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It is given that $$f(x,y)=\begin{cases} 864(2x-y)(y-x), & x \le y\le 2x, x+y \le 1\\ 0, & \text{otherwise} \end{cases}$$

Find the density $f_z(z)$ for $Z=X+Y$

I have used the formula for convolution that says

$$f_z(z)=\int_{-\infty}^\infty f(x,x-z) \, dx$$

The problem is i have no idea which limits to use on the integral. According to the solution, we have that $ z/3\leq x\leq z/2$ and $z \in [0,1]$

Anyone who could explain to me how these limits are found?

Michael Hardy
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Wstate123
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1 Answers1

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You need $\displaystyle f_Z(z)=\int_{-\infty}^\infty f_{X,Y}(x,z-x) \, dx$, not the version with $x-z$. The point here is that in the expression $(x,z-x)$ the sum of the $x$ and $y$ components should be $z$, so the point $(x,y)$ moves along the line $x+y=z$ with $z$ constant.

For any fixed value of $z$, consider the line $x+y=z$. This line has slope $-1$ in the $(x,y)$-plane. It intersects the line $y=2x$ at $(x,y)=(z/3,\,2z/3)$ and it intersects the line $y=x$ at $(x,y)=(z/2,\,z/2)$.

Michael Hardy
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  • I was using the right integral, just misswrote in my question, my bad!

    How exactly do you determine where z intersects with y?

     – Wstate123 Dec 16 '15 at 22:07
  • Say for example $z=5$. Then the line $x+y=z$ is the line $x+y=5$. You need the intersection of $x+y=5$ with $y=2x$. If $y=2x$ then $2x$ can be put in place of $y$ in the other equation, getting $x+2x=5$. That gives us $3x=5$, so $x=5/3$. If $x=5/3$ then $y=2x=10/3$. With some other number $z$ rather than $5$, instead of $5/3$ we have $z/3$. Similarly for the intersection of $y=x$ with $x+y=5$. ${}\qquad{}$ – Michael Hardy Dec 16 '15 at 22:38