茆诗松高等数理统计课后习题第三次作业.pdf

ppp OOO 111nnnggg YYY JJJ 1 an 0 bn 0 yyy eee 1 Op 1 op 1 Op 1 2 Op 1 op 1 op 1 3 Op an op bn op anbn 4 Op an op bn Op an bn 5 op an r op ar n 1 n Op 1 and n op 1 then we have n Op 1 Mn P n Mn 0 as n Also P n n Mn P n Mn 2 P n Mn 2 Since n op 1 P n Mn 2 0 as n we have limsupn P n n Mn 0 i e limn P n n Mn 0 By the necessary and suffi cient condition mentioned above we have n n Op 1 2 Suppose M satisfi es M s t P n M P n n P n n n M P n n n 0 M s t P n an n bn Then we have P n n an bn M P n an n bn M 0 P nb 1n r P nb 1n 1 r 0 as n goes to infi nity which means nb 1n r P 0 2 Xn X yyy Xn L X a a Xn L a X Suffi ciency If a a Xn L a X then for t a we have E it a Xn E it a X Let t 1 We get for a E i a Xn E i a X 1 Necessity By Xn s convergence E i t Xn E i t X for t Let t t a Then we have Ei t a Xn Ei t a X That is to say Eit a Xn Eit a X 3 Xn L X Yn P c yyy 1 Xn Yn L X c 2 c 6 0 XnY 1 n L Xc 1 3 XnYn L Xc 4 Xn Op 1 1 By conclusion of the last problem we only need to prove that for a a Xn Yn L a X c Let X n a Xn Y n a Yn X a X c a c We prove X n Y n L X c For z P X n Y n z P X n Y n z Y n c P X n Y n z Y n c P Y n c P X n z c Thus we have limsupnP X n Y n z F z c Also P X n Y n z P X n Y n z Y n c P X n z c Y n c P X n z c P Y n c Thus liminfnP X n Y n z F z c Let 0 we have limnP X n Y n z F z c So far we have proved that for a a Xn Yn L a X c 2 Suppose that c 0 otherwise Yn c z P XnY 1 n z P XnY 1 n z Yn c P XnY 1 n z Yn c P Yn c P Xn z c limsupnP XnY 1 n z F z c Similarly P XnY 1 n z P XnY 1 n z Yn c P Xn z c P Yn c Thus liminfnP XnY 1 n z F z c is arbitrary limnP XnY 1 n z F zc The right hand side of the equation is the c d f of Xc 1 Thus c 6 0 XnY 1 n L X 1 3 The proof is similar to 2 4 A sequence of random variables Xn with respective distribution function Fn is said 2 to be bounded in probability if for every 0 there exist M and N such that Fn M Fn M 1 n N The notation Xn Op 1 will be used We only need to prove that each element of the vector is Op 1 Since limnFn