Results 1 to 3 of 3

Math Help - correlation

  1. #1
    Junior Member
    Joined
    Jan 2008
    Posts
    29

    correlation

    Suppose (X_1, X_2) has the bivariate normal distribution with density function
    f(x_1, x_2) = \frac{1}{2\pi \sqrt{1-p^2}} exp[- \frac{x_1^2 - 2p x_1 x_2 + x_2^2}{2(1-p^2)}]

    Show that Corr(X_1^2, X_2^2) = p^2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by graticcio View Post
    Suppose (X_1, X_2) has the bivariate normal distribution with density function
    f(x_1, x_2) = \frac{1}{2\pi \sqrt{1-p^2}} exp[- \frac{x_1^2 - 2p x_1 x_2 + x_2^2}{2(1-p^2)}]

    Show that Corr(X_1^2, X_2^2) = p^2
    First you should recognise from the joint pdf that E(X_1) = E(X_2) = 0 and Var(X_1) = Var(X_2) = 1.


    Therefore \text{Corr} (X_1^2, X_2^2) = \frac{\text{Cov} (X_1^2, X_2^2)}{\sigma_1 \sigma_2} = \frac{E(X_1^2 \cdot X_2^2) - E(X_1^2) \cdot E(X_2^2)}{\sigma_1 \sigma_2} = E(X_1^2 \cdot X_2^2) - 1. Mr F Edit: The mistake I mention below is here. See post #3.

    Note: Var(X_i) = E(X_i^2) - [E(X_i)]^2 \Rightarrow 1 = E(X_i^2) - 0^2 \Rightarrow E(X_i^2) = 1.


    So the task is to calculate E(X_1^2 \cdot X_2^2):


    E(X_1^2 \cdot X_2^2) = \int_{x_1 = -\infty}^{+\infty} \int_{x_2 = -\infty}^{+\infty} x_1^2 \, x_2^2 \, \frac{1}{2 \pi \sqrt{1 - \rho^2}} \exp \left( -\frac{(x_1^2 - 2 \rho x_1 x_2 + x_2^2)}{2 (1 - \rho^2)}\right) \, dx_2 \, dx_1


    Now complete the square and re-arrange:


    = \frac{1}{\sqrt{2 \pi}} \int_{x_1 = -\infty}^{+\infty} x_1^2 \, \exp \left( - \frac{x_1^2}{2} \right) \left[ \int_{x_2 = -\infty}^{+\infty} x_2^2 \, g(x_1, x_2) \, dx_2 \right] \, dx_1


    where g(x_1, x_2) = \frac{1}{\sqrt{2 \pi (1 - \rho^2)}} \exp \left( -\frac{(x_2 - \rho x_1)^2}{2 (1 - \rho^2)}\right) is the pdf of N(\mu = \rho x_1, \, \sigma^2 = 1 - \rho^2).


    Now note that \int_{y = -\infty}^{+\infty} y^2 \, g(x_1, y) \, dy represents E(Y^2) and so:

    Var(Y) = E(Y^2) - [E(Y)]^2 \Rightarrow 1 - \rho^2 = E(Y^2) - \rho^2 x_1^2 \Rightarrow E(Y^2) = (1 - \rho^2) + \rho^2 x_1^2.

    Therefore:


    E(X_1^2 \cdot X_2^2) = \frac{1}{\sqrt{2 \pi}} \int_{x_1 = -\infty}^{+\infty} x_1^2 \, \exp \left( - \frac{x_1^2}{2} \right) [(1 - \rho^2) + \rho^2 x_1^2] \, dx_1


    = (1 - \rho^2) \, \frac{1}{\sqrt{2 \pi}} \int_{x_1 = -\infty}^{+\infty} x_1^2 \, \exp \left( - \frac{x_1^2}{2} \right) \, dx_1 + \rho^2 \, \frac{1}{\sqrt{2 \pi}} \int_{x_1 = -\infty}^{+\infty} x_1^4 \, \exp \left( - \frac{x_1^2}{2} \right) \, dx_1


    = (1 - \rho^2) E(X_1^2) + \rho^2 E(X_1^4)

    where the moments are for a standard normal distribution. Cheating and refering to a table of moments (see Normal distribution - Wikipedia, the free encyclopedia):


    = (1 - \rho^2) + 3 \rho^2 = 1 + 2 \rho^2.


    Therefore \text{Corr} (X_1^2, X_2^2) = E(X_1^2 \cdot X_2^2) - 1 = 1 + 2 \rho^2 - 1 = 2 \rho^2.


    This is different from the result that was given to be proved so there's a mistake somewhere (that I don't have time to look for). Nevertheless, I'm sure you get the idea.
    Last edited by mr fantastic; December 14th 2008 at 03:40 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    First you should recognise from the joint pdf that E(X_1) = E(X_2) = 0 and Var(X_1) = Var(X_2) = 1.


    Therefore \text{Corr} (X_1^2, X_2^2) = \frac{\text{Cov} (X_1^2, X_2^2)}{\sigma_1 \sigma_2} = \frac{E(X_1^2 \cdot X_2^2) - E(X_1^2) \cdot E(X_2^2)}{\sigma_1 \sigma_2} = E(X_1^2 \cdot X_2^2) - 1.

    [snip]
    The above is wrong. It should be:

    Therefore \text{Corr} (X_1^2, X_2^2) = \frac{\text{Cov} (X_1^2, X_2^2)}{ sd(X_1^2) \, sd(X_2^2) } = \frac{E(X_1^2 \cdot X_2^2) - E(X_1^2) \cdot E(X_2^2)}{sd(X_1^2) \, sd(X_2^2)} = \frac{E(X_1^2 \cdot X_2^2) - 1}{sd(X_1^2) \, sd(X_2^2)}.

    I leave it for you to calculate the required standard deviations. The technique is in post #2.


    Edit: E(X_i^4) = 3 and E(X_i^2) = 1 therefore Var(X_i^2) = 2 and all is well.
    Last edited by mr fantastic; December 14th 2008 at 05:23 AM. Reason: Fixed an error
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Correlation
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 20th 2010, 06:44 AM
  2. Correlation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: April 16th 2010, 02:29 PM
  3. Correlation
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: December 6th 2009, 07:21 PM
  4. MIN, MAX from correlation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: September 28th 2009, 06:10 AM
  5. sample correlation and population correlation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 23rd 2008, 04:31 PM

Search Tags


/mathhelpforum @mathhelpforum