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I have $X$, $Y$ random variables with joint density function $$f_{X,Y}(x,y)=\begin{cases}8xy,& 0<x<y<1\\0,& \text{otherwise}.\end{cases}$$ The region is a triangle.

a) Find $f_X$, $f_Y$ the marginal density functions.

I got this:

$f_X(x)=\int_{-\infty}^{\infty} f_{X,Y}(x,y)dy=\int_{x}^{1}8xydy=4x(1-x^2)$

$f_{Y}(y)=\int_{-\infty}^{\infty} f_{X,Y}(x,y)dx=\int_{0}^{y}8xydx=4y^3$

Question I

Are right the integral limits in $f_{Y}$? I'm not sure because I don't know if I should keep the condition of $x<y<1$ or should I just take $0<y<1$

b) Find $F_{X,Y}(x,y)$, $F_X$, $F_Y$.

REALLY ANNOYING QUESTION

I got $F_{X,Y}(x,y)= 2x^2y^2-x^4$, and I know that $F_{X}(x)=lim_{y \rightarrow \infty}F_{X,Y}(x,y) $

but $F_{X,Y}(x,y)=0$ if $(x,y)$ is not in the triangle and when y tends to infinity, (x,y) is not in the triangle so F=0!

I get a result by integrating $f_y$ but I want to know how to do it with the limit method.

Thanks in advance.

Math1000
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HeMan
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  • You are extrapolating the formula you got way outside the range of it's validity when taking the limit. $f_{X,Y}$ is zero for $y>1$ so when you integrate up to get $F_{X,Y}$ then this formula only holds for $y<1$ and equal to it's value at $y=1$ for larger values of $y$. – Winther Nov 25 '15 at 04:10
  • The marginals are right. Where these hold should be specified. Indeed $F_{X,Y}(x,y)$ is very annoying, since there are many cases. It is not true that the joint cdf is $0$ for $(x,y)$ outside the triangle. – André Nicolas Nov 25 '15 at 04:14
  • @Winther but why $y=1$ for larger values of $y$ formally? I start thinking that I can't take the infinite limit because $F $ is zero outside the triangle, in particular for bigger values of $y$ but I see that it MUST take the value when $y=1$ but I can't justify why. This is my main headache. – HeMan Nov 25 '15 at 04:37
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    You can see this from the definition of the joint CDF $F_{X,Y}(x,y) = \int_{-\infty}^x\int_{-\infty}^yf_{X,Y}(x',y'){\rm d}x'{\rm d}y'$. Since $f_{X,Y} = 0$ outside the triangle as we increase $x$ or $y$ outside this region the integral does not grow any more. It is perhaps easier to understand this by thinking what it really means in terms of probabillity: the probabillity of having, say, $Y\leq 10$ is just the same as having $Y\leq 1$ as the probabillity of having $Y\geq 1$ is $0$. – Winther Nov 25 '15 at 04:46

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