Under (3.39), take θ = 0 without loss of generality. Then we can show that                        Ln(X1,... ,Xn) = −Ln(−X1,... , −Xn) D= −Ln(X1,... ,Xn), since the distribution function F0 is...

Under (3.39), take θ = 0 without loss of generality. Then we can show that

                       Ln(X1,... ,Xn) = −Ln(−X1,... , −Xn) D= −Ln(X1,... ,Xn),

since the distribution function F0 is symmetric about 0. hence, Ln has a distribution function symmetric about 0. Using the result in Problem 3.2.3, we can claim that Ln is median unbiased estimator of θ.