1) If Var(X-Y) = 2, Var(X+Y) = 4, E(X) = 3 and E(Y) = 1, than what is E(XY)? 2) If X and Y are independent random variables with, E(X)=0= E(Y), show that Var(XY) = Var(X) x Var(Y).
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- Suppose that the random variables X and Y are independent with Var(X)=8 and Var(Y)=6. Calculate Var(5X−7Y+17)the random variable Y is such that : E(2Y+3)=6 and Var(2-3Y)=11 find E(Y^2)Suppose that X and Y are random variables with E(XY) =E(X)E(Y). Then X and Y + independent. Also %3D Var(X+Y) = Var(X) + Var(Y) true. must be cannot be may or may not be Previous page Next page
- Suppose X and Y are independent random variables with E(X) =2, E(Y)=3,V(X)=4,V(Y)=16. Finda)E(5X-Y) b)V(5X-Y) c)COV(3X+Y,Y) d)COV(X,5X-Y)Suppose X and Y are two random variables with covariance Cov(X, Y) = 3 and Var(X) = 16. Find the correlation coefficient between X and Y.Suppose that X and Y are random variables with E(XY) = E(X)E(Y). Then X and Y * independent. Also Var(X + Y) = Var(X) + Var(Y) true.
- Suppose X, Y and Z are independent random variables with var (X) = 4 , var (Y) = 4 , var (Z) = 1/n Find: var (X/2 - Y + the square root of nZ) (divided by -2)If X, Y are standardized random variables and r(aX+bY,bX+aY)= 1+2ab a²+b² I Find r(X,Y), coefficient of correlation between X and Y.An ordinary (fair) coin is tossed 3 times. Outcomes are thus triple of “heads” (h) and tails (t) which we write hth, ttt, etc. For each outcome, let R be the random variable counting the number of tails in each outcome. For example, if the outcome is hht, then R (hht)=1. Suppose that the random variable X is defined in terms of R as follows X=6R-2R^2-1. The values of X are given in the table below. A) Calculate the values of the probability distribution function of X, i.e. the function Px. First, fill in the first row with the values X. Then fill in the appropriate probability in the second row.
- Suppose X~ Binom(6, p) and define an estimator T ==to estimate p. Then, variance 6. of T is, p(1–p) a) Var (T)= 6. b) Var (T) = 6 p(1 – p) c) Var (T)~ p(1-p) d) Var (T) - 36р2. Let Y,,., Y, be independent random variables such that Y, (Yı., Yp)" and 0 = (0,.,0p)". Let = 0(Y) = (0,(Y),... , @p(Y))" be an estimator of 0, and let g(Y) = (g(Y),... , gp(Y))" = – Y. Denote by || - || the Euclidean norm, ||Y° = Y} + .. + Y. N(8), 1). Write Y = %3D Suppose that D(Y) = @g(Y)/ay, exists. Then it is known that %3D R(Ô(Y)} = +2 D(Y) + É19(Y)² =1 is an unbiased estimator of the risk of 0, under squared error loss L(0, ê) = ||0 – e|P. [You are not required to show this]. %3D (i) The James-Stein estimator is 6.s(Y) = (1 – )Y. _P-2y ||Y? Show that an unbiased estimator of the risk of d Js(Y) is Řlójs(Y)) = p – (p - 2) /||YII°. Deduce that Y is inadmissible as an estimator of 0. Is ô js(Y) admissible? Justify your answer.Let X be a random variable with the function, -(x -A) "for x > 2, Let f (x) = { elsewhere. Derive u and show that T = X is biased estimator of 2 . Can you modify T =X in order to get an unbiased estimator of 2.