Correlation Co-efficient : Bivariate Contineous Probability Distribution?

**mrsenim** · July 31st 2010, 10:47 AM

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

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

Find var(x), var(y)

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

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

$ var(x) = E[x-E(X)]^2 $
$ = E(x-E(X))^2 $

$ = \int^{\infty}_{-\infty} (x-E(x))^2 . g(x) dx $

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

I am unable to understand how $ 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.

**frm** · July 31st 2010, 11:51 AM

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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