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Math Help - Correlation Co-efficient : Bivariate Contineous Probability Distribution?

  1. #1
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    Correlation Co-efficient : Bivariate Continuous Probability Distribution?

    I am facing a problem to understand the formula to compute var(x) (variance of x). Any help in this regard would be highly appreciate

    <br />
f(x,y) = x^{2}+\frac{xy}{3}, 0 < x < 1, 0 < y < 2, 0 \ elsewhere<br />

    Find var(x), var(y)

    Solution:
    <br />
g(x) = \int^{\infty}_{-\infty} f(x,y) dy = 2x^{2}+\frac{2}{3}x <br />
    <br />
h(y) = .......<br />

    <br />
E(x) = \int^{\infty}_{-\infty} x.g(x) dx = \frac{13}{18}<br />
    <br />
E(y) = ......<br />

    <br />
 var(x) = E[x-E(X)]^2<br />
    <br />
            = E(x-E(X))^2<br />

    <br />
= \int^{\infty}_{-\infty} (x-E(x))^2 . g(x) dx<br />

    <br />
 = \int^{1}_{0} (x-\frac{13}{18})^2 . (2x^{2}+\frac{2x}{3}) dx<br />

    I am unable to understand how <br />
E(x-E(X))^2 equals \int^{\infty}_{-\infty} (x-E(x))^2 . g(x) dx ?

    Any detail, derivation even a link would be greatly helpful.

    Thank you.
    Last edited by mrsenim; July 31st 2010 at 11:42 AM. Reason: Some more details entered
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  2. #2
    frm
    frm is offline
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    Hi,
    it looks like your doubt is not related to the bivariate but rather to the definition of the variance?
    Once you marginalize the joint PDF f(x,y) respect the r.v. X you obtain your univariate PDF g(x), from which you can compute the mean value of the r.v. X and then just compute the variance (see Variance - Wikipedia, the free encyclopedia).

    The fact that E(x-E(X))^2 equals \int^{\infty}_{-\infty} (x-E(X))^2 g(x) dx is by definition of the expectation operator (Expected value - Wikipedia, the free encyclopedia - see the part related to E{g(x)}).
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