see attached Let X1, X2,...,Xn be random variables each one having the following distribution (this is one of the Pareto family of distributions). Suppose the parameter theta > 0 is unknown. Consider the estimator theta = X(1) = min(X1, X2...Xn). Show that the probability density function of . Show that the expected value of the estimator above is . Show that the bias of theta is . Solution First, find F(x) .P(x min <= x) = 1 - P(x1, x2, ...xn) >= x P(x1 >= x) = the integral from x to infinity of 3theta^3x^-4dx = -theta^3x^-3 evaluated from x to infinity = 0 - -theta^3/x^3 = theta^3/x^3. Then, F(x) = (theta^3/x^3)^n = theta^3n/x^3n, and 1-P(x1, x2, ...xn) = 1 - theta^3n/x^3n Then, g1(x) = 3ntheta^3n/x^(3n+1) b) Integrate x 3ntheta^3n/x^(3n+1) from theta to infinity = the integral of 3ntheta^3n/x^(3n) from theta to infinity = -1/(3n-1) 3ntheta^3n/x^(3n-1) evaluated at theta and infinity = 0 - - 1/(3n-1) 3ntheta^3n/theta^(3n-1) = 1/(3n-1) 3ntheta^3n/theta^3ntheta^-1 = 3n theta/(3n-1) As the parameter is theta, the bias is 3n theta/(3n-1) - theta = (3n-1+1)theta/(3n-1) - theta= theta + theta/(3n-1) - theta = theta/(3n-1).