在 10000 B(t1) X1 1951 11000 B(t2) X2 〉 三 11100 ::::: B(tn) Xn 11111 正态随机向量的线性变换服从正态分布 12/41 GoBack FullScreen Close Quit
12/41 kJ Ik J I GoBack FullScreen Close Quit → B(t1) B(t2) ... B(tn) = 1 0 0 0 0 1 1 0 0 0 1 1 1 0 0 ... ... ... ... ... 1 1 1 1 1 X1 X2 ... Xn ëÅï˛Ç5CÜ—l©Ÿ
数传在更 四、维纳过程的数字特征 1951 ·E{B(t)}=0; ●RB(s,t):=E(B(s)B(t)=σ2min(s,t) 例4.1.3设{B(t),t≥0}是参数为σ2的维纳过程，求 下列过程的均值函数和相关函数。 ·X(t)=B2(t),t≥0； 13/41 ●X(t)=tB(1/t),t>0 Proof 1) mx(t)=E[X(t)】=E[B] =D[B(t)]+{E[B(t)]2=σ2t GoBack FullScreen Close Quit
13/41 kJ Ik J I GoBack FullScreen Close Quit o!ëBLßÍiA • E{B(t)} = 0; • RB(s, t) := E(B(s)B(t)) = σ 2min(s, t) ~ 4.1.3 {B(t), t ≥ 0}¥ÎÍèσ 2 ëBLß߶ eL߲äºÍ⁄É'ºÍ" • X(t) = B2 (t), t ≥ 0; • X(t) = tB(1/t), t > 0 Proof 1) mX(t) = E[X(t)] = E[B 2 ] = D[B(t)] + {E[B(t)]} 2 = σ 2 t
在文☑ Rx(s,t)=E[X(s)X(t)]=E[B2(s)B2(t)] 1951 =E{B(s)[B(t)-B(s)+B(s)]2}(s<t) =E{B2(s)[B(t)-B(s)]2}+E[B4(s)川 +2E{B3(s)[B(t)-B(s)]} =E[B2(s)]E{[B(t)-B(s)]2)+E[B4(s)】 =o2Sa2(t-s)+3o4s2=o4(st+2s2) 14/41 故 Rx(s,t)=o4(st +2min2(s,t)) 其中因B(t)~N(0,σ2t): B(t)-B(s)~N(0,o21t-s) 另因：若X心N(0,σ2)，有 GoBack FullScreen Close Quit
14/41 kJ Ik J I GoBack FullScreen Close Quit RX(s, t) = E[X(s)X(t)] = E[B 2 (s)B 2 (t)] = E{B 2 (s)[B(t) − B(s) + B(s)]2 } (s < t) = E{B 2 (s)[B(t) − B(s)]2 } + E[B 4 (s)] +2E{B 3 (s)[B(t) − B(s)]} = E[B 2 (s)]E{[B(t) − B(s)]2 } + E[B 4 (s)] = σ 2Sσ2 (t − s) + 3σ 4 s 2 = σ 4 (st + 2s 2 ) RX(s, t) = σ 4 (st + 2min2 (s, t)) Ÿ•œ B(t) ∼ N(0, σ2 t); B(t) − B(s) ∼ N(0, σ2 |t − s|) ,œ: eX ∼ N(0, σ2 ),k
在园 9=合-n到1-h0 n=1,3,5,… 1951 2) mx(t)=EX(t】=tE ()】 0,t>0 R,=Fa(目)*a()-e时 15/41 -sto"min -o'min(s,). GoBack FullScreen Close Quit
15/41 kJ Ik J I GoBack FullScreen Close Quit E(Xn ) = ( 0, n = 1, 3, 5, · · · ; σ n (n − 1)(n − 3) · · · 1, n = 2, 4, 6, · · · . 2) mX(t) = E[X(t)] = tE B 1 t = 0, t > 0 RX(s, t) = E sB 1 s ∗ tB 1 t = stE B( 1 s )B( 1 t ) = stσ2min 1 s , 1 t = σ 2min(s, t)