Z= X+Y, то z ≥ 0
Применяем формулу
g(z)= ∫ z0f1xf2(z–x)dx
Используя данные задачи
g(z)= ∫ z0(1/3)·e–x/3·(1/5)·e–(z–x)/5dx=
=(–1/2)·e–z/5·∫ z0e–2x/15d(–2x/15)=
=(–1/2)·e–z/5·(e–2z/15–e0)=
=(1/2)e–z/3–(1/2)e–z/5
О т в е т.
g(z)= (1/2) · (e–z/3–e–z/5) при z ∈ [0;+ ∞ )