# Thread: Correlation Co-efficient : Bivariate Contineous Probability Distribution?

1. ## 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

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

Find var(x), var(y)

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

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

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

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

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

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

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

Thank you.

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