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Random Variables X and Y have the joint probability density function f(x,y)= 8xy/81 for 0 < x < 3 and 0 < y < x Determine the marginal probability density function of X.

I am confused if I should be using integration of 0 to X or 0 to 3?
fx(X)=$$\frac{8}{81}$$ $$\int_{0}^{3} xy dy$$

: $$\frac{8X}{81}$$ $$\int_{0}^{3} [y^2] dy$$

$$\frac{8X}{81} (3^2) - (0^2)$$ =$$\frac{8X}{27}$$

Brian
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1 Answers1

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Draw a picture of the triangle on which the given joint density function "lives." This is the triangle with corners $(0,0)$, $(3,0)$, and $(3,3)$. ' For the (marginal) density $f_X(x)$ of $X$, we want to "integrate out" $y$. Take any fixed value of $x$ between $0$ and $3$, and draw the vertical line through $(x,0)$. Then $y$ travels from $0$ to $x$. So for $0\lt x\lt 3$ we have $$f_X(x)=\int_0^x \frac{8xy}{81}\,dy.$$

André Nicolas
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  • Thank you for explaining to draw a picture. This makes much more sense now. – Brian Oct 09 '15 at 19:11
  • Despite the fact that I have taught this material for many years, I still always make a sketch. If I don't, I am not fully confident I am doing things correctly. – André Nicolas Oct 09 '15 at 19:15