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$X$ and $Y$ are jointly continuous random variables. Their probability density function is: $$f(x,y) = \begin{cases}2x & \mbox{if } x\in [0,1], y\in[0,1] \\ 0 & \mbox{ otherwise }\end{cases}$$

Calculate the co-relation coefficient between $X$ and $Y$.

I know that $\mathsf{Corr}(X,Y) = \dfrac{\mathsf{Cov}(X,Y)}{\sqrt{\mathsf{Var}(x)\,\mathsf{Var}(y)\,}}$ but I'm not sure how to get those values.

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    $\begingroup$ and where do you have difficulties? $\endgroup$ – Seyhmus Güngören Oct 29 '14 at 20:29
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    $\begingroup$ I have the formula Corr(x,y) = Cov(x,y)/(sqrt(Var(x)Var(Y)) but im not sure how to calculate those with the given information. $\endgroup$ – StatsHelp Oct 29 '14 at 20:32
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    $\begingroup$ so first $x$ and $y$ is different from $X$ and $Y$. Second, if you are given the joint density you can calculate the marginal densities right? $\endgroup$ – Seyhmus Güngören Oct 29 '14 at 20:33
  • $\begingroup$ @StatsHelp, just a bit of advice: it's good you have a starting point with the Corr(x,y) formula. You might want to add that to your question, so others know you started somewhere. $\endgroup$ – daOnlyBG Oct 29 '14 at 20:37
  • $\begingroup$ @SeyhmusGüngören I calculated the marginals f(x) = 2x and f(y) = 1. $\endgroup$ – StatsHelp Oct 29 '14 at 20:47
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$\begin{align} f(x,y) & = \begin{cases}2x & \mbox{if } (x,y)\in [0,1]\times[0,1] \\ 0 & \mbox{ otherwise }\end{cases} \\[2ex] f_X(x) & = \int_0^1 2x\operatorname d y \\ & = 2x \\[2ex] f_Y(y) & = \int_0^1 2x\operatorname d x \\ & = 1 \\[3ex] \therefore f(x,y) & = f_X(x)f_Y(y) & \text{so what does this mean?} \\[4ex] \mathsf {Cov}(X,Y) & = \mathsf E(XY)-\mathsf E(X)\mathsf E(Y) \\[1ex] & = (\text{what?}) & \text{and why?} \end{align}$

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Have a look at this web-pages:

http://en.wikipedia.org/wiki/Independence_(probability_theory)

you will find the answer to your question.

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