随机过程matlab实验

1.程序如下：

n=10;x=0:n;X=zeros(10,10);q=zeros(1,11);s=0;p=0;

y=binopdf(x,n,0.6)

z=rand(10,10)

t=[0 1 2 3 4 5 6 7 8 9 10];

for i=1:11

for j=1:i

q(i)=q(i)+y(j);

end

end

q

for i=2:11

for k=1:100

if z(k)<=q(1)

X(k)=t(1);

end

if q(i-1)

X(k)=t(i);

end

end

end

X

for i=1:100

s=s+X(i);

E=s/100;

end

E

for i=1:100

p=p+(X(i)-E)^2;

Var=p/99;

end

Var

运行结果如下：

y =

0.0001 0.0016 0.0106 0.0425 0.1115 0.2007 0.2508 0.2150 0.1209 0.0403 0.0060

z =

0.1048 0.0198 0.2672 0.2501 0.7960 0.9173 0.0919 0.5508 0.4050 0.0348

0.8584 0.9643 0.7537 0.9277 0.2334 0.5098 0.4021 0.8709

0.1736 0.2928

0.6982 0.9704 0.8984 0.0686 0.6008 0.9742 0.2952 0.0423 0.5752 0.8014

0.7337 0.1239 0.7284 0.2994 0.1125 0.1973 0.3065 0.9047 0.6062 0.3465

0.6505 0.4674 0.4068 0.5916 0.5158 0.1112 0.1056 0.1310 0.2144 0.0833

0.5163 0.6567 0.9383 0.2033 0.8378 0.2974 0.5938 0.8337 0.5199 0.5111

0.3264 0.2902 0.2554 0.6359 0.9208 0.3964 0.2827 0.8005 0.9892 0.3668

0.6618 0.7545 0.5332 0.7984 0.4982 0.4208 0.1552 0.9179 0.4899 0.7395

0.1176 0.5581 0.9548 0.5017 0.2776 0.3115 0.0007 0.1373 0.6949 0.5247

0.1478 0.4278 0.2677 0.6508 0.6525 0.6938 0.2836 0.5047 0.4114 0.8045

q =

0.0001 0.0017 0.0123 0.0548 0.1662 0.3669 0.6177 0.8327 0.9536 0.9940 1.0000

X =

4 3

5 5 7 8 4

6 6 3

8 9 7 8 5 6 6 8 5 5

7 9 8 4 6 9 5 3 6 7

7 4 7 5 4 5 5 8 6 5

7 6 6 6 6 4 4 4 5 4

6 7 8 5 8 5 6 8 6 6

5 5 5 7 8

6 5

7 9 5

7 7 6 7 6 6 4 8 6 7

4 6 9 6

5 5 1 4 7 6

4 6

5 7 7 7 5

6 6 7

E =

5.9400

Var =

2.3600

由计算知原均值为6，原方差为2.4，可见模拟过程近似度很好。

2.程序如下：

function f=possion(d1,d2,d3,n)

R1=rand(1,d1);R2=rand(1,d2);R3=rand(1,d3);

disp('[ i x(i) t(i) N(t(i))]');

for i=1:d1

x(i)=-(log(1-R1(i)))/n;

end

for i=1:d2

y(i)=-(log(1-R2(i)))/n;

end

for i=1:d3

z(i)=-(log(1-R3(i)))/n;

end

for i=2:d1

t1(1)=x(1);

t1(i)=t1(i-1)+x(i);

end

for i=2:d2

t2(1)=y(1);

t2(i)=t2(i-1)+y(i);

end

for i=2:d3

t3(1)=z(1);

t3(i)=t3(i-1)+z(i);

end

for i=1:d1

X=[i,x(i),t1(i),i];

disp(X);

end

for j=1:d2

Y=[j,y(j),t2(j),j];

disp(Y);

end

for k=1:d3

Z=[k,z(k),t3(k),k];

disp(Z);

end

m1=d1/t1(d1)