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$f_{X,Y}(x,y)=2$ for $0 < x < y < 1$ & 0 otherwise.

Find the density function of $Z$ where $Z = X + Y$.

My textbook has this formula: $f_Z(z)=\int f_{X,Y}(u,z-u)\;du$.

Apologise for the formatting. The solution sets the limits as $0 < u < \frac{z}{2}$ if $0 < z < 1$ and $z-1 < u < \frac{z}{2}$ if $1 < z < 2$. I would like to know how and why the limits are so. Thank you.

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    $\begingroup$ What is $u$ here? You will have to show us the solution given to you so that we understand what is going on. $\endgroup$ – Kavi Rama Murthy Mar 25 at 23:53
  • $\begingroup$ My textbook has this formula: fZ(z)=∫fX,Y(u,z-u)du $\endgroup$ – Rachel12 Mar 25 at 23:59
  • $\begingroup$ See this introduction to posting mathematical notation, e.g. subscripts and integrations. $\endgroup$ – hardmath Mar 26 at 3:13
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If $0<u<z-u<1$ then $2u <z$ so $u <\frac z 2$. Also, $z-u<1$ gives $z-1<u$. Remember that we also need $u >0$. The condition $z-1<u$ is automatically satisfied if $z-1<0$ or $z<1$. In case $z >1$ we get two conditions: $z-1 <u$ and $u <\frac z 2$ so the integral is from $z-1$ to $\frac z 2$.

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