Stats - normal distribution question?

Unbiased estimator %26amp; theta - Statistics - Normal distribution?





In class we said that S is an unbiased estimator of 胃 if E[S] = 胃, otherwise it was biased, with bias = E[S] 鈥?胃. We would like to find an unbiased estimator of the area (A) of a square plate with sides of length L. Suppose that we make two independent measurements L(1), L(2) of the length of the sides of the plate. Suppose that L (I) is Normally distributed with mean L and variance 蟽^2. Let your first estimator of A be the square of the average length of the two measurements and let your second estimator be the average of the squares of the first measurement and the second measurement. That is,


S(1) = [0.5(L(1)+ L(2))]^2 and S(2) = [0.5(L^2(1) + L^2(2))]. Which of these estimators has the least bias? If both are biased, can you think of a third estimator that is unbiased?





Well, I know that if X and Y are two independent random variables then E[XY] = E[X]E[Y], and that VAR[X] = E[X2]-[E[X]]^2





Any ideas?





Thanks!|||Just apply the formulas you gave and the definitions.


Note that the mean of the measurements (usually mu) is L here.





The expected value of S(1) is, expanding and using the independence of L(1) and L(2),





ES(1) = E[0.5(L(1)+ L(2))]^2


= (1/4){E[L(1)^2] + 2E[L(1)L(2)] + E[L(2)^2]}


= (1/4){...............+ 2E[L(1)]E[L(2)] + ........}


= (1/4)[(sigma^2+L^2) + 2L^2 + (sigma^2+L^2)]


= L^2 + sigma^2/2 = A + (1/2)sigma^2





so S(1) is biased upwards (as an estimator of A).





Similarly





ES(2) = E[0.5(L^2(1) + L^2(2))] = (1/2){E[L(1)^2] + E[L(2)^2]}


= (1/2)[(sigma^2+L^2) + (sigma^2+L^2)] = L^2 + sigma^2


= A + sigma^2





which is more biased tahn S(1).


How about S(3) = L(1)L(2)? Using the independence of L(1) and L(2),





E[L(1)L(2)] = E[L(1)] E[L(2)] = L^2





so S(3) is unbiased.