Convolution Method of Random Variable

Question

I have read about convolution method in the Karlin & Taylor's book. I'm confused, why the domain of integral changed from $-\infty$ in to $z$?

Is it because $X$ and $Y$ non negative, so $Z=X+Y\geq 0$? So, the domain of integral changed?

Answer 1

If $\Pr[X < 0] = 0$ and $\Pr[Y < 0] = 0$, then $f_X(x) = 0$ and $f_Y(y) = 0$ for $x < 0$ and $y < 0$. Consequently $f_Y(z - \xi) = 0$ if $\xi > z$, and $f_X(\xi) = 0$ if $\xi < 0$. Equation 2.13 then becomes $$f_Z(z) = \int_{\xi = -\infty}^\infty f_Y(z-\xi) f_X(\xi) \, d\xi = \int_{\xi = 0}^z f_Y(z - \xi) f_X(\xi) \, d\xi,$$ and if $z < 0$, this integral is $0$ since no $\xi \in \mathbb R$ satisfies the inequality $0 \le \xi \le z$.

Answer 2

If $\xi <0$ the $f_{X} (\xi)=0$ and if $\xi >z$ then $f_Y(z-\xi)=0$. So you only have integrate over the interval $0 <\xi <z$.

Answer 3

Since $X$ and $Y$ are nonnegative functions $f_X$ and $f_Y$ take value $0$ on $(-\infty,0)$.

So the integrand $f_X(z-\eta)f_Y(\eta)$ can only have positive value on interval $[0,z]$.

Same story for integrand $f_Y(z-\xi)f_X(\xi)$.

Answer 4

Since $X,Y$ are non-negative.

For all $\eta \le 0, f_Y(\eta) = 0$ and for all $\eta \ge z, f_X(z-\eta) = 0$