I am confused about this convolution question

Question

I am confused about the two cases with $z>1$ and $z<1$ in part a. By following the formula for convolution you get integral $([0,1],\quad 1e^{-(z-x)}=e^{-z}(e-1)$, which is the case for $z>1$. How is there another case?

Answer 1

To follow your approach that

$$f_Z(z) = \int_0^1 1\cdot f_Y(z-x)\ dx$$

But the RHS is not always simply $\int_0^1 1\cdot e^{-(z-x)}\ dx$; that would be missing the other case of $f_Y(y)$ when $y<0$.

Case if $z\ge 1$: The RHS integration bounds are $0\le x \le 1$, so $z-x \ge 0$, so $f_Y(z-x) = e^{-(z-x)}$ as you expected.

$$f_Z(z) = \int_0^1 1\cdot f_Y(z-x)\ dx = \int_0^1 e^{-(z-x)}\ dx$$

---

Case if $0\le z \le 1$: Break the RHS into $y=z-x\ge 0$ or $y=z-x<0$, i.e. into $x\le z$ or $x>z$ respectively,

$$\begin{align*} f_Z(z) &= \int_0^1 1\cdot f_Y(z-x)\ dx\\ &= \int _0^z 1\cdot f_Y(z-x)\ dx + \int_z^1 1\cdot f_Y(z-x)\ dx\\ &= \int _0^z 1\cdot e^{-(z-x)}\ dx + \int_z^1 1\cdot 0\ dx\\ &= \int _0^z e^{-(z-x)}\ dx\\ \end{align*}$$

---

Case if $z \le 0$: For the integration bounds $0\le x \le 1$, $z-x\le 0$, so $f_Y(z-x) = 0$.

$$f_Z(z) = \int_0^1 1\cdot f_Y(z-x)\ dx = 0$$

This case is missing in your quoted image. Your image instead implies that when $e^{-z} > 1$, $p(z) < 0$.