# Limits for the density function

$$f_{X,Y}(x,y)=2$$ 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.

• What is $u$ here? You will have to show us the solution given to you so that we understand what is going on. – Kavi Rama Murthy Mar 25 at 23:53
• My textbook has this formula: fZ(z)=∫fX,Y(u,z-u)du – Rachel12 Mar 25 at 23:59
• See this introduction to posting mathematical notation, e.g. subscripts and integrations. – hardmath Mar 26 at 3:13

If $$0 then $$2u so $$u <\frac z 2$$. Also, $$z-u<1$$ gives $$z-1. Remember that we also need $$u >0$$. The condition $$z-1 is automatically satisfied if $$z-1<0$$ or $$z<1$$. In case $$z >1$$ we get two conditions: $$z-1 and $$u <\frac z 2$$ so the integral is from $$z-1$$ to $$\frac z 2$$.