5 Solutions No credit will be given without supporting work. 1, (a) E(X) = 0 x 1 3 e x x dx = 3. Set Mn = X1 X2 Xn n . By WLLN, we have Mn P 3, i.e. for all > 0, lim n P (|Mn 3| ) = 0. For = 1, {Mn > 1} {|Mn 3| 1} lim n P (Mn > 1) lim n P (|Mn 3| 1) = 1 lim n P (|Mn 3| > 1) = 1. (b) E(Y ) = 0 y3e3y dy = 1 3 . Set Mn = Y1 Y2 Yn n . By WLLN, we have Mn P 1 3 , i.e. for all > 0, lim n P (|Mn 1 3 | ) = 0. For = 1 3 , {Mn > 1} {|Mn 1 3 | > 1 3 } lim n P (Mn > 1) lim n P (|Mn 1 3 | > 1 3 ) = 0