《《随机过程及其在金融领域中的应用》习题六答案.docx》由会员分享，可在线阅读，更多相关《《随机过程及其在金融领域中的应用》习题六答案.docx（12页珍藏版）》请在三一文库上搜索。
1、第六章习题 61、在下列矩阵的空白处填上适当的数字，使矩阵 Q 成为一个转移强度矩阵。 -523 Q= 00022-42、一连续时间 Markov 链有 0 和 1 两个状态，在状态 0 和 1 的逗留时间分别服从参数为 l 0 及 m 0 的指数分布。试求在时刻 0 从状态 0 起始， t 时刻后过程处于状态 0 的概率 P00 (t )。答：设 x (t )为 t 时刻所处状态，记P00 (t ) = P (x ( t ) = 0 x (0 ) = 0 ), P01 (t ) = P (x ( t ) = 1 x (0 ) = 0)易知： P00 (t )+ P01 (t ) =1 ，采用
2、无穷小分析法P00 (t + Dt ) = P (x (t + Dt ) = 0 x (0 ) = 0)= P (x (t + Dt ) = 0, x (t ) = 0 x (0 ) = 0 )+ P (x (t + D t ) = 0, x (t ) = 1 x (0 ) = 0)= P00 (t ) P (x (t + Dt ) = 0 x (t ) = 0 )+ P01 (t ) P (x (t + Dt ) = 0 x (t ) = 1)= P00 (t ) (1 - l Dt + o ( Dt )+ (1- P00 (t ) m DtP00 (t + DDtt) - P00 (t )
3、 = - ( l + m ) P00 (t )+ m + o (DDtt ) P00 (t ) = - ( l + m ) P00 (t )+ m P00 (t ) = l l m e-(l +m)t + l m m+3、在上题中如果 l = m 定义 N (t )为过程在0, t 中改变状态的次数，试求 N (t )的概率分布。答：P01 (t ) = 1 - P00 (t )=l-le - (l +m)t=1(1 - e -2lt ), (l = m )l +ml +m2Pt)=11+e -2 l t)，P(t)=11+e -2 l t)，P(t)=11- e-2lt)00 (2(112(
4、102(记 Pk (t ) = P (N (t ) = k x (0 ) = 0)kPk (t + t ) = P (N (t + t ) = kx (0 ) = 0 ) = P (N (t + t ) = k , N (t ) = jx (0) = 0)j =0= P (N (t + t ) = k N (t ) = k )Pk (t )+ P (N (t + t ) = k N (t ) = k -1)Pk -1 (t )= (1 - 2l t ) Pk (t ) + 2l DtPk -1 (t )+o (t ) Pk ( t ) = -2l Pk (t )+ 2l Pk -1 (t ),
5、 P0 (t ) = -2lP0 (t ), P0 (t ) = e -2lt , P0 (0 ) =1P (t ) = -2l P (t )+ 2l e -2 l t , P (t ) = 2lte-2lt ,111P(t ) = (2lt )k e -2lt , k = 0,1, 2,kk !N (t )P (lt )4、一质点在 1,2,3 点上作随机游动。若在时刻 t 质点位于这 3 个点中的一个上，则在t , t + h) 内，它以概率1h +o (h) 分别转移到其他两点中的一个上。试求质点2随机游动的 Kolmogorov 方程、转移概率 Pij (t )及平稳分布。答：依题意，
6、由 limPij (Dt)= qij (i j ) ，得 qij =1(i j )，Dt2Dt 0Kolmogorov 向前方程为 Pij (t ) = - Pij (t )+ 12 Pi , j -1 (t )+ 12 Pi , j +1 (t ) 由于状态空间 I = 1, 2, 3，故 Pij (t )+ Pi , j -1 (t )+ Pi , j +1 (t ) = 12所以 Pij (t ) = -2 Pij (t )+ 12 - Pij (t ) = -3Pij (t )+ 12P (t ) = ce- 3 t + 1解上述一阶线性微分方程得：ij231, i = j由初始条件
7、Pij (0) = 0, i j11-2t-, i je333确定常数 c ，得转移概率 Pij (t ) = 123+-t, i = je233故其平稳分布p j = lim Pij (t ) = 1 , j =1, 2, 3t 35、设某车间有 M 台车床，由于各种原因车床时而工作，时而停止。假设时刻 t 一台正在工作的车床在时刻 t + h 停止工作的概率为 m h +o (h)，而时刻 t 不工作的车床在时刻 t + h 开始工作的概率为 l h +o (h)，且各车床工作情况是相互独立的。若 N (t )表示时刻 t 正在工作的车床数，求：（1）齐次马尔可夫过程N (t ), t 0
8、的平稳分布；（2）若 M = 10, l = 60, m = 30 ，系统处于平稳状态时有一半以上车床在工作的概率。答：（1）p01 (Dt ) = CM1 (lDt + o ( Dt )1 (1- lDt - o (Dt )2limp01 (Dt )= M l = qDt 0Dt01同理： q12= ( M - 1)l , q23= ( M - 2 )l , qM -1,M = lqM , M -1 = M m , q21 = 2 m , q10 = m可得 Q 矩阵： -M lM l000m- m +(M - 1 l (M -1 l00) ) 02 m- 2 m +M - 2)l00Q =
9、 (000- M - 1 m + l l ()000M m-M m 得 (p 0 , p 1 , , p M -1, pM )Q = 0mM解得p0= ，p jl + m （2）l j mM - j= CMj, j =1, 2, , Ml + m l + m 30101j60 j 30 10 - jj2 j 110- jp 0= =, p j= C10= C10, j =1, 2, , M90103390 90 352j110- jP (t 5 ) = 1 - p+C j 0.7809010j =1 3 36、设X (t ), t 0是纯生过程，且有P (X (t + h ) - X (t )
10、 = 1 X (t )为奇数) = a h +o (h)P (X (t + h ) - X (t ) = 1 X (t )为偶数) = b h +o (h)取初始条件 X (0 ) = 0 ，求下列概率p1 (t ) = P (X (t ) = 奇数), p2 (t ) = P (X (t ) = 偶数)答：记 p1 (t ) = P (X (t ) = 奇数 X (0 ) = 0 ), p2 (t ) = P (X (t ) = 偶数 X (0 ) = 0)易知： p1 (t ) = 1 - p2 (t )p2 (t + h ) = P (X (t + h ) = 偶数 X (0 ) = 0
11、)= P (X (t + h ) = 偶数,X (t ) = 奇数 X (0 ) = 0 )+ P (X (t + h ) = 偶数,X (t ) = 偶数 X (0 ) = 0)= P (X (t + h ) = 偶数 X (t ) = 奇数) p1 (t )+ P (X (t + h ) = 偶数 X (t ) = 偶数) p 2 (t )a h (1 - p2 (t )+ (1- b h ) p2 (t )+o (h) p2 (t ) = - ( a + b ) p2 (t )+a p1 (t ) = a b+ b - a b+ b e - ( a + b ) t , p2 (t ) =
12、 a a+ b + a b+ b e- ( a +b )t7、考虑状态 0, 1,N 上的纯生过程 X (t ) ，假定 X (0 ) = 0 及 lk = ( N - k )l ，k = 0,1, N 。其中 lk 满足 P (X (t + h ) - X (t ) = 1 X (t ) = k ) = lk h +o (h) 。试求 Pn (t ) = P (X (t ) = n) 。这是新生率受群体总数反馈作用的例子。答：设 Pk (t ) = P (X (t ) = k X (0 ) = 0)Pk (t + h ) = P (X (t + h ) = k X (0 ) = 0)N= P
13、 (X (t + h ) = k , X (t ) = jX (0) = 0)j =0N= P (X (t + h ) = k X (t ) = j ) Pj (t )j =0= P (X (t + h ) = k X (t ) = k ) Pk (t ) + P (X (t + h ) = k X (t ) = k - 1) Pk -1 (t )+o (h)= (1- lk h ) Pk (t ) + lk -1 hPk -1 (t )+o (h) Pk (t ) = - lk Pk (t )+ lk -1 Pk -1 (t )P0 (t ) = - l0 P0 (t ), P0 (t )
14、= e - l Nt = e-l 0t()1 ( )1 1 ( )01 ( )P t= - l P t+ l e - l 0 t, P t= Ne - l 1t1- e-lt可类推测 Pn (t ) Nn= 0,1,= e - N l t (e l t - 1), n, N n8、证明：Furry-Yule 过程（即 li = il, i =1, 2,的纯生过程）的概率母函数GGG (t , z ) = Pik (t ) zk 满足- lz (z - 1)= 0 。tk =1z证明：依题意 G (t , z ) = Pik (t ) zkk =1建立母函数的微分方程：G (t , z )Pik
15、(t ) zkPik (t ) zk=k =0= tttk =0G (t , z ) = - l0 (t )Pi 0 (t )+ m1 (t )Pi1 (t )t+ lk -1 (t )Pik -1 (t )- ( lk (t )+ m k (t )Pik (t )+ mk +1 (t )Pik +1 (t ) zkk =1G (t, z ) = -l0 (t )Pi0(t )- lk(t )Pik (t ) zk - mk (t )Pik (t ) zktk =1k =1+lk -1 (t )Pik -1 (t ) zk + mk +1 (t )Pik +1 (t ) zk + m1 (t
16、)Pi1 (t ) *k =1 k =1将*式化简l (t ) Pik (t ) zk- l0 (t )Pi 0 (t )- lk(t )Pik(t ) z k = - k l (t )Pik (t ) z k = -zk =1zk =1k =1G (t , z )= - l ( t ) z Pik(t) z k = -l(t ) zzk =1- m k (t)Pik (t ) z k = - k m k (t )Pik (t ) z k = -zmk(t )Pik (t ) zkzk =1k =1G (t , z )k =1= - z m ( t )Pik (t) z k = -z m (t
17、 )zz lk -1(t )Pik -1 (t ) z k = ( k - 1)l ( t )Pik -1 (t ) z k = ml (t )Pim (t )z m+1k =1k =1m=0= zml (t )Pim (t ) zm = z2l (t ) Pim (t ) zm= z 2l (t ) G (t, z )m = 0z m=0z1m1 (t )Pi1 (t )+ m k +1(t )Pik +1 (t ) z k =mk +1(t )Pik +1 (t ) zk +1z=k =1k =0= m (t ) G (t, z )1 mm (t )Pim (t ) zm = m(t )
18、Pim (t ) zmz m =1z m=1z+得G (t , z )= l(t)z(z - 1G (t , z )+ mt 1- z)G (t , z )zz)z( )(=(z - 1z lt)+ mt)G (t , z )z)(考虑齐次条件下的微分方程：G (t , z )= lz(z -1G (t , z )，即G (t , z )- lz(z - 1G (t , z )= 0zzz)z)9、有无穷多个服务员的排队系统：假定顾客以参数为 l 的 Poisson 过程到达，而服务员的数量巨大，可理想化为无穷多个。顾客一到就与别的顾客相互独立地接受服务，并在时间 h 内完成服务的概率近似为a
19、h 。记 X (t )为在时刻 t 正接受服务的顾客总数，试建立此过程的转移机制的模型。答：X (t ), t 0是生灭过程。记 Pn (t ) = P (x ( t ) = n x (0 ) = 0)P0 (t + h ) = P (x (t + h ) = 0 x (0 ) = 0)= P (x (t + h ) = 0, x (t ) = 0 x (0 ) = 0 )+ P (x (t + h ) = 0, x (t ) = 1 x (0 ) = 0)= (1- l h - a h ) P (t )+ a hP (t )+o (h)01 P (t ) = - ( l0+ a) P (t )0+aP (t )1同理有： Pn (t ) = - ( l + a ) Pn (t )+ l Pn -1 (t )+ aPn +1 (t ), n 1初始条件： P0 (0 ) = 1, Pn (0 ) = 0, n 1