Exercise 6. Suppose a random variable has support: {−2,−1,0, 1, 2} and its pmf is: X -2 -1 0 1 2 f(x) 2/7 1/7 1/7 1/7 2/7 (a) Calculate E(X) and E(X²). (b) Using the calculated values of E(X) and E(X²) and properties of the expectation operator, calculate the following quantities: (i) E(2X – 3), (ii) E(X² – 3X), (iii) E((X− 1)²). (c) Calculate the variance of X.
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- Let X1, X2, ..., X, be independent random variables and Y = min{X1, X2, ..., Xm}. Fy (y) = 1 – || (1 – Fx,(y)) i=1 (a) A certain electronic device uses 5 batteries, with each battery to have a life that is exponentially distributed with mean of 48 hours and is independent of the life of other batteries. If the device fails as soon as at least one of its batteries fail, what is the expected life of the device?The random variable X is given and we define a new random variable Y = g(X)Og(x) = a + bx + cx². In this problem, you will do,showing detail work, and the definition of expectation of a function of a random variable (i.e., definition of E(g(X))): Show that E(Y) = a+ bE(X)+ c[E(x)]² + cVar(X) when X is a discrete random variable. You must use the definition of Expectation of a function of a discrete random variable, (summation operators and indicating at each step why you substitute)as we did in all the proofs of this lecture. O Repeat (a) but assuming that X is continuous, in which case you will be using the integration operator. (b)The index model has been estimated for stocks A and B with the following results: RA = 0.03 + 0.8RM + eA. RB = 0.01 + 0.9RM + eB. σM = 0.35; σ(eA) = 0.20; σ(eB) = 0.10. The covariance between the returns on stocks A and B is A) 0384. B) 0.0406. C) 0.0882. D) 0.0772. E) 0.4000. 2) Analysts may use regression analysis to estimate the index model for a stock. When doing so, the slope of the regression line is an estimate of A) the α of the asset. B) the β of the asset. C) the σ of the asset. D) the δ of the asset. Choose correct answer with justification.
- The moment generating function of the random variable X is given by mx(s) = e2e-2 and the moment generating function of the random variable Y is my (s) = (et + )10. If it is assumed that the random variables X and Y are independent, find the following: (a) E(XY); (b) E[(X – Y)²] (c) Var(2X – 3Y).By Laws of Expected Value and Variance, determine E(Z) and V(Z) if Z = 4X + 1. E(Z) = 4E(X) + 1 and V(Z) = 16V(X) E(Z) = 4E(X) + 1 and V(Z) = 4VX) E(Z) = 4E(X) + 4 and V(Z) = 4V(X) E(Z) = 4E(X) + 4 and V(Z) = 16V(X)Let X = (X1, X, )" be a bivariate random variable with variance-covariance matrix 4 -1.5 E(X – E(X))(X – E(X))") = ( -1.5 1 You are given that X1 + aX2 is independent of X1. Find the number a. Give your answer in 2 decimal places. Answer:
- (a) Use the rules of expected value to show that Cov(ax + b, cY + d) = ac Cov(X, Y). Cov(ax + b, CY + d) = E - E(ax + b)E ])-) + adx + bcY + bd - (aE(X) + b) CE = = acE + adE(X) + bcE(Y) + bd + adE(X) + bcE = acE acE(X)E 1)) = — асCov(X, Y) (b) Use part (a) along with the rules of variance and standard deviation to show that Corr(ax + b, cY + d) = Corr(X, Y) when a and c have the same sign. co( b, cY + Corr(ax + b, cY + d) = Oax + bº cY + d Cov(X, Y) la||clox0y Corr(X, Y) lallc| = Corr(X, Y) (c) What happens if a and c have opposite signs? ac O When a and c differ in sign, ac is negative and = -1. Therefore, Corr(ax + b, CY + d) = -Corr(X, Y). la||c| ac O When a and c differ in sign, ac is positive and = 1. Therefore, Corr(ax + b, cY + d) = -Corr(X, Y). Jal|c ас O when a and c differ in sign, ac is positive and = 1. Therefore, Corr(ax + b, cY + d) = Corr(X, Y). lal|c ac O when a and c differ in sign, ac is positive and = -1. Therefore, Corr(ax + b, cY + d) = -Corr(X, Y). la||c| ас O…If Z = X+2Y, find P(Z) and find E(Z) (the expected value)Let X be a discrete random variable with support {0,2,3}, and suppose E(X) =1 and P(X=0)=P(X=2). Find Var(X)