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For the random variables X, Y with joint density function

$f(x,y)=cx^2y(1-x)(y-1) \space, \space \space \space 0 \leq x \leq 1,\space \space0 \leq y \leq 1$$

$f(x,y)=0 ,\space \space \space \space otherwise$

(a) For what value of c is this a joint density function?

(b) Using this value of c, compute the density functions of X and Y .

I'm not sure of my answer in part a), I took the double integral as such: $$\int_{0}^{1} \int_{0}^{1} cx^2y(1-x)(y-1)dxdy = 1$$

Am I right by doing it this way?

Also if I understand part b) correctly, am I right to say that I would need to compute the following: $$\int_{0}^{1} \int_{0}^{1} -72x^2y(1-x)(y-1)dxdy $$

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    $\begingroup$ For part $a)$ what you are doing sounds correct, you want to integrate $f(x,y)$ over the domain and you will get some value $Z$ say, now if $Z = 1$ you are done, else take $c = 1/Z$. For part $b)$ I understand that to mean calculate the marginal densities of $X$ and $Y$, that is to say from a new function $f(x) = \int c f(x,y) dy$ and if necessary rescale so that $f(x)$ is a valid pdf, and do the same for $f(y)$ $\endgroup$ – Nadiels Feb 19 '17 at 22:39
  • $\begingroup$ I don't quite get it for part b)... is my reasoning ok above? $\endgroup$ – JKawa Feb 20 '17 at 1:22
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    $\begingroup$ No. You just want: $f_X(x) = \int_0^1 -72x^2 y(1-x)(y-1)\operatorname d y = 12x^2(1-x)$ (when $x\in(0;1)$) and similarly for $f_Y(y)$,,, $\endgroup$ – Graham Kemp Feb 20 '17 at 2:28
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Part a is correct.

Part b:

$ f_X(x) = \int f(x,y)\operatorname d y = \int_0^1 -72x^2 y(1-x)(y-1)\operatorname d y = 12x^2(1-x) $ and $ f_Y(y) = \int f(x,y)\operatorname d x = \int_0^1 -72x^2 y(1-x)(y-1)\operatorname d x = -6y(y-1). $

Moreover $X$ and $Y$ are independent since $f(x,y) = f_X(x)f_Y(y).$

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