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Random variables X,Y have joint pdf as f_X,Y ^ (x,y) = 24 x.y and x,y>0 x+y<1 find the marginal pdf's of U=X+Y and V = X/Y i tried to solve it and got that: g_V,U (v,u) = 24 . v . u^3 / (1+v)^4 g_U^(u) = 4.u^3 g_V^(v) = 12 . v / (1+v)^4 and hence the two random variables U and V are not independent Are these answers right ??

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If I understand, you obtain that $$g_{U,V} (u,v) = \frac{24 . v . u^3}{(1+v)^4} = \frac{1}{2} 4.u^3 \times \frac{12 . v }{ (1+v)^4} = \frac{1}{2} g_U(u) g_V(v).$$

First, the factor $\frac{1}{2}$ should not be here and is probably due to a mistake in the calculation. Second, forgetting about the $\frac{1}{2}$, the relation you obtain show that the random variable are independent, because the joint probability of $(V,U)$ is $g_U(u) g_V(v)$. Isn't it what you should prove?

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the factor 1/2 makes me confused and i tried it more than three times and i got the same result. By the way i tried to prove that the two random variables are independent ! – Ahmed samy Mar 17 '14 at 9:47

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