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I don't know what I'm doing wrong as my answer doesn't match the answer our prof gave us and the exam is tomorrow please help!!!

Joint pdf of two random variables is

$f(x,y)= 6x$ if $0<x<y<1$ and $0$ otherwise

find the covariance between x and y

My solution: Cov(x,y)= E(XY)-E(X)E(Y)

E(XY)= $\int _0^1\int _0^{\frac{y}{6}}xy\cdot \:6xdxdy=\frac{1}{540}$

E(X)=$\int _0^1\int _0^{\frac{y}{6}}x\cdot \:6xdxdy=\frac{1}{432}$

E(Y)= $\int _0^1\int _0^{\frac{y}{6}}y\cdot \:6xdxdy=\frac{1}{48}$

so cov(x,y)= $\frac{1}{540}$-($\frac{1}{432}$)($\frac{1}{48}$)= $1.8\cdot \:10^{-3}$

the answer our prof gave is $\frac{1}{40}$

thanks in advance!

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Since you have $x \lt y$, try $$\int\limits_0^1 \int\limits_0^{y} \cdots$$ rather than $\int\limits_0^1 \int\limits_0^{y/6} \cdots$. Done for all three integrals, this will get you to $\frac1{40}$

As a check, $\int\limits_0^1 \int\limits_0^{y} 6x\, dx\, dy = 1$ as you would hope

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  • $\begingroup$ thank you so much that check will really help me tomorrow! $\endgroup$
    – user520403
    Apr 14 '19 at 17:11
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If $a+2,\,a+b+3$ are both positive, $$\Bbb EX^aY^b=\int_0^1 dy\int_0^y 6x^{a+1}y^b dx=\int_0^1\frac{6}{a+2}y^{a+b+2}dy=\frac{6}{(a+2)(a+b+3)}.$$(Sanity check: $a=b=0$ gives $\Bbb E 1=1$.) Hence $$\operatorname{Cov}(X,\,Y)=\frac{6}{3\times 5}-\frac{6}{3\times 4}\times\frac{6}{2\times 4}=\frac{1}{40}.$$

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