3. To control the variance of 8,, we determine À from the following minimization min M(X). A>0 where M(X) = maxf(y+a)= 20 1 √2TOX exp min d+ = 1 √2πOX V20

max 20² 2x { exp (_ (1 + 2 = 14)² + xy) } -xy}). • If 0 < x≤, the minimization in M(A) is attained at y = 0 and (-(a2-4)²). (y+a-µ)² 20² 1 M(X) = = √27/01 exp(-(a2-4)²) whose roots are equal to ≥ whose equality holds at A= (a-µ)/0². • If X>, the minimization in M(A) is attained at y = Ao² - (a-μ) and (2=(a − µ) a−µ± √(a−µ)² + 4ơ² 20² exp M(X) = = √²=₁X EXP (²x² - (a − μµ)x). exp √250X In this case, differentiating M(A) with respect to À yields dM o²x² - (a−µ)λ - 1 dX A - M(X) (but A ≤ 2 < is ruled out a-μ 20² It is not difficult to show that the global minimum point of M(A) over > Hence, we could choose A = A4 and in this case, Var(,) ≤0(M(X+) — 0). >0 (but not required) and thus λ, is