Again, let X1,…,Xn∼iidU([a,a+1]) where a is an unknown parameter. In terms of n , what is the variance of the estimator X¯¯¯¯n ?
Var[X¯¯¯¯n]=
To find the variance of the estimator X¯¯¯¯n, we can use the fact that the variance of a sum of random variables is equal to the sum of their variances, and the variance of a constant times a random variable is equal to the constant squared times the variance of the random variable.
First, let's find the variance of a single random variable X. Since X follows a uniform distribution with range [a, a+1], its variance can be calculated as (b - a)^2 / 12, where b is the upper bound of the range.
In this case, b = a+1, so the variance of X is ((a+1) - a)^2 / 12 = 1/12.
Now, let's find the variance of the sample mean X¯¯¯¯n. The sample mean is the sum of n identical and independently distributed random variables divided by n. Since the X's are identically distributed and independent, the variance of X¯¯¯¯n can be calculated as the variance of a single X divided by n.
Therefore, Var[X¯¯¯¯n] = Var[X] / n = (1/12) / n = 1 / (12n).
So, in terms of n, the variance of the estimator X¯¯¯¯n is 1 / (12n).