, X 2,..., X k be random variables and let F i be a function of X i for i = 1, 2,..., k then E (F 1 (X 1 ) + + F k (X k )) = E (F 1 (X 1 )) + + E (F k (X k )).

17 Expected Value of Binomial Random Variable If X i is a Bernoulli random variable then E (X i ) = (1)(p) + (0)(1 p) = p.

18 Expected Value of Binomial Random Variable If X i is a Bernoulli random variable then E (X i ) = (1)(p) + (0)(1 p) = p. If the binomial random variable X = X X n where each X i is a Bernoulli random variable, then E (X) = E (X 1 ) + + E (X n ) = n p.

19 Expected Value of a Geometric RV The probability mass function of a geometric RV is given by P (X = n) = f X (n) = (1 p) n 1 p for n N. Recall the geometric series summation, n=0 z n = 1 1 z for z < 1. Differentiate the summation and replace z by 1 p.

20 Expected Value of a Geometric RV The probability mass function of a geometric RV is given by P (X = n) = f X (n) = (1 p) n 1 p for n N. Recall the geometric series summation, n=0 z n = 1 1 z for z < 1. Differentiate the summation and replace z by 1 p. n z n 1 = n=1 n (1 p) n 1 p = n=1 1 (1 z) 2 p (1 (1 p)) 2 E (X) = 1 p

21 Independent Random Variables (1 of 2) If X and Y are independent random variables then f X,Y (x, y) = f X (x)f Y (y) where f X (x) is the marginal probability distribution of X and f Y (y) is the marginal probability distribution of Y.

22 Independent Random Variables (1 of 2) If X and Y are independent random variables then f X,Y (x, y) = f X (x)f Y (y) where f X (x) is the marginal probability distribution of X and f Y (y) is the marginal probability distribution of Y. Theorem Let X 1, X 2,..., X k be pairwise independent random variables, then E (X 1 X 2 X k ) = E (X 1 ) E (X 2 ) E (X k ).

23 Independent Random Variables (2 of 2) Proof. Now let X and Y be independent random variables with joint probability distribution f X,Y (x, y). E (X Y ) = = x (x,y) Ω x y f X,Y (x, y) x y f X (x)f Y (y) = x f X (x) x y = E (X) E (Y ) The general case holds by induction on k. y y f Y (y)

24 Variance (1 of 2) Definition If X is a random variable, the variance of X is denoted Var (X) and ( Var (X) = E (X E (X)) 2). The standard deviation of X is denoted σ(x) = Var (X).

25 Variance (2 of 2) Theorem Let X be a random variable, then the variance of X is ( Var (X) = E X 2) E (X) 2.

26 Variance (2 of 2) Theorem Let X be a random variable, then the variance of X is ( Var (X) = E X 2) E (X) 2. Proof. Var (X) = E ((X E (X)) 2) ( = E X 2) E (2XE (X)) + E (E (X) 2) ( = E X 2) 2 E (X) E (X) + E (X) 2 ( = E X 2) E (X) 2.

27 Example (1 of 2) What is the variance in the number of female children in a family of 5 children?

28 Example (1 of 2) What is the variance in the number of female children in a family of 5 children? If X represents the number of female children in a family of five children, then ( Var (X) = E X 2) (E (X)) 2 = 7.5 (2.5) 2 = 1.25.

29 Example (2 of 2) Find the variance of a Bernoulli random variable for which the probability of success is p.

30 Example (2 of 2) Find the variance of a Bernoulli random variable for which the probability of success is p. If X represents the outcome of a Bernoulli trial, then ( Var (X) = E X 2) (E (X)) 2 [ ] = 1 2 (p) (1 p) p 2 = p p 2 = p(1 p).

31 Variance of a Sum (1 of 3) Theorem Let X 1, X 2,..., X k be pairwise independent random variables, then Var (X 1 + X X k ) = Var (X 1 ) + Var (X 2 ) + + Var (X k ).

32 Variance of a Sum (2 of 3) Take the case when k = 2. Var (X + Y ) = E (((X + Y ) E (X + Y )) 2) = E (((X E (X)) + (Y E (Y ))) 2) ( = E (X E (X)) 2) ( + E (Y E (Y )) 2) + 2E ((X E (X))(Y E (Y ))) = Var (X) + Var (Y ) + 2E ((X E (X))(Y E (Y )))

33 Variance of a Sum (2 of 3) Since we are assuming that random variables X and Y are independent, then E ((X E (X))(Y E (Y ))) = E (X E (X)) E (Y E (Y )) = (E (X) E (X))(E (Y ) E (Y )) = 0, and thus Var (X + Y ) = Var (X) + Var (Y ). The result can be extended to any finite value of k by induction.

34 Example Find the variance of a Binomial random variable with n trials when the probability of success on a single trial is p.

35 Example Find the variance of a Binomial random variable with n trials when the probability of success on a single trial is p. Each trial of a binomial experiment is a Bernoulli experiment with probability of success p. Since the n trials of a binomial experiment are independent Var (X) = n p(1 p).

36 Variance of a Geometric RV (1 of 2) The probability mass function of a geometric RV is given by P (X = n) = f X (n) = (1 p) n 1 p for n N. Recall the geometric series summation, n=0 z n = 1 1 z Differentiate the summation twice. n(n 1) z n 2 = n=1 for z < 1. 2 (1 z) 3

37 Variance of a Geometric RV (2 of 2) n(n 1) z n 2 = n=1 Replace z by 1 p. n=1 n=1 2 (1 z) 3 n 2 (1 p) n 2 n (1 p) n 2 = n=1 n 2 (1 p) n 1 p n (1 p) n 1 p = n=1 ( E X 2) 1 p ( E X 2) E (X) 2 = 2 (1 (1 p)) 3 2(1 p)p (1 (1 p)) 3 = 2(1 p) p 2 2(1 p) p 2 Var (X) = 1 p p p ( ) 1 2 p

38 Variance of a Product (1 of 2) Theorem Let X 1, X 2,..., X k be pairwise independent random variables, then ( ) ( ) ( ) Var (X 1 X 2 X k ) = E X1 2 E X2 2 E Xk 2 (E (X 1 ) E (X 2 ) E (X k )) 2.

39 Variance of a Product (2 of 2) Proof. ( Var (X 1 X 2 X k ) = E (X 1 X 2 X k ) 2) (E (X 1 X 2 X k )) 2 ( ) = E = E X1 2 X 2 2 X k 2 ) ( E X2 2 ( X 2 1 (E (X 1 ) E (X 2 ) E (X k )) 2 ( ) Xk 2 ) E (E (X 1 ) E (X 2 ) E (X k )) 2

40 Homework Read Sections 2.6, 2.7 Exercises: 16, 18 25

41 Credits These slides are adapted from the textbook, An Undergraduate Introduction to Financial Mathematics, 3rd edition, (2012). author: J. Robert Buchanan publisher: World Scientific Publishing Co. Pte. Ltd. address: 27 Warren St., Suite , Hackensack, NJ ISBN: