homework1.1.9

# homework1.1.9 - figure of x2,y2 figure of x3,y3 figure of...

This preview shows pages 1–8. Sign up to view the full content.

Peng Huang ID:U02078166 Problem 1.9 (c)If we can not get the inverse function of PY(y) but given the aiming PDF is with a Gaussion distribution whose mean is zero and the variance is σ , we can use the Central Limit Theorem that provides us a statement that the sum of IID variables can be the N(0,1) distribution. Thus, we are sure that we can generate such a Gaussion distribution random variables by a uniform random variables. Here, I choose the most straightforward one. Because the Rayleigh random variables are the magnitude vectors of pairs of independent, zero-mean, variance σ random variables. We can therefore use the Rayleih random variable R and a uniform random θ to generate a variable which equals to R*sin θ and satisfies the N(0,1) distribution. I use such an M-file: a=0;b=2*pi; theata=unifrnd(a,b,1,100); R=raylrnd(1,1,100); for i=1:1:100; X(i)=R(i)*sin(theata(i)); end y=0:1:99; pdf1d(X,100) figure(2) stem(y,X), xlabel( 'Sample Numbers' ),ylabel( 'samples' )

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
And use the given pdf1d.m, I got the results as following: The results show that the method is work.
Peng Huang ID:U02078166 Problem 1.10 (a)1.figure of x1,y1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: figure of x2,y2 figure of x3,y3 figure of x4,y4 2.marginal distributions: I used such an M-file: py=zeros(1,100); py1=zeros(100,1); [p,x,y,P]=pdf2d(x1,y1,100,100); for i=1:1:100 for j=1:1:100 py=py+p(j,:); end py(i)=py(i); end for j=1:1:100; for i=1:1:100; py1=py1+p(:,i); end py1(j)=py1(j); end subplot(211), plot(y,py); subplot(212), plot(y,(py1).'); subplot the resultes are shown as followings: the upper is the figure of X serious and the other is Y serious. px1(x) and py1(y) px2(x) and py2(y) px3(x) and py3(y) px4(x) and py4(y) 3. cov(x1,y1) ans = 0.4309 -0.0003-0.0003 3.2938 cov(x2,y2) ans = 0.0835 -0.0003-0.0003 0.0828 cov(x3,y3) ans = 2.4706 1.4910 1.4910 2.4953 cov(x4,y4) ans = 0.0829 0.0011 ρ (x1,y1)=-0.0002, ρ (x2,y2)=-0.0041, ρ (x3,y3)=0.6005, ρ (x4,y4)=0.0039 (b)Based on the correlation coefficient values, I think the sets of NO.1, NO.2 and NO.4 is uncorrelated. Because the coefficient values are extremely small. Using the pdm2d to do the analysis the joint probability density function of X and Y, I found that the NO1 and NO4 are independent....
View Full Document

{[ snackBarMessage ]}

### Page1 / 8

homework1.1.9 - figure of x2,y2 figure of x3,y3 figure of...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online