We have n independent pairs $(X_i,Y_i)$, which are uniformly distributed over the triangle bounded by $$X_i+Y_i<\lambda$$ $$Y_i-X_i<\lambda$$ $$Y_i>0$$
Find the density of $A=|X_1|+|X_2|+Y_1+Y_2$
My attempt:
First, we find the density of $B=|X_1|+Y_1$, which is $$f_B(b)=\frac{2b}{\lambda^2}$$ Similarly, due to the independence of the pair, we have the density of $C=|X_2|+Y_2$ is $$f_C(c)=\frac{2c}{\lambda^2}$$, and their joint density $$f_{B,C}(b,c)=\frac{4bc}{\lambda^4}$$ Now, the density of A is: $$f_A(a)=\int_0^\lambda \frac{4(a-c)b}{\lambda^4}db=\frac{4}{\lambda^4}\left[\frac{1}{2}a\lambda^2-\frac{1}{3}\lambda^3\right]=\frac{2a}{\lambda^2}-\frac{4}{3\lambda}$$
My question:
I checked if it is a density function, but it does not equal $1$ when integrated. Hence, I don't know where I made the mistake. I hope someone can point it out or suggest another way to solve this problem. Thank you!
First of all you have obviously misprinted and the integrated expression should be $\frac{4(a-\color{red}{b})b}{\lambda^4}$.
The main issue is however the limits of integration. Obviously the following inequalities have to be satisfied: $$ 0\le b\le\lambda,\quad\text{and}\quad 0\le a-b\le\lambda. $$ Therefore the correct expression is: $$ f_A(a)=\int_{\max(0,a-\lambda)}^{\min(\lambda,a)}\frac{4(a-b)b}{\lambda^4}\;{\rm d}b=\frac4{\lambda^4}\times\begin{cases} \displaystyle\int_0^a(a-b)b\;{\rm d}b,& 0\le a\le \lambda\\ \displaystyle\int_{a-\lambda}^\lambda(a-b)b\;{\rm d}b,& \lambda\le a\le 2\lambda\\ \end{cases}. $$ You can check that $$ \int_0^{2\lambda}f_A(a)\;{\rm d}a=1. $$