Suppose that E (0₂) = E(0₂) = = 0, var(0₂) = ₁, var(8₂) = ₂. Consider the following estimator ₂ = a₁ + (1 - a). Find all values of a such that the variance of estimator , is minimal, a. when estimators, and are independent. b. when cov(,,,) = c = 0.
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- If X1 and X2 constitute a random sample of size n = 2from an exponential population, find the efficiency of 2Y1relative to X, where Y1 is the first order statistic and 2Y1and X are both unbiased estimators of the parameterLet X1, . . . , Xn, . . . ∼ iid Bern(θ). Consider the Bayes estimator under squared error loss with the Unif(0,1) prior. Show that this estimator is consistent.Consider the following two formulations of the bivariate PRF, where ui and εi are both mean-0 stochastic disturbances (i.e random errors): yi = β0 + β1xi + u yi = α0 + α1(xi − x¯) + ϵ a) Write the OLS estimators of β1 and α1. Are the two estimators the same? b) What is the advantage, if any, of the second model over the first?
- Let X1, . . . , Xn ∼ iid Unif(θ1, θ2), where both θ1 and θ2 are unknown. Find the MOM estimator and compare them to the MLE.2) Let G and H be two independent unbiased estimators of θ. Assume that the variance of G is two times the variance of H. Find the constants a and b so that aG + bH is an unbiased estimator with the smallest possible variance for such a linear combination.Suppose X and Y are random variables with E[XY ] = 6, E[Y ] = 4 and E[X] = 5 Find Cov(X; Y )
- If X1, X2, ... , Xn constitute a random sample of size nfrom a geometric population, show that Y = X1 + X2 +···+ Xn is a sufficient estimator of the parameter θ.For every ?, ? > 0. 1. Derive the variance of ?.If X1, X2, and X3 constitute a random sample of sizen = 3 from a Bernoulli population, show that Y =X1 + 2X2 + X3 is not a sufficient estimator of θ. (Hint:Consider special values of X1, X2, and X3.)
- If the PDF of X is f(x)=2x/k2 for 0<x<k, for what value of k is the variance of X equal to 2?Based on the following values: Mathematical expectation: E(X) = ∑Xn1 = 1204 = 30 E(Y) = ∑Yn2 = 1884 = 47 E(Z) = ∑Zn3 = 1784 = 44.5 Variance: var(X) = 1n1-1∑X2-1n∑X2 = 14-13678-12024 =26 var(Y) = 1n2-1∑Y2-1n∑Y2 = 14-18890-18824 =18 var(Z) = 1n3-1∑Z2-1n3∑Z2 = 14-18322-17824 =133.6667 Root mean square deviation sd(X)= var(X) =26 = 5.0990 sd(Y)= var(Y) =18 = 4.2426 sd(Z)= var(Z) =133.6667 = 11.5614 Correlation coefficients X Y Z 25 44 28 28 43 48 37 49 55 30 52 47 1. Expectation 120 188 178 2. Variance 26 18 133.6667 3. root mean square deviation 5.099 4.2426 11.5614 Check the hypotheses in the image below, in accordance with the tablic values.Based on the following values: Mathematical expectation: E(X) = ∑Xn1 = 1204 = 30 E(Y) = ∑Yn2 = 1884 = 47 E(Z) = ∑Zn3 = 1784 = 44.5 Variance: var(X) = 1n1-1∑X2-1n∑X2 = 14-13678-12024 =26 var(Y) = 1n2-1∑Y2-1n∑Y2 = 14-18890-18824 =18 var(Z) = 1n3-1∑Z2-1n3∑Z2 = 14-18322-17824 =133.6667 Root mean square deviation sd(X)= var(X) =26 = 5.0990 sd(Y)= var(Y) =18 = 4.2426 sd(Z)= var(Z) =133.6667 = 11.5614 Correlation coefficients X Y Z 25 44 28 28 43 48 37 49 55 30 52 47 1. Expectation 120 188 178 2. Variance 26 18 133.6667 3. root mean square deviation 5.099 4.2426 11.5614 Check the hypothesis in the image below. The table values are also there