I am taking a probability class and I am not fully getting these two problems below. I do know solutions and I don't need the whole calculation but I want to grasp the idea a little bit better.
Suppose again that Z = X + Y. Find $f(z)$.
$f_X(x)=\frac{1}{2}$ when $0<x<2$; $0$ otherwise.
I know that $f_Y(y)$ has the same shape as $f_X(x)$.QUESTION 1: So what is exactly $f_Y(y)$? Is it $\frac{1}{2}$ or ...?
$f_X(x)=f_Y(x)=\frac{1}{2}(x-3)$ when $3<x<5$; $0$ otherwise.
I know that the answer is: for $6 <z <8$, we integrate $\frac{1}{4}(x-3)(z-x-3)$ from x=3 to x=z and get $f_Z(z) = \frac{1}{24}(z^3 – 18z^2 + 81z – 108)$.
For $8 <z <10$, we integrate $\frac{1}{4}(x-3)(z-x-3)$ from x=z to x=5 and get $f_Z(z) = \frac{1}{24}(-z^3 + 18z^2 – 69z +20)$.QUESTION 2: Why do we consider only $6<z<8$ and $8<z<10$? What happens between $3<z<6$?
QUESTION 3: Why do we integrate from x=z to to x=5 for $8 <z <10$?
Questions and solutions: enter image description here
