随机过程matlab程序.doc

% PPT 例2 一维正态密度与二维正态密度syms x y;s=1; t=2; mu1=0; mu2=0; sigma1=sqrt(1+s2); sigma2=sqrt(1+t2); x=-6:0.1:6;f1=1/sqrt(2*pi*sigma1)*exp(-(x-mu1).2/(2*sigma12);f2=1/sqrt(2*pi*sigma2)*exp(-(x-mu2).2/(2*sigma22);plot(x,f1,r-,x,f2,k-.) rho=(1+s*t)/(sigma1*sigma2); f=1/(2*pi*sigma1*sigma2*sqrt(1-rho2)*exp(-1/(2*(1-rho2)*(x-mu1)2/sigma12-2*rho*(x-mu1)*(y-mu2)/(sigma1*sigma2)+(y-mu2)2/sigma22);ezsurf(f)% % The daily log returns on the stock have a mean of 0.05/year and a standard deviation of 0.23/year. These can be converted to rates per trading day by deviding by 253 and sqrt(253), respectively.Question 1: What is the probability that the value of the stock will be below $950,000 at the close day of at least one of the next 45 trading days? clear;niter=1.0E5; % number of iterationsbelow=repmat(0,1,niter); % set up storagerandn(seed,0);for i=1:niter r=normrnd(0.05/253,0.23/sqrt(253),1,45); % generate random numbers logPrice=log(1.0E6)+cumsum(r); minlogP=min(logPrice); % minmum price over next 45 days below(i)=sum(minlogPlog(950000); endPro=mean(below)% P29 随机相位正弦波仿真% 1 time simulationw=2; N=1000; mu=2; sigma=3;s=rand(state);A=mu+sigma*randn(1,N); % A=normrnd(mu,sigma,1,N)theta=-pi+2*pi*rand(1,N);t=1:N;x=A.*cos(w*t+theta); capmu=mean(x) tao=1x1=A.*cos(w*(t+tao)+theta);capgamma=mean(x-capmu).*(x1-capmu) % m time simulationclear; w=2; N=1000; mu=2; sigma=3;m=500;capmu1=;capgamma1=;for i=1:ms=rand(state);A=mu+sigma*randn(1,N);theta=-pi+2*pi*rand(1,N);t=1:N;x=A.*cos(w*t+theta);capmu=mean(x);capmu1=capmu1,capmu; tao=1;x1=A.*cos(w*(t+tao)+theta);capgamma=mean(x-capmu).*(x1-capmu);capgamma1=capgamma1,capgamma;endplot(1:m,capmu1,*,1:m,capgamma1,o) capmu=mean(capmu1);capgamma=mean(capgamma1); err1=mean(capmu1-0).2);gamma=(sigma2+mu2)*cos(w*tao)/2;err2=mean(capgamma1-gamma).2);capmu,capgamma; err1, err2 % 输出： 0.0058 -2.7005 0.0065 0.0736% P37 例3.1.1p1=poisscdf(5,10)p2=poisspdf(0,10)p1,p2%输出p1 =0.0671p2 =4.5400e-005ans =0.0671 0.0000 % P43 例3.2.1p3=poisspdf(9,12)% 输出p3 = 0.0874 % P43 例3.2.2p4=poisspdf(0,12)% 输出p4 = 6.1442e-006% P39-40(Th3.1.1) Solve the difference equation system, find the solution% 输入：syms p0 p1 p2 ;S=dsolve(Dp0=-lamda*p0,Dp1=-lamda*p1+lamda*p0,Dp2=-lamda*p2+lamda*p1,p0(0) = 1,p1(0) = 0,p2(0) = 0);S.p0,S.p1,S.p2% 输出：ans =exp(-lamda*t), exp(-lamda*t)*t*lamda, 1/2*exp(-lamda*t)*t2*lamda2% P43 泊松过程仿真% simulate 10 timesclear;m=10; lamda=1; x=; for i=1:ms=exprnd(lamda,seed,1);x=x,exprnd(lamda);t1=cumsum(x);endx,t1 %输出：ans = 0.6509 0.6509 2.4061 3.0570 0.1002 3.1572 0.1229 3.2800 0.8233 4.1033 0.2463 4.3496 1.9074 6.2570 0.4783 6.7353 1.3447 8.0800 0.8082 8.8882%输入：N=;for t=0:0.1:(t1(m)+1)if tt1(1) N=N,0;elseif tt1(2) N=N,1;elseif tt1(3) N=N,2;elseif tt1(4) N=N,3;elseif tt1(5) N=N,4;elseif tt1(6) N=N,5;elseif tt1(7) N=N,6;elseif tt1(8) N=N,7;elseif tt1(9) N=N,8; elseif tt1(10) N=N,9;else N=N,10;endendplot(0:0.1:(t1(m)+1),N,r-) %输出：% simulate 100 timesclear;m=100; lamda=1; x=; for i=1:ms= rand(seed);x=x,exprnd(lamda);t1=cumsum(x);endx,t1 N=;for t=0:0.1:(t1(m)+1)if t=t1(i) & tt1(m) N=N,m;endendplot(0:0.1:(t1(m)+1),N,r-) % 输出：% P48 非齐次泊松过程仿真% simulate 10 timesclear;m=10; lamda=1; x=; for i=1:ms=rand(seed); % exprnd(lamda,seed,1); set seedsx=x,exprnd(lamda);t1=cumsum(x);endx,t1 N=; T=;for t=0:0.1:(t1(m)+1)T=T,t.3; % time is adjusted, cumulative intensity function is t3. if t=t1(i) & tt1(m) N=N,m; endendplot(T,N,r-) % outputans = 0.4220 0.4220 3.3323 3.7543 0.1635 3.9178 0.0683 3.9861 0.3875 4.3736 0.2774 4.6510 0.2969 4.9479 0.9359 5.8838 0.4224 6.3062 1.7650 8.0712 10 times simulation 100 times simulation% P50 复合泊松过程仿真% simulate 100 timesclear;niter=100; % iterate numberlamda=1; % arriving ratet=input(Input a time:,s)for i=1:niter rand(state,sum(clock); x=exprnd(lamda); % interval time t1=x; while t1t x=x,exprnd(lamda); t1=sum(x); % arriving time end t1=cumsum(x); y=trnd(4,1,length(t1); % rand(1,length(t1); gamrnd(1,1/2,1,length(t1); frnd(2,10,1,length(t1); t2=cumsum(y);endx,t1,y,t2 X=; m=length(t1);for t=0:0.1:(t1(m)+1) if t=t1(i) & tt1(m) X=X,t2(m); endendplot(0:0.1:(t1(m)+1),X,r-) 跳跃度服从0,1均匀分布情形 跳跃度服从分布情形跳跃度服从t（10）分布情形 % Simulate the probability that sales revenue falls in some interval. (e.g. example 3.3.6 in teaching material)clear; niter=1.0E4; % number of iterationslamda=6; % arriving rate (unit:minute)t=720; % 12 hours=720 minutesabove=repmat(0,1,niter); % set up storage for i=1:niter rand(state,sum(clock); x=exprnd(lamda); % interval time n=1; while x=t n=n;else n=n+1; end end z=binornd(200,0.5,1,n); % generate n sales y=sum(z); above(i)=sum(y432000); end pro=mean(above)Output: pro =0.3192% Simulate the loss pro. For a Compound Poisson processclear; niter=1.0E3; % number of iterationslamda=1; % arriving ratet=input(Input a time:,s) below=repmat(0,1,niter); % set up storage for i=1:niter rand(state,sum(clock); x=exprnd(lamda); % interval time n=1; while x=t n=n;else n=n+1; end end r=normrnd(0.05/253,0.23/sqrt(253),1,n); % generate n random jumps y=log(1.0E6)+cumsum(r); minX=min(y); % minmum return over next n jumps below(i)=sum(minXlog(950000); end pro=mean(below)Output: t=50, pro=0.45% P86-87 (Example 5.1.5) markov chainchushivec0=0 0 1 0 0 0 P=0,1/2,1/2,0,0,0;1/2,0,1/2,0,0,0;1/4,1/4,0,1/4,1/4,0;0,0,1,0,0,0,;0,0,1/2,0,0,1/2;0,0,0,0,1,0jueduivec1=chushivec0*Pjueduivec2=chushivec0*(P2)% find 1 to n absolute-vectorchushivec0=0 0 1 0 0 0;P=0,1/2,1/2,0,0,0;1/2,0,1/2,0,0,0;1/4,1/4,0,1/4,1/4,0;0,0,1,0,0,0,;0,0,1/2,0,0,1/2;0,0,0,0,1,0;n=10t=1/6*ones(1 6);jueduivec=repmat(t,n 1); for k=1:n jueduiveck=chushivec0*(Pk); jueduivec(k,1:6)=jueduiveckend% Comparing two neighbour absolute-vectorsn=70;jueduivecn=chushivec0*(Pn)n=71;jueduivecn=chushivec0*(Pn)% Replace the first distribution, Comparing two neighbour absolute-vectors once morechushivec0=1/6 1/6 1/6 1/6 1/6 1/6;P=0,1/2,1/2,0,0,0;1/2,0,1/2,0,0,0;1/4,1/4,0,1/4,1/4,0;0,0,1,0,0,0,;0,0,1/2,0,0,1/2;0,0,0,0,1,0;n=70;jueduivecn=chushivec0*(Pn)n=71;jueduivecn=chushivec0*(Pn)% 赌博问题模拟（带吸收壁的随机游走：结束1次游走所花的时间及终止状态）a=5; p=1/2;m=0; while m0 & a0 & a15 m=m+1; r=2*binornd(1,p)-1; if r=-1 a=a-1; else a=a+1; endendif a=0 t1(1,n)=m; m1=m1+1;else t2(1,n)=m; m2=m2+1;endendfprintf(The average times of arriving 0 and 10 respectively are %d,%d.n,sum(t1,2)/m1,sum(t2,2)/m2);fprintf(The frequencies of arriving 0 and 10 respectively are %d,%d.n,m1/N, m2/N); % verify: fprintf(The probability of arriving 0 and its approximate respectively are %d,%d.n, (p10*(1-p)5-p5*(1-p)10)/(p5*(p10-(1-p)10), m1/N); fprintf(The expectation of arriving 0 or 10 and its approximate respectively are %d,%d.n,5/(1-2*p)-10/(1-2*p)*(1-(1-p)5/p5)/(1-(1-p)10/p10), (sum(t1,2)+sum(t2,2)/N); % 应用随机过程（04年第一版）P125(example 5.29) 连续时间马尔可夫链PPT P29(例2)Solve the Kolmogorov difference equation,find the transation probabilities输入：clear;syms p00 p01 p10 p11 lamda mu; P=p00,p01;p10,p11;Q=-lamda,lamda;mu,-muP*Q输出：ans = -p00*lamda+p01*mu, p00*lamda-p01*mu -p10*lamda+p11*mu, p10*lamda-p11*mu输入：p00,p01,p10,p11=dsolve(Dp00=-p00*lamda+p01*mu,Dp01=p00*lamda-p01*mu,Dp10=-p10*lamda+p11*mu,Dp11=p10*lamda-p11*mu,p00(0)=1,p01(0)=0,p10(0)=0,p11(0)=1)输出：p00 =mu/(mu+lamda)+exp(-t*mu-t*lamda)*lamda/(mu+lamda)p01 =(lamda*mu/(mu+lamda)-exp(-t*mu-t*lamda)*lamda/(mu+lamda)*mu)/mup10 =mu/(mu+lamda)-exp(-t*mu-t*lamda)*mu/(mu+lamda)p11 =(lamda*mu/(mu+lamda)+exp(-t*mu-t*lamda)*mu2/(mu+lamda)/mu% BPATH1 Brownian path simulation: forend randn(state,100) % set the state of randnT = 1; N = 500; dt = T/N;dW = zeros(1,N); % preallocate arrays .W = zeros(1,N); % for efficiency dW(1) = sqrt(dt)*randn; % first approximation outside the loop .W(1) = dW(1); % since W(0) = 0 is not allowedfor j = 2:N dW(j) = sqrt(dt)*randn; % general increment W(j) = W(j-1) + dW(j); end plot(0:dt:T,0,W,r-) % plot W against t xlabel(t,FontSize,16) ylabel(W(t),FontSize,16,Rotation,0)% BPATH2 Brow