My Math Forum Covariance and correlation coefficient

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November 23rd, 2010, 04:35 PM   #1
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Covariance and correlation coefficient

Hi all,

If anybody knows how to work with joint probability function of multiple random variables, please take a look at this problem and see if you can give me any pointers.

I know you first need to find the marginal pdf of both U and V to get their respective means and variances. Then you need to find the expected value of X and Y (EXY) by double integration. Following, you use all the obtained results in the covariance formula in the second attachment below.

Then finally, you use the obtained covariance of X and Y to find the correlation coefficient in the third attachment below.

I hope my explanation is somewhat clear.

What I need help with is the mechanics of the problem. Is anybody familiar?

Thanks for looking.

knp
Attached Images
 Correlation Coefficient.jpg (9.5 KB, 160 views) Covariance.jpg (10.6 KB, 160 views) C and CC.jpg (18.1 KB, 160 views)

November 23rd, 2010, 07:31 PM   #2
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Re: Covariance and correlation coefficient

The domain on the u-v plane is shown as the following:

[attachment=0:238ndci0]1.JPG[/attachment:238ndci0]

As the joint probability function is

$f(u,v)=\begin{cases}\frac{1}{64}e^{-\frac{v}{8}} &0\leq u\leq v\leq +\infty\\0=&\text{otherwise}\end{cases} \text{ (i)}=$

the marginal probability functions are

$f_{\small{U}}(u)=\begin{cases}\int_u^{+\infty}\fra c{1}{64}e^{-\frac{v}{8}}\;d v &0\leq u\leq +\infty\\0=&\text{otherwise}\end{cases}=\begin{cases}\frac{1} {8}e^{-\frac{u}{8}} &0\leq u\leq +\infty\\\text{ }\\0=&\text{otherwise}\end{cases}\text{ (ii)}=$

$f_{\small{V}}(v)=\begin{cases}\int_0^{v}\frac{1}{6 4}e^{-\frac{v}{8}}\;d u &0\leq v\leq +\infty\\0=&\text{otherwise}\end{cases}=\begin{cases}\frac{1} {64}ve^{-\frac{v}{8}} &0\leq v\leq +\infty\\\text{ }\\0=&\text{otherwise}\end{cases}\text{ (iii)}=$ .

Use (ii) and (iii), the expectations are

$E(U)=\int_0^{+\infty}u.\frac{1}{8}e^{-\frac{u}{8}}\;d u=8$

$E(V)=\int_0^{+\infty}v.\frac{1}{64}ve^{-\frac{v}{8}}\;d v=16$

Also, the variances are

$D(U)=E(U^2)-[E(U)]^2=\int_0^{+\infty}u^2.\frac{1}{8}e^{-\frac{u}{8}}\;d u-8^2=64$

$D(V)=E(V^2)-[E(V)]^2=\int_0^{+\infty}v^2.\frac{1}{64}ve^{-\frac{u}{8}}\;d v-16^2=128$

And the covariance is

$\text{Cov}(U,V)=E(UV)-E(U)E(V)=\iint_{\small{0\leq u\leq v\leq+\infty}}uv.\frac{1}{64}e^{-\frac{v}{8}}\;d ud v -8\times16=\int_0^{+\infty}\frac{1}{64}ve^{-\frac{v}{8}}dv\int_0^vudu-128=64$

The correlation cofficient is

$\rho_{\small{UV}}=\frac{\text{Cov}(U,V)}{\sqrt{D(U )D(V)}}=\frac{64}{\sqrt{64\times128}}=\frac{\sqrt{ 2}}{2}\approx 0.7071$
Attached Images
 1.JPG (7.3 KB, 151 views)

 November 23rd, 2010, 08:13 PM #3 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Covariance and correlation coefficient I have corrected some mistakes in the above. Please have a check
 November 23rd, 2010, 08:15 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Covariance and correlation coefficient I vaguely thought the correlation coefficient should be 0 < r ? 1, but I wasn't gonna question you, when I wasn't sure.
 November 23rd, 2010, 08:27 PM #5 Newbie   Joined: Nov 2010 Posts: 11 Thanks: 0 Re: Covariance and correlation coefficient I'm not quite sure how you obtained 0.7071 for the correlation coefficient. Am I missing something here? The rest of the problem looks great.
November 23rd, 2010, 08:42 PM   #6
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Re: Covariance and correlation coefficient

Quote:
 Originally Posted by knp I'm not quite sure how you obtained 0.7071 for the correlation coefficient. Am I missing something here? The rest of the problem looks great.
the standard deviation $\sigma_{\small{X}}=\sqrt{D(X)}$, this is from my text book

November 23rd, 2010, 09:20 PM   #7
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Re: Covariance and correlation coefficient

Quote:
 Originally Posted by MarkFL I vaguely thought the correlation coefficient should be 0 < r ? 1, but I wasn't gonna question you, when I wasn't sure.

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